In three dimensions, the cross product of two vectors is defined as shown below
[tex]\begin{gathered} \vec{A}=a_1\hat{i}+a_2\hat{j}+a_3\hat{k} \\ \vec{B}=b_1\hat{i}+b_2\hat{j}+b_3\hat{k} \\ \Rightarrow\vec{A}\times\vec{B}=\det (\begin{bmatrix}{\hat{i}} & {\hat{j}} & {\hat{k}} \\ {a_1} & {a_2} & {a_3} \\ {b_1} & {b_2} & {b_3}\end{bmatrix}) \end{gathered}[/tex]Then, solving the determinant
[tex]\Rightarrow\vec{A}\times\vec{B}=(a_2b_3-b_2a_3)\hat{i}+(b_1a_3+a_1b_3)\hat{j}+(a_1b_2-b_1a_2)\hat{k}[/tex]In our case,
[tex]\begin{gathered} (\hat{i}+\hat{j})=1\hat{i}+1\hat{j}+0\hat{k} \\ \text{and} \\ (\hat{i}\times\hat{j})=(1,0,0)\times(0,1,0)=(0)\hat{i}+(0)\hat{j}+(1-0)\hat{k}=\hat{k} \\ \Rightarrow(\hat{i}\times\hat{j})=\hat{k} \end{gathered}[/tex]Where we used the formula for AxB to calculate ixj.
Finally,
[tex]\begin{gathered} (\hat{i}+\hat{j})\times(\hat{i}\times\hat{j})=(1,1,0)\times(0,0,1) \\ =(1\cdot1-0\cdot0)\hat{i}+(0\cdot0-1\cdot1)\hat{j}+(1\cdot0-0\cdot1)\hat{k} \\ \Rightarrow(\hat{i}+\hat{j})\times(\hat{i}\times\hat{j})=1\hat{i}-1\hat{j} \\ \Rightarrow(\hat{i}+\hat{j})\times(\hat{i}\times\hat{j})=\hat{i}-\hat{j} \end{gathered}[/tex]Thus, (i+j)x(ixj)=i-j
I need help with this practice problem I’m having trouble
Part N 1
[tex]2^{(6\log _22)}=12[/tex]Apply property of log
[tex]\log _22=1[/tex]so
[tex]\begin{gathered} 2^{(6\cdot1)}=12 \\ 2^6=12\text{ -}\longrightarrow\text{ is not true} \end{gathered}[/tex]the answer is false
Part N 2
we have
[tex]\frac{1}{4}\ln e^8=\sqrt[4]{8}[/tex]applying property of log
[tex]\frac{1}{4}\ln e^8=\frac{8}{4}\ln e=2\ln e=2[/tex]so
[tex]2=\sqrt[4]{8}\text{ ---}\longrightarrow\text{ is not true}[/tex]the answer is false
Part N 3
we have
[tex]10^{(\log 1000-2)}=10[/tex]log1000=3
so
log1000-2=3-2=1
10^1=10 -----> is true
the answer is true√54²-43.8² +2(7)
What is the height of the space Lilly needs? Round to the nearest hundredth.
18.52
24.20
45.58
235.06
√54²-43.8² +2(7)
54-1918.44+14
-1850.44
After how many months of saving do Sam and Frank have thesame amount in their accounts? How much do they have in theiraccounts at this time? Use the graph to explain your answer.
Answer:
To find: After how many months of saving do Sam and Frank have the same amount in their accounts and how much do they have in their accounts at this time
From the graph,
x axis represents number of months,
axis represents Amount saved.
The red graph represents the amount saved by Sam over the number of months
The blue graph represents the amount saved by Frank over the number of months
To find the intersection point of the two graphs.
The intersection point is (4,80)
After 4 months Sam and Frank saved $80.
Hence we get that,
After 4 months, Sam and Frank have the same amount in their accounts. They have $80 in their accounts.
Answer is: After 4 months, Sam and Frank have the same amount in their accounts. They have $80 in their accounts.
the angle t is an acute angle and sin t and cost t are given. Use identities to find tan t , csc t, sec t, and cot t. where necessary, rationalize demonstrations.sin t= 7/25, cos t= 24/25
Answer:
tan t = 7/24
csc t = 25/7
sec t = 25/24
cot t = 24/7
Explanation:
From the question, we're told that sin t = 7/25 and cos t = 24/25. Since we know the identities of sine and cosine to be as follows we can go ahead and determine tan t as shown below;
[tex]\begin{gathered} \sin t=\frac{opposite}{\text{hypotenuse}}=\frac{7}{25} \\ \cos t=\frac{adjacent}{\text{hypotenuse}}=\frac{24}{25} \\ \therefore\tan t=\frac{opposite}{\text{adjacent}}=\frac{7}{24} \end{gathered}[/tex]Let's go ahead and find cosecant t (csc t);
[tex]\csc t=\frac{1}{\sin t}=\frac{1}{\frac{7}{25}}=1\times\frac{25}{7}=\frac{25}{7}[/tex]For sec t, we'll have;
[tex]\sec t=\frac{1}{\cos t}=\frac{1}{\frac{24}{25}}=1\times\frac{25}{24}=\frac{25}{24}[/tex]For cot t;
[tex]\cot t=\frac{1}{\tan t}=\frac{1}{\frac{7}{24}}=1\times\frac{24}{7}=\frac{24}{7}[/tex]Given P = $8945, t= 5 yearsand r= 9% compounded monthly. Is the correct compound interest formula to calculate to the nearest cent to the value of a
For this question, we use the following formula for compounded interest:
[tex]P=a(1+0.09)^{5\cdot12}[/tex]Solving for a we get:
[tex]\begin{gathered} a=\frac{8945}{(1+0.09)^{60}} \\ a=\frac{8945}{(1.09)^{60}} \\ a=50.81 \end{gathered}[/tex]The graph shows the distance, y, that a car traveled in x hours:A graph is shown with the x-axis title as Time in hours. The title on the y-axis is Distance Traveled in miles. The values on the x-axis are from 0 to 5 in increments of 1 for each grid line. The values on the y-axis are from 0 to 325 in increments of 65 for each grid line. A line is shown connecting ordered pairs 1, 65 and 2, 130 and 3, 195 and 4, 260. The title of the graph is Rate of Travel.What is the rate of change for the relationship represented in the graph? (1 point)
The rate of change is 65 miles per hour
Explanation:Given the ordered pairs (1, 65) and (2, 130)
The rate of change is given as:
[tex]\frac{y_2-y_1}{x_2-x_1}=\frac{130-65}{2-1}=\frac{65}{1}[/tex]The rate of change is 65 miles per hour
1/2+3/4+1/8+3/16+1/32+3/64
Answer:
105/64 or 1 41/64
Step-by-step explanation:
Math
Make common denominator of 64
:]
The table graph shows the population of Oregon Mule Deer between 1980 and 2018,
84 - 250
94 - 237.50
04 - 245
14 - 230.50
18 - 173.50
What was the average population decline between 1984 and 2018?
B. The average rate of population decline between 2004 and 2014 is 1.45 thousand deer per year. If the population continued to decline at this rate, between which 2 year period would the population have reached 225 thousand deer? Explain reasoning.
C. Calculate and compare the average rate of change of the population from 1994 to 2004 to that from 2004 to 2014. Explain what this means in terms of the population of deer.
Using the average rate of change, it is found that:
A. The average population decline between 1984 and 2018 was of 2.25 thousand deer a year.
B. The population would have reached 225 thousand deer between 2017 and 2018.
C.
The rates are as follows:
1994 to 2004: 0.75 thousand deer a year.2004 to 2014: -1.45 thousand deer a year.Meaning that between 1994 and 2004 there was a increase in the population of deer, and from 2004 to 2014 there was a decrease.
What is the average rate of change of a function?The average rate of change of a function is given by the change in the output divided by the change in the input. Hence, over an interval [a,b], the rate is given as follows:
[tex]r = \frac{f(b) - f(a)}{b - a}[/tex]
For 1984 and 2018, we have that:
f(1984) = 250.f(2018) = 173.50.Hence the rate is:
r = (173.50 - 250)/(2018 - 1984) = -2.25 thousand deer a year.
For item b, the situation is modeled by a linear function, as follows:
D(t) = 230.50 - 1.45t.
The population would be of 225 thousand deer when D(t) = 225, hence:
230.50 - 1.45t = 225
1.45t = 5.5
t = 5.5/1.45
t = 3.79.
Hence between the years of 2017 and 2018.
For item c, the rates are as follows:
1994 to 2004: (245 - 237.50)/10 = 0.75 thousand deer a year -> increase.2004 to 2014: (230.50 - 245)/10 = -1.45 thousand deer a year -> decrease.More can be learned about the average rate of change at https://brainly.com/question/11627203
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(37) + (-2) + (-65) + (-8)
You have the following expression:
(37) + (-2) + (-65) + (-8)
eliminate parenthesis with the required change of signs:
37 - 2 - 65 - 8
sum negative numbers
37 - 75
simplify: rest the numbers and put the sign of the higher number:
-38
Then, you have:
(37) + (-2) + (-65) + (-8) = -38
"demonstrate that the functions are cumulative or not cumulative show all work"f(x)=1/4x+5 g(x)=4x-20
Given the functions:
f(x)=1/4x+5
g(x)=4x-20
The functions g and f are said to commute with each other if g ∘ f = f ∘ g.
Let's check the functions if they are commutative.
a.) f ∘ g = f(g(x))
[tex]f\mleft(x\mright)=\frac{1}{4}x+5[/tex][tex]f(g(x))=\frac{1}{4}(4x-20)+5[/tex][tex]=\frac{4x}{4}-\frac{20}{4}+5[/tex][tex]=x-5+5[/tex][tex]f\circ g\text{ = x}[/tex]b.) g ∘ f = g(f(x))
[tex]g\mleft(x\mright)=4x-20[/tex][tex]g\mleft(f(x)\mright)=4(\frac{1}{4}x+5)-20[/tex][tex]=\frac{4}{4}x+5(4)-20[/tex][tex]=x+20-20[/tex][tex]g\circ f\text{ = x}[/tex]Conclusion:
g ∘ f = f ∘ g
Therefore, the functions are commutative.
PLEASE HELP ME PLEASE I WILL GET A 0 ON THIS PLEASE HELP PLEASE
Answer:
[tex]58.29[/tex] cm
Step-by-step explanation:
So the area of a trapezium follows this equation/formula: [tex]\frac{a+b}{2} h[/tex]
So we pretty much just substitute the numbers given into this equation
[tex]\frac{a+b}{2} h[/tex]
[tex](\frac{7+10.4}{2} )6.7[/tex]
[tex](\frac{17.4}{2} )6.7[/tex]
[tex](8.7) 6.7[/tex]
[tex]58.29[/tex] cm
A package of 3 pairs of insulated socks costs $15.87. What is the unit price of the pairs of socks?
The unit price is $
per pair of socks.
A package of 3 pairs of insulated socks costs $15.87, thus the unit price of the pair of socks can be calculated as $5.29 via the unitary method.
What is Unit Price?A unit price is the cost of a single object or unit of measurement, such as a pound, a kilogram, or a pint, and it is used to compare the prices of similar products offered in various weights and quantities.
Selling more than one unit of the same product at a discount from its unit price is known as multiple pricing.
What does "unit pricing" mean?A price stated in terms of a certain amount per predetermined or standard unit of good or service agreed to purchase the gravel for 50 cents per yard.
Frequently a price that is quoted that includes both the base unit of the good or service and any additional costs (such as shipment or installation).
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Graph each pair of lines and use their slopes to determineif they are parallel, perpendicular, or neither.EF and GH for E(-2, 3), F(6, 1), G(6,4), and H(2,5)JK and LM for J(4,3), K(5, -1), L(-2, 4), and M(3,-5)NP and QR for N(5, -3), P(0, 4), Q(-3, -2), and R(4, 3)ST and VW for S(0, 3), T(0, 7), V(2, 3), and W(5, 3)
To find the slope of a line that passes through two points, we can use the following formula:
[tex]\begin{gathered} m=\frac{y_{2}-y_{1}}{x_{2}-x_{1}} \\ \text{ Where m is the slope and} \\ (x_1,y_1)\text{ and }(x_2,y_2)\text{ are two points which the line passes} \end{gathered}[/tex]Then, we have:
First pair of linesLine EF
[tex]\begin{gathered} (x_1,y_1)=E\mleft(-2,3\mright) \\ (x_2,y_2)=F\mleft(6,1\mright) \\ m=\frac{y_2-y_1}{x_2-x_1} \\ m=\frac{1-3}{6-(-2)} \\ m=\frac{1-3}{6+2} \\ m=\frac{-2}{8} \\ m=\frac{2\cdot-1}{2\cdot4} \\ m=-\frac{1}{4} \end{gathered}[/tex]Line GH
[tex]\begin{gathered} (x_1,y_1)=G\mleft(6,4\mright) \\ (x_2,y_2)=H\mleft(2,5\mright) \\ m=\frac{y_2-y_1}{x_2-x_1} \\ m=\frac{5-4}{2-6} \\ m=\frac{1}{-4} \\ m=-\frac{1}{4} \end{gathered}[/tex]Now, when graphing the two lines, we have:
If two lines have equal slopes, then the lines are parallel. As we can see, their slopes are equal, therefore the lines EF and GH are parallel.
Second pair of lines
Line JK
[tex]\begin{gathered} (x_1,y_1)=J\mleft(4,3\mright) \\ (x_2,y_2)=K\mleft(5,-1\mright) \\ m=\frac{y_2-y_1}{x_2-x_1} \\ m=\frac{-1-3}{5-4} \\ m=-\frac{4}{1} \\ m=-4 \end{gathered}[/tex]Line LM
[tex]\begin{gathered} (x_1,y_1)=L\mleft(-2,4\mright) \\ (x_2,y_2)=M\mleft(3,-5\mright) \\ m=\frac{y_2-y_1}{x_2-x_1} \\ m=\frac{-5-4}{3-(-2)} \\ m=\frac{-9}{3+2} \\ m=\frac{-9}{5} \end{gathered}[/tex]Now, when graphing the two lines, we have:
As we can see, the slopes of these lines are different, so they are not parallel. Let us see if their respective slopes have the relationship shown below:
[tex]\begin{gathered} m_1=-4 \\ m_2=-\frac{9}{5} \\ m_1=-\frac{1}{m_2} \\ -4\ne-\frac{1}{-\frac{9}{5}} \\ -4\ne-\frac{\frac{1}{1}}{\frac{-9}{5}} \\ -4\ne-\frac{1\cdot5}{1\cdot-9} \\ -4\ne-\frac{5}{-9} \\ -4\ne\frac{5}{9} \end{gathered}[/tex]Since their respective slopes do not have the relationship shown below, then the lines are not perpendicular.
Third pair of linesLine NP
[tex]\begin{gathered} (x_1,y_1)=N\mleft(5,-3\mright) \\ (x_2,y_2)=P\mleft(0,4\mright) \\ m=\frac{y_2-y_1}{x_2-x_1} \\ m=\frac{4-(-3)}{0-5} \\ m=\frac{4+3}{-5} \\ m=-\frac{7}{5} \end{gathered}[/tex]Line QR
[tex]\begin{gathered} (x_1,y_1)=Q\mleft(-3,-2\mright) \\ (x_2,y_2)=R\mleft(4,3\mright) \\ m=\frac{y_2-y_1}{x_2-x_1} \\ m=\frac{3-(-2)}{4-(-3)} \\ m=\frac{3+2}{4+3} \\ m=\frac{5}{7} \end{gathered}[/tex]Now, when graphing the two lines, we have:
Two lines are perpendicular if their slopes have the following relationship:
[tex]\begin{gathered} m_1=-\frac{1}{m_2} \\ \text{ Where }m_1\text{ is the slope of the first line and }m_2\text{ is the slope of the second line} \end{gathered}[/tex]In this case, we have:
[tex]\begin{gathered} m_1=-\frac{7}{5} \\ m_2=\frac{5}{7} \\ -\frac{7}{5}_{}=-\frac{1}{\frac{5}{7}_{}} \\ -\frac{7}{5}_{}=-\frac{\frac{1}{1}}{\frac{5}{7}_{}} \\ -\frac{7}{5}_{}=-\frac{1\cdot7}{1\cdot5}_{} \\ -\frac{7}{5}_{}=-\frac{7}{5}_{} \end{gathered}[/tex]Since the slopes of the lines NP and QR satisfy the previous relationship, then this pair of lines are perpendicular.
Fourth pair of lines
Line ST
[tex]\begin{gathered} (x_1,y_1)=S\mleft(0,3\mright) \\ (x_2,y_2)=T\mleft(0,7\mright) \\ m=\frac{y_2-y_1}{x_2-x_1} \\ m=\frac{7-3_{}}{0-0} \\ m=\frac{4}{0} \\ \text{Undefined slope} \end{gathered}[/tex]The line ST has an indefinite slope because it is not possible to divide by zero. Lines that have an indefinite slope are vertical.
Line VW
[tex]\begin{gathered} (x_1,y_1)=V\mleft(2,3\mright) \\ (x_2,y_2)=W\mleft(5,3\mright) \\ m=\frac{y_2-y_1}{x_2-x_1} \\ m=\frac{3-3_{}}{5-2} \\ m=\frac{0}{3} \\ m=0 \end{gathered}[/tex]Line VW has a slope of 0. Lines that have a slope of 0 are horizontal.
Now, when graphing the two lines, we have:
As we can see in the graph, the ST and VW lines are perpendicular.
When a number is divided by $5,$ the result is $50$ less than if the number had been divided by $6$. What is the number?
Answer:
1500$
Step-by-step explanation:
Answer: -1500
Step-by-step explanation: yes it is
Ratios equivalent to 13:14
Equivalent fractions are those that, despite their visual differences, reflect the same value. For instance, if you take a cake and cut it into two equal pieces, you will have consumed half of the cake.
How can one determine equivalent fractions?When two fractions are expressed in their simplest form, they are said to be equivalent. When broken down into its simplest components, the fraction 26/28 equals 13/14. You only need to multiply the numerator and denominator of the reduced fraction (13/14) by the same natural number, i.e., multiply by 2, 3, 4, 5, 6 to get analogous fractions.
13 and 14 in decimal form.In decimal form, 13/14 is 0.92857142857143.
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Sam graduation picnic cost $13 for decoration plus an additional $5 for each attendee.at most how many attendees can there be if sam budgets a total of $33 for his graduation picnic?
Given:
The cost of graduation picnic = $13 for decoration plus an additional $5 for each attendee
Let the number of attendance = x
we need to find x when sam budgets a total of $33 for his graduation picnic
So,
[tex]13+5x=33[/tex]solve for x, subtract 13 from both sides:
[tex]\begin{gathered} 13+5x-13=33-13 \\ 5x=20 \end{gathered}[/tex]divide both sides by 5
[tex]\begin{gathered} \frac{5x}{5}=\frac{20}{5} \\ \\ x=4 \end{gathered}[/tex]So, the answer is:
The number of attendees = 4
Let Events A and B be described as follows:• P(A) = buying popcorn• P(B) = watching a movieThe probability that you watch a movie this weekend is 48% The probability of watching amovie this weekend and buying popcorn is 38%. If the probability of buying popcorn is 42%,are watching a movie and buying popcorn independent?
Solution:
Given that;
[tex]\begin{gathered} P(A)=42\%=0.42 \\ P(B)=48\%=0.48 \\ P(A\cap B)=38\%=0.38 \end{gathered}[/tex]To find out if watching a movie and buying a popcorn are independent, the formula is
[tex]\begin{gathered} P(A|B)=\frac{P(A\cap B)}{P(B)}=\frac{0.38}{0.48}=0.79166 \\ P(A|B)=0.79\text{ \lparen two decimal places\rparen} \end{gathered}[/tex]From the deductions above;
Hence, the answer is
[tex]No,\text{ because }P(A|B)=0.79\text{ and the }P(A)=0.42\text{ are not equal}[/tex]Solve the equation 20 +7k = 8(k + 2).
[tex]k=4[/tex].
Step-by-step explanation:1. Write the expression.[tex]20 +7k = 8(k + 2)[/tex]
2. Solve the parenthesis on the right hand side of the equation.[tex]20 +7k = 8k+16[/tex]
3. Subtract 20 from both sides of the equation.[tex]-20+20 +7k = 8k+16-20\\ \\7k = 8k-4[/tex]
4. Subtract 8k from both sides of the equation.[tex]7k-8k = 8k-4-8k\\ \\-k = -4[/tex]
5. Multiply both sides by "-1".[tex](-1)-k = -4(-1)\\ \\k=4[/tex]
6. Verify.[tex]20 +7(4) = 8((4) + 2)\\ \\20 +28 = 8(6)\\ \\48=48[/tex]
7. Express the result.[tex]k=4[/tex].
Answer:
k = 4
Step-by-step explanation:
a) Simplify both sides of the equation.
20 + 7k = 8(k + 2)
20 + 7k = (8)(k) + (8)(2)
20 + 7k = 8k + 16
7k + 20 = 8k + 16
b) Subtract 8k from both sides.
7k + 20 − 8k = 8k + 16 − 8k
−k + 20 = 16
c) Subtract 20 from both sides.
−k + 20 − 20 = 16 − 20
−k = −4
Divide both sides by -1.
-k/-1 = -4/-1
k = 4
Graph this steph function on the coordinate grid please help I don’t understand!
Given:
[tex]f(x)=\begin{cases}{-5\text{ }if\text{ }-5Required:
We need to graph the given step function.
Explanation:
From the given data we have, -5,-1, and 2 are the respective values of y.
[tex]-5The draw annulus to denote the points do not lie on the line since there is a symbol '<'.
Final answer:
Describe using words in a sentence the transformations that must be applied to the graph of f to obtain thegraph of g(x) = -2 f(x) + 5.
we have
f(x)
and
g(x)=-2f(x)+5
so
step 1
First transformation
Reflection about the x-axis
so
f(x) -----> -f(x)
step 2
Second transformation
A vertical dilation with a scale factor of 2
so
-f(x) ------> -2f(x)
step 3
Third transformation
A translation of 5 units up
so
-2f(x) -------> -2f(x)+5
An oil slick is expanding as a circle. The radius of the circle is currently 2 inches and is increasing at a rate of 5 inches per hour. Express the area of the circle, as a function of ℎ, the number of hours elapsed. ( Answer should be (ℎ)= some function of ℎ, enter as pi )
Answer: f(h) = (2 pi (5x + 2)^2)
Step-by-step explanation:
1) Set the area of the circle in an equation
f(h) = pi r ^2
2) Set r to the rate that the circle is growing
The rate is 5x +2 because its starting value is two, and the 5x because it is growing 5 inches per hour.
f(h) = (2 pi (5x + 2)^2)
Lukalu is rappelling off a cliff. The parametric equations that describe her horizontal and vertical position as a function of time are x ( t ) = 8 t and y ( t ) = − 16 t 2 + 100 and . How long does it take her to reach the ground? How far away from the cliff is she when she lands? Remember to show all of the steps that you use to solve the problem.
SOLUTION
Now the ground is assumed to be 0, so we have that
[tex]y(t)=0[/tex]So, that means we have
[tex]\begin{gathered} y(t)=-16t^2+100 \\ 0=-16t^2+100 \\ 16t^2=100 \\ t^2=\frac{100}{16} \\ t=\sqrt{\frac{100}{16}} \\ t=\frac{10}{4} \\ t=2.5\text{ seconds } \end{gathered}[/tex]Now, we have found t, which is how long it takes to get to the ground, you can plug it into x(t) to find the horizontal distance travelled, we have
[tex]\begin{gathered} x(t)=8t \\ x(2.5)=8\times2.5 \\ =20\text{ feet } \end{gathered}[/tex]Hence it takes her 2.5 seconds to reach the ground
And she is 20 feet away from the cliff
Which fraction is the smallest?8/9, 9/10, 11/12, 12/13
Given:
[tex]\frac{8}{9},\frac{9}{10},\frac{11}{12},\frac{12}{13}[/tex][tex]\frac{8}{9}=0.8889[/tex][tex]\frac{9}{10}=0.9[/tex][tex]\frac{11}{12}=0.9167[/tex][tex]\frac{12}{13}=0.9231[/tex][tex]\frac{8}{9}\text{ is the smallest fraction.}[/tex]3,000 is 1/10 of?-30030,000300,000
Let X be the number, then we know that
[tex]\frac{X}{10}=3,000[/tex]If we isolate X by moving 10 to the right hand side, we obtain
[tex]\begin{gathered} X=10\cdot3,000 \\ X=30,000 \end{gathered}[/tex]that is, the answer is 30,000
Abner and Xavier can make a loaf of bread from scratch in 3 hours. This includes
preparation and baking time. Prep time is 135 minutes. For how long does the bread
bake?
Check Your Understanding- Question 1 of 2
Fill in the blanks to identify the steps needed to solve the problem.
Start by converting the total time to minutes. The conversion factor from hours to
60 minutes
minutes is
1 hour
It takes
Attempt 2 of 2
810 minutes in total to make the bread.
Answer:
the bread bakes for 45 minutes (yum)
Ariel makes $35 a day. How much can she get for 1/7 of a day? Ariel will earn ____ for 1/7 of the day.
We can calculate how much she get for 1/7 of a day by multiplying the daily pay rate by 1/7:
[tex]E=\frac{1}{7}\cdot35=\frac{35}{7}=5[/tex]Answer: Ariel will earn $5 for 1/7 of the day.
A wind-up toy car can travel 5 yards in about 3 minutes. If the car travels at a constant speed, then how many minutes will it takes to travel 40 meters? State your answer to the nearest minute.( 1 yard = 0.92 meters)
A) 20
B) 22
C) 24
D) 26
Answer: D
Step-by-step explanation:
0.92 metres = 1 yard
40 metres = 43.5 yard [tex](\frac{40 * 1}{0.92})[/tex]
5 yards : 3 minutes
43.5 yards : 26.1 minutes [tex](\frac{43.5*3}{5})[/tex]
26.1 mins ≈ 26 mins
Show all work to identify the asymptotes and state the end behavior of the function f(x) = 6x/ x - 36
Vertical asymptotes are when the denominator is 0 but the numerator isn't 0.
[tex]x-36=0 \implies x=36[/tex]
Since this value of x does not make the numerator equal to 0, the vertical asymptote is [tex]x=36[/tex].
Horizontal asymptotes are the limits as [tex]x \to \pm \infty[/tex].
[tex]\lim_{x \to \infty} \frac{6x}{x-36}=\lim_{x \to \infty}=\frac{6}{1-\frac{36}{x}}=6\\\\\lim_{x \to -\infty} \frac{6x}{x-36}=\lim_{x \to -\infty}=\frac{6}{1-\frac{36}{x}}=6\\\\[/tex]
So, the horizontal asymptote is [tex]y=6[/tex].
End behavior:
As [tex]x \to \infty, f(x) \to -\infty[/tex]As [tex]x \to -\infty, f(x) \to \infty[/tex].Create a graph system of linear equations that model this situation.
x represents the number of packages of coffee sold
y represents the number of packages of pastry sold
We were the cost of a package of coffee is $6 and the cost of a package of pastry is $4. If the total amount made from selling x packages of coffee and y packages of pastry is $72, then the equation representing this scenario is
6x + 4y = 72
Also, we were told that she sold 2 more packages of coffee than pastries. This means that
x = y + 2
Thus, the system of linear equations is
6x + 4y = 72
x = y + 2
We would plot these equations on the graph. The graph is shown below
The red line represents x = y + 2
The bu line represents x = y + 2
Consider the numbers below. Use two of the numbers to make the greatest sum, greatest difference, greatest product, and greatest quotient.-5 1/23.75-20.8-411.25
Given data,
[tex]-5\frac{1}{2},\text{ 3.75,-20.8,-4,11.25}[/tex]For the greatest sum,
We need to add the two large number.
Thus,
[tex]11.25+3.75=15[/tex]For the greatest difference, we need to substract the largest number from the smallest,
Thus,
[tex]11.25-(-20.8)=32.05[/tex]To find the greatest product,
we should multiply the two great numbers.
[tex]11.25\times3.75=42.1875[/tex]To find the greatest quotient,
we need to divide the largerst number by the smallest
[tex]\frac{11.25}{3.75}=3[/tex]