The inequality that represents the domain of the function is [tex]\frac 12x - 10 \ge 0[/tex]
How to determine the domain?The function is given as:
[tex]f(x) = \sqrt{\frac 12x - 10} + 3[/tex]
Set the radicand greater than or equal to 0
[tex]\frac 12x - 10 \ge 0[/tex]
Multiply through by 2
x - 20 ≥ 0
Add 20 to both sides
x ≥ 20
Hence, the domain of the function is x ≥ 20
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x^2+3x-18 use factoring to determine the zeros
Answer:
x = 3
x = -6
Step-by-step explanation:
Hello!
Let's first set the equation to 0.
x² + 3x - 18 = 0To factor, first think about two numbers that multiply to -18, but add up to 3. The two numbers are 6 and -3.
Now, expand 3x to 6x and -3x. Then, factor by grouping.
Factorx² + 3x - 18 = 0x² + 6x - 3x - 18 = 0(x² + 6x) - (3x + 18) = 0x(x + 6) -1(3)(x + 6) = 0(x - 3)(x + 6) = 0Now, set each factor to 0 and solve for x
Solve(x - 3) = 0x = 3, x = -6
Are the equations 5x=24+2x and 3x=24 equivalent?
Answer:
Yes they are.
Step-by-step explanation:
By simply moving 2x from the LHS to the RHS we'll get it's equivalent to be 3x.
i.e 5x - 2x = 3x
what is the GCF of (8x-6)
sos T-T
Hi student, let me help you out! :)
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We are asked to find the G.C.F. of 8x-6.
[tex]\triangle~\fbox{\bf{KEY:}}[/tex]
G.C.F. stands for Greatest Common Factor.So what is the G.C.F. of 8x-6? We can list all of their common factors and then select the greatest one, like so:
[tex]\longmapsto\bf{Factors\;of\;8x:}[/tex] 1, 2, 4, 8, x
[tex]\longmapsto\bf{Factors\;of\;-6:}[/tex] -1, 1, 2, -3, 3, -6, 6
So the GCF is:
2
Now, we can also factor it out by placing it outside the parentheses ():
2(4x-3)
See, we factored it out by dividing both terms of the expression by the G.C.F.
[tex]\ddot\bigstar[/tex] Remember this...
[tex]\fbox{\bf{We\;factor\;out\;the\;G.C.F.\;by\;dividing\;all\;the\;terms\;of\;the\;expression\;by\;it}}[/tex]
Hope this helps you out! :D
Ask in comments if any queries arise.
~Just a smiley person helping fellow students :)
Answer: 2, you have to use math apps man it will save your life
Match each value with its formula for ABC.
The solution to the question is:
c is 6 = [tex]\sqrt{a^{2} + b^{2} -2abcosC }[/tex]
b is 5 = [tex]\sqrt{a^{2} + c^{2} -2accosB }[/tex]
cosB is 2 = [tex]\frac{a^{2} + c^{2} - b^{2} }{2ac}[/tex]
a is 4 = [tex]\sqrt{b^{2} + c^{2} -2bccosA }[/tex]
cosA is 3 = [tex]\frac{b^{2} + c^{2} -a^{2} }{2bc}[/tex]
cosC is 1 = [tex]\frac{b^{2} + a^{2} - c^{2} }{2ab}[/tex]
What is cosine rule?it is used to relate the three sides of a triangle with the angle facing one of its sides.
The square of the side facing the included angle is equal to the some of the squares of the other sides and the product of twice the other two sides and the cosine of the included angle.
Analysis:
If c is the side facing the included angle C, then
[tex]c^{2}[/tex] = [tex]a^{2}[/tex] + [tex]b^{2}[/tex] -2ab cos C-----------------1
then c = [tex]\sqrt{a^{2} + b^{2} -2abcosC }[/tex]
if b is the side facing the included angle B, then
[tex]b^{2}[/tex] = [tex]a^{2}[/tex] + [tex]c^{2}[/tex] -2accosB-----------------2
b = [tex]\sqrt{a^{2} + c^{2} -2accosB }[/tex]
from equation 2, make cosB the subject of equation
2ac cosB = [tex]a^{2}[/tex] + [tex]c^{2}[/tex] - [tex]b^{2}[/tex]
cosB = [tex]\frac{a^{2} + c^{2} - b^{2} }{2ac}[/tex]
if a is the side facing the included angle A, then
[tex]a^{2}[/tex] = [tex]b^{2}[/tex] + [tex]c^{2}[/tex] -2bccosA--------------------3
a = [tex]\sqrt{b^{2} + c^{2} -2bccosA }[/tex]
from equation 3, making cosA subject of the equation
2bcosA = [tex]b^{2}[/tex] + [tex]c^{2}[/tex] - [tex]a^{2}[/tex]
cosA = [tex]\frac{b^{2} + c^{2} -a^{2} }{2bc}[/tex]
from equation 1, making cos C the subject
2abcosC = [tex]b^{2}[/tex] + [tex]a^{2}[/tex] - [tex]c^{2}[/tex]
cos C = [tex]\frac{b^{2} + a^{2} - c^{2} }{2ab}[/tex]
In conclusion,
c is 6 = [tex]\sqrt{a^{2} + b^{2} -2abcosC }[/tex]
b is 5 = [tex]\sqrt{a^{2} + c^{2} -2accosB }[/tex]
cosB is 2 = [tex]\frac{a^{2} + c^{2} - b^{2} }{2ac}[/tex]
a is 4 = [tex]\sqrt{b^{2} + c^{2} -2bccosA }[/tex]
cosA is 3 = [tex]\frac{b^{2} + c^{2} -a^{2} }{2bc}[/tex]
cosC is 1 = [tex]\frac{b^{2} + a^{2} - c^{2} }{2ab}[/tex]
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how many 4 digit numbers n have the property that the 3 digit number obtained by removing the leftmost digit is one ninth of n
Answer:
7
Step-by-step explanation:
We want to find the number 4-digit of positive integers n such that removing the thousands digit divides the number by 9.
__
Let the thousands digit be 'd'. Then we want to find the integer solutions to ...
n -1000d = n/9
n -n/9 = 1000d . . . . . . add 1000d -n/9
8n = 9000d . . . . . . . . multiply by 9
n = 1125d . . . . . . . . . divide by 8
The values of d that will give a suitable 4-digit value of n are 1 through 7.
When d=8, n is 9000. Removing the 9 gives 0, not 1000.
When d=9, n is 10125, not a 4-digit number.
There are 7 4-digit numbers such that removing the thousands digit gives 1/9 of the number.
Three pupils are given some books to share among themselves. Molly receives 772 books, which is 345 more than Marie. Marie receives six times more books than Melanie. There are some books left over. If Melanie's share is 1/25 of the total number of books, how many books are left over?
Using proportions, it is found that the number of books that was left over was of 505.
What is a proportion?A proportion is a fraction of a total amount, and the measures are related using a rule of three.
The number of books received by Marie is given by:
M = 772 - 345 = 427.
Melanie received one-sixth of Marie, hence:
Me = 427/6 = 71.
The total amount is 25 times Melanie's amount, hence:
T = 25 x 71 = 1775.
The amount left over is:
L = 1775 - (772 + 427 + 71) = 505
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Write a equation of a line through (4, 6) and (-10, 2)
Answer:
y=2/7x+34/7
Step-by-step explanation:
the slope of this line is 2/7.
y=2/7x+?
? is the y-intercept.
Looking at the line that we have graphed, we can see that it intersects at 34/7 y.
the equation for the line that goes through 4,6 and -10, 2 is
y=2/7x + 34/7
What is the ratio of the length of DE to the length of BC?
OA 1/4
OB.1/3
OC.2/5
OD.1/5
Answer: OD, 1/5
Step-by-step explanation:
Well what I did was take DE and seen how many time it could fit into BC. BC would take up a total of 5 DE's. So since we already have one which is DE then we would have 1/5.
HOPE THIS HELPS! ^_^
The mass of a dust particle is approximately
7.5 x 10-10 kilograms and the mass of an electron
is 9.1 x 10-31 kilograms. Approximately how many
electrons have the same mass as one dust particle?
Total electrons
Mass of dust/Mass of electron7.5×10^{-10}\9.1×10^{-31}0.824×10²¹8.25×10²⁰Answer:
Given:
[tex]\textsf{Mass of a dust particle} = \sf 7.5 \times 10^{-10}[/tex][tex]\textsf{Mass of an electron} = \sf 9.1 \times 10^{-31}[/tex]To calculate how many electrons have the same mass as one dust particle, divide the mass of a dust particle by the mass of an electron:
[tex]\begin{aligned}\implies \dfrac{\textsf{Mass of a dust particle}}{\textsf{Mass of an electron}} & = \sf \dfrac{7.5 \times 10^{-10}}{9.1 \times 10^{-31}}\\\\& = \sf \dfrac{7.5}{9.1} \times \dfrac{10^{-10}}{10^{-31}}\\\\& = \sf 0.8241... \times 10^{(-10-(-31))}\\\\& = \sf 0.8241... \times 10^{21}\\\\& = \sf 8.2 \times 10^{20}\end{aligned}[/tex]
Therefore, approximately [tex]\sf 8.2 \times 10^{20}[/tex] electrons have the same mass as one dust particle.
What is the ratio of the wavelengths of these two notes? a. 5 to 4 b. 3 to 2 c. 6 to 5 d. 2 to 1
The ratio of the wavelengths of the two notes is 2:1 option fourth 2 to 1 is correct.
What is wavelength?The wavelength is the distance between two successive troughs or crests. The highest point on the wave is the crest, while the lowest point is the trough.
We have two notes
Number of cycle for blue = 8
Number of cycle for red = 4
Ratio = 8:4 = 2:1
Or
2 to 1
Thus, the ratio of the wavelengths of the two notes is 2:1 option fourth 2 to 1 is correct.
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Write the ratio of squares to shapes.
I’m really struggling! Please help me!!
Answer:
8.0
Step-by-step explanation:
The triangle is a right triangle, so we are able to use trigonometric functions. Relative to the angle of 37°, we have the opposite side which is 6 and the adjacent side which is x. The trig function that uses the opposite and adjacent is tangent(SohCahToa). We can set up the following equation:
[tex]tan(37) = \frac{opposite}{adjacent}[/tex]
[tex]tan(37) = \frac{6}{x}[/tex]
tan(37) evaluates to about , so we can plug it in and solve for x:
[tex]0.7536 = \frac{6}{x} \\0.7536x = 6\\x = 7.962[/tex]
To the nearest tenth, x rounds to 8.0.
If f(x) = 3x + 2 and g(x) = x² + 1, which expression is equivalent to (fog)(x)?
O (3x + 2)(x2 + 1)
Answer:
Hi,
Step-by-step explanation:
[tex](fog)(x)=g(f(x))=g(3x+2)=(3x+2)^2+1=9x^2+6x+5[/tex]
Answer:
hi
Step-by-step explanation:
Enter your answer and show all the steps that you use to solve this problem in the space provided. Use the binomial expression ( p + q ) n (p+q)n to calculate a binomial distribution with n = 5 and p = 0.3.
The binomial expression (p + q ) n (p+q)n to calculate the binomial distribution given will give a value of 10.
How to calculate the binomial expression?From the information given, the expression (p + q ) n (p+q)n is given to calculate a binomial distribution with n = 5 and p = 0.3.
p = 0.3
q = 1 - 0.3 = 0.7
n = 5
Therefore, expression ( p + q ) n (p+q)n will b:
= (0.3 + 0.7)× 5 + (0.3 + 0.7)× 5
= 5 + 5
= 10
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PLEASE HELPPP!
Consider triangle ABC. What is b?
Answer: 18.7346314730578
This value is approximate.
=================================================
Explanation:
Use the law of sines to find side b
sin(A)/a = sin(B)/b
sin(112)/37 = sin(28)/b
b*sin(112) = 37*sin(28)
b = 37*sin(28)/sin(112)
b = 18.7346314730578
Round that value however you need to.
The portion on your screenshot says "round to the nearest", but it doesn't say to the nearest what. Is it nearest tenth? Hundredth? Something else? That part is cut off. So I'll just write down all of the decimal digits my calculator is showing, and let you do the final rounding step.
Using the curve of best fit, what would be the output (y) when the input (x) is 7?
Answer:
y = 260
Step-by-step explanation:
locate x = 7 on the x- axis, go vertically up to meet the graph at (7, 260 )
when input is 7 then output y = 260
What is a pair of overlapping triangles?
△EDB and △DEC
△ADE and △FCB
△BDE and △EDA
can someone help me please
Replace x with -2 and solve:
A) -2 + -2 = -4 which is not less than or equal to -8
B) 2 + -2 = 0, this is greater than or equal to -8
C) -2 + -2 = -4, this is not less than -8
D) 2 + -2 = 0, this is not greater than 8
Answer: B is true
In the statement 10 +(-10) =0, how would you describe the -10?
Answer:
-10 is the opposite of 10
Step-by-step explanation:
-10 is the opposite of 10 ,because The sum of a number and its opposite equals to zero.
7. The sum of 42 and a certain number is
divided by 4. The result is twice the
number. Find the number.
Answer:
Let the number be X
42 + 2x/4 =2X
cross multiply
42+2X=8x
42=8X-2x
42=6X
6X=42
X=7
Step-by-step explanation:
I'm just a smart guy
Which expression simplifies to 5√3?
A. √30
B.
√45
OC. √75
OD. 15
OO
Answer:
Square root 75
Step-by-step explanation:
You can just enter it in the calculator to see which is equal to 5 square root 3
Given the circle below secant kjI and tangent hi find the length of hi round to the nearest tenth if necessary.
The length of the segment HI in the figure is 32.9
How to determine the length HI?To do this, we make use of the following secant-tangent equation:
HI² = KI * JI
From the figure, we have:
KI = 21 + 24 = 45
JI = 24
So, we have:
HI² = 45 * 24
Evaluate the product
HI² = 1080
Take the square root of both sides
HI = 32.9
Hence, the length of the segment HI is 32.9
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Whoever answers correctly gets BRAINLIEST and 100 points
A regular-size box of crackers measures 214 inches by 912 inches by 14 inches. The manufacturer also sells a snack-size box that has a volume that is 15 of the volume of the regular-size box.
What is the volume of the snack-size box of crackers?
Enter your answer as a mixed number in simplest form by filling in the boxes.
$$
in³
Answer:
[tex]\sf 59 \frac{17}{20} \: in^3[/tex]
Step-by-step explanation:
Regular-size box of crackers
Dimensions: [tex]\sf 2 \frac{1}{4} \: in \times 9 \frac{1}{2}\:in \times 14\: in[/tex]
Snack-size box of crackers
Volume = [tex]\sf \frac{1}{5}[/tex] of the volume of a regular-size box
To calculate the volume of the snack-size box of crackers, calculate the volume of the regular-size box then multiply it by [tex]\sf \frac{1}{5}[/tex].
[tex]\begin{aligned}\textsf{Volume of a rectangular prism} & = \textsf{width x length x height}\\\\\implies \textsf{Volume of regular-size box} & = \sf 9 \frac{1}{2} \times 14 \times 2 \frac{1}{4}\\\\& = \sf \dfrac{19}{2} \times 14 \times \dfrac{9}{4}\\\\& = \sf \dfrac{19 \times 14 \times 9}{2 \times 4}\\\\& = \sf \dfrac{1197}{4}\: in^3\end{aligned}[/tex]
[tex]\begin{aligned}\implies \textsf{Volume of snack-size box} & = \sf \dfrac{1}{5} \times \textsf{volume of regular-size box}\\\\& = \sf \dfrac{1}{5} \times \dfrac{1197}{4}\\\\& = \sf \dfrac{1 \times 1197}{5 \times 4}\\\\& = \sf \dfrac{1197}{20}\\\\& = \sf 59 \dfrac{17}{20}\: in^3\end{aligned}[/tex]
Therefore, the volume of the snack-size box is [tex]\sf 59 \frac{17}{20} \: in^3[/tex]
Find the volume and round to the nearest tenth if necessary 7.5 In 4in 2in
[tex]\huge\huge\bold{\green a \blue n \pink s \purple w \orange e \red r}[/tex]
volume of rectangular prism
Formula to calculate:
[tex]\large\bold{\green V\green = \green l \green * \green w \green * \green h}[/tex]
[tex]\large\bold{V= 7.5*4*2}[/tex]
[tex]\huge\boxed{\red V\red = \red 6 \red 0 \: {\red i \red n}^{\red 3}}[/tex]
length= 7.5 in
width= 2 in
height= 4 in
To find:The volume of the rectangular prism or cuboid.
Solution:Formula: volume= length × width × height
so, V= 7.5 × 2 × 4
[tex]\bold{V= 60 \: {\: in}^{3}}[/tex]
hope this helps!
What percent of 72 is 9? 1/8% 1 1/4% 12 1/2% 125%
Answer:
[tex]12\dfrac 12\%[/tex]
Step by step explanation:
[tex]\text{Let it be}~ x\%,\\\\~~~~~72 \cdot x \% = 9\\\\\\\implies 72\cdot \dfrac{x}{100} = 9\\\\\\\implies x = \dfrac{9 \times 100}{72}\\\\\\\implies x = \dfrac{100}{8}\\\\\\\implies x = \dfrac{25}2\\\\\\\implies x = 12 \dfrac 12\\\\\\\text{Hence 9 is}~ 12\dfrac 12\% ~ \text{of}~ 72[/tex]
Use a net to find the surface area of the cylinder. Use 3.14 for .
Answer:
Formula = 2nr
= 2(3,14 + 3 + 4)
= 2(10,14)
= 20,28
plsssss. Quadrilateral IJKL is similar to quadrilateral MNOP. Find the measure of side MN. Figures are not drawn to scale.
[tex]\\ \rm\Rrightarrow \dfrac{IJ}{IL}=\dfrac{MN}{MP}[/tex]
[tex]\\ \rm\Rrightarrow \dfrac{24}{12}=\dfrac{MN}{3.8}[/tex]
[tex]\\ \rm\Rrightarrow 2=\dfrac{MN}{3.8}[/tex]
[tex]\\ \rm\Rrightarrow MN=3.8(2)[/tex]
[tex]\\ \rm\Rrightarrow MN=7.6[/tex]
Given the diagram below, what is cos(45")?
45° 6
A. √2
B. 6/√2
C. 1/√2
D. 3√2
HELP ME PLEASEEEE ANSWER MY MATH
The water consumption based on the information for three Months will be 22267m³.
How to calculate the values?The total water consumption for the three months will be:
= 7219 + 7390 + 7658
= 22267m³
In conclusion, the information simply want you to add the values of the water given.
Therefore, the water consumption based on the information for three Months will be 22267m³.
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Find the largest interval centered about x = 0 for which the given initial-value problem has a unique solution. (enter your answer using interval notation. ) y'' (tan(x))y = ex, y(0) = 1, y'(0) = 0
The largest interval centered about x = 0 is ([tex]-\frac{\pi}{2}, \frac{\pi}{2}[/tex]) for which the given initial-value problem has a unique solution: [tex]y'' + tan(x)y = e^x[/tex], y(0) = 1, y'(0) = 0.
An initial-value problem (IVP) is one kind of mathematical problem that contains obtaining a solution to a differential equation along with a set of initial conditions. It is commonly encountered in the field of differential equations.
Since it is mentioned in the theorem that in an initial-value problem y' + p(t)y = g(t), y([tex]t_o[/tex]) = [tex]y_o[/tex] has a unique solution if p(t) and g(t) are continuous functions on the interval (a, b) containing [tex]t_o[/tex].
Since [tex]e^x[/tex] is continuous everywhere and tanx is continuous on ([tex]-\frac{\pi}{2}, \frac{\pi}{2}[/tex]) containing 0.
Since R, a set of real numbers, and ([tex]-\frac{\pi}{2}, \frac{\pi}{2}[/tex]), both contain 0.
Thus the largest interval centered about x = 0 is ([tex]-\frac{\pi}{2}, \frac{\pi}{2}[/tex]).
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