Help me please I need help

Answer:

Domain: All real numbers
Range: f(x) ≥ - 10

Step-by-step explanation:

I had the same test. Completed it and got it right.
Sorry for the lack of explanation, I can't put it into words at the moment.

Answer:

can be written as  without exponent.

Given:

To find:

Write the given expression without exponent.

Solution:

Formula to be used:

Step 1:

Write the expanded form of given expression.

Step 2:

simplify

or

125×y×y×y without exponent.

Answer:

5 × y × y × y

Step-by-step explanation:

5y³ ( only y is being cubed with 5 a multiplier ) , so

= 5 × y × y × y

Answer:

Step-by-step explanation:

Answer:

B) y = -5x + 7

Step-by-step explanation:

Pick 2 points on the line:  (10, 5) and (15, 4)

slope = change in y ÷ change in x = (15 - 10) ÷ (4 - 5) = -5

from inspection, the y-intercept is (0, 7)

Therefore, using slope-intercept formula:  y = mx + b

(where m is the slope and b is the y-intercept)

y = -5x + 7

Answer:

A) 9. B) 16

Step-by-step explanation:

Range is

12, 14, 15, 16, 17, 17, 21

so 21-12= 9

add all them up 12+14+15+16+17+17+21 = 112/7 = 16

Function defines relationship between variables. The value of the f[g(x)] when the value of f(x)=6x+11 and g(x)=x²+6 is f[g(x)]= 36x²+47.

A function assigns the value of each element of one set to the other specific element of another set.

Given to us

f(x) = 6x + 11

g(x) = x² +  6

As we know the two functions, given to us f(x) = 6x + 11, therefore substitute the value of x as g(x) in order to find the value of f[g(x)] ,

[tex]f(x) = 6x + 11\\\\f[g(x)] = 6(x^2 + 6) + 11\\\\f[g(x)] = 6x^2 + 36 + 11\\\\f[g(x)] = 6x^2 + 47[/tex]

Hence, the value of the f[g(x)] when the value of f(x)=6x+11 and

g(x)=x²+6 is f[g(x)]= 36x²+47.

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Answer:

x = 4

Step-by-step explanation:

15x + 8x -2 = 90

23x - 2 = 90

23x = 90+2

23x = 92

92/23 = 4

check answer:

4 x 15 = 60

4 x 8 = 32

32-2=30

60+30=90

Answer:x=4

Step-by-step explanation: we know,

90 + 8x-2+15x=180

90+23x-2=180

88+23x=180

23x=180-88

x=92/23

x=4

Answer:

B = 150

Step-by-step explanation:

Since angles A and B are opposite interior angles, they are congruent.

8x - 10 = 3x + 90

8x - 3x = 90 + 10

5x = 100

x = 20

Now that we have the value of x, we have to solve for angle B.

B = 3x + 90

B = 3*20 + 90

B = 60 + 90

B = 150

Answer: the right answer on khan academy is 150° (with PROOF)

(x - 7)(x + 12)

Answer:

[tex] {x}^{2} + 5x - 84 \\ {x}^{2} + 12x - 7x - 84 \\ x(x + 12) - 7(x + 12) \\ (x - 7)(x + 12)[/tex]

Answer:

You would expect to win the game 16 times.

Step-by-step explanation:

I set up a proportion for this :

2/7 = x/56

2(56) = 7x (cross multiply)

112 = 7x (multiply 2 and 56)

x = 16 (divide 7 from both sides)

:)

Answer:

y = -3x + 5

Step-by-step explanation:

Answer: A

Step-by-step explanation:

Divide top and bottom by 4

Answer:

[tex]\frac{1}{2}[/tex]

Step-by-step explanation:

The simplest form of the fraction is simply finding a common number to divide both the numerator and denominator by.

4/8 , we can divide both sides by 4 since it's possible to do so.

[tex]\frac{4/4}{8/4}[/tex]

[[tex]\frac{1}{2}[/tex]]

A. is the simplest form

Answer:

Step-by-step explanation:

here you have answer.....

Answer:

(8.2 x 10^9) x (4.5 x 10^-5) = (8.2 x 4.5) x 10 ^(9 - 5) = 36.9 x 10^4 = 3.69 x 10^5

Step-by-step explanation:

The ordinates of the  linear system  is mathematically given as

C=(4,-6)

Question Parameters:

The equations

x - 4y = 28

2x + y = 2

Generally the equation for the linear system is mathematically given as

x - 4y = 28

2x + y = 2

Resolving the equations

x=4 and y=-6

Therefore

co ordinates are

C=(4,-6)

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Answer:

4π/7 radians is 102.86°

Step-by-step explanation:

changing radian to degrees: 1 radian → [tex]\frac{180}{\pi }[/tex]

changing radians to degrees:

[tex]\dfrac{4\pi }{7} *\frac{180}{\pi }[/tex]

[tex]\dfrac{4*180}{7}[/tex]

[tex]\dboxed{102.86}[/tex]

[tex]\\ \rm\longmapsto \dfrac{4\pi}{7}\times \dfrac{180)(\pi}[/tex]

[tex]\\ \rm\longmapsto \dfrac{180(4)}{7}[/tex]

[tex]\\ \rm\longmapsto 720/7[/tex]

[tex]\\ \rm\longmapsto 102.8°[/tex]

Answer: Option 2

The line starts at an x of -4, so the domain is from -4 to infinity since the arrow goes off the graph, meaning it keeps going. The line starts on a y of 2, and the line continues up with the arrow pointing slightly up, so the range is from 2 to infinity.

The domain is from left to right, and the range is down to up

Answer:

2.  {x | x ≥ −4}; {y | y ≥ 2}

Step-by-step explanation:

Let's try to determine the domain first. Look at the x values. The function begins at -4, so our possible domain values also begin at -4, and the values continue positively after -4.

Therefore, our domain is: Domain: {x ≥ -4}.

Or, we could assign our domain using interval notation: [-4,infinity).

Our range, or y values, begin at 2 and continue positively after 2.

Therefore, our range is: Range: {y ≥ 2}.

Again, we could use interval notation to assign our range: [2,infinity)

Therefore, the answer would be

2. {x | x ≥ −4}; {y | y ≥ 2}

Look at the picture to help you better understand

Answer:

29.52cm sqauared

Step-by-step explanation:

length x width = area of rectangle

Answer: 29.52cm2

Step-by-step explanation:

8.2 times 3.6=...

Answer:

1: x = 19 y = 16 ; 2. x = 5 y = -4

Step-by-step explanation:

To check your work plug the numbers back into one of the equations and see if it's true.

For example:
10(19)-9(16)=46

46=46

1. I distributed 3 to all the numbers so that 9y and -9y would cancel then you add like terms.

2. I distributed -4 to all the numbers so that 4x and -4x would cancel. Then add all like terms.

Once you get x or y plug it back into one problem to get the other variable.

Hope this helps!

#1

(2)×3

[tex]\\ \tt\hookrightarrow -6x+9y=30[/tex]--(3)

Subtracting eq(1) from (3)

[tex]\\ \tt\hookrightarrow -16x=-16[/tex]

[tex]\\ \tt\hookrightarrow x=1[/tex]

Put in eq(2)

[tex]\\ \tt\hookrightarrow -2+3y=10[/tex]

[tex]\\ \tt\hookrightarrow y=4[/tex]

#2

(2) ×4

[tex]\\ \tt\hookrightarrow 4x-8y=52--(3[/tex]

Subtracting eq(1) from (3)

[tex]\\ \tt\hookrightarrow -11y=44[/tex]

[tex]\\ \tt\hookrightarrow y=-4[/tex]

Put in eq(2)

[tex]\\ \tt\hookrightarrow x+8=13[/tex]

[tex]\\ \tt\hookrightarrow x=5[/tex]

The number of races needed to determine the champion sprinter = 39 races.

The number of sprinters = 271

The number of sprinters per race= 7

The area to be covered for each race = 200m

For every race covered, a champion is selected and the rest is eliminated.

Therefore, to determine the remaining champion from 271 sprinters,

271/ 7 = 38.7

Which is approximately 39 races.

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Answer:

D of course:

Step-by-step explanation:

Think about it,

50.54 is 50 estimated to the roundest digit, and 20.10 is 20 estimated. So we'll just do 50-20 which is 30

I hope this answers your concern :) Mark me brainliest if you want to (:

Answer:

-30

Step-by-step explanation:

50.54 - 20.10 = 30.44

Answer:

1. 5

2. 3 adult tickets; 2 child tickets

3. See below

Step-by-step explanation:

1.

2 adult tickets cost 2 × $12 = $24

Subtract $24 from $64:

$64 - $24 = $40

Divide $40 by $8:

$40 ÷ $8 = 5

Answer: 5 child tickets

2.

x = number of adult tickets

y = number of child tickets

The total number of tickets is 5, so the first equation is:

x + y = 5

The cost of x adult tickets is 12x. The cost of y child tickets is 8y. The total cost is 12x + 8y. We are told the total cost is $52. The second equation is:

12x + 8y = 52

The system of equation is:

x + y = 5

12x + 8y = 52

Let's solve it by substitution.

Solve the first equation for x.

x = 5 - y

Now we substitute x in the second equation by 5 - y.

12(5 - y) + 8y = 52

Distribute:

60 - 12y + 8y = 52

Combine like terms on the left side:

-4y + 60 = 52

Subtract 60 from both sides:

-4y = -8

Divide both sides by 4.

y = 2

Substitute y = 2 in the first original equation and solve for x.

x + y = 5

x + 2 = 5

x = 3

Answer: 3 adult tickets; 2 child tickets

3.

The customer's claim is incorrect. His sum is $71. 71 is an odd number. Since all ticket prices are even numbers, it is impossible to add only even numbers and get an odd sum. A sum of several amounts of 12 and several amounts of 8 can never equal 71.

We can also show his claim is false using a system of equations.

x + y = 7

12x + 8y = 71

Use substitution.

x = 7 - y

12(7 - y) + 8y = 71

84 - 12y + 8y = 71

84 - 4y = 71

-4y = -13

y = 3.25

This means 3.25 child tickets. The problem states that tickets can be purchased only in whole numbers, so 3.25 tickets cannot be purchased, so the customer's claim is false.

Answer:

They are both right

Step-by-step explanation:

This is because you can do it either way.

For example, if you had a rectangle with the side length 10 and a width 5,

-Korey: you could add them to get 15 and them double them to get 30

or

-Ian: double both (5*2=10 and 10*2=20) and then add and 10+20 = 30

So they are the same

Answer:

The answer is B.

w + m > 224.

11w + 15m ≤ 2,876

Step-by-step explanation:

w + m > 224. There were more than (>) 224 who donated.

11w + 15m ≤ 2,876 No more than (≤) $2,876 was donated.

Hope that helps!

Answer:

the other person is wrong

Step-by-step explanation:

I looked it up and its HEXAHEDRON

Tetrahedron is a pyramid

Answer:

Step-by-step explanation:

A: -1/3x+1

B: 1/4x -2

Answer:

Given Below.

question (a)

coordinates:

(0,1), (3,0)

slope:

[tex]\frac{0-1}{3-0}[/tex]

[tex]-\frac{1}{3}[/tex]

equation:

[tex]y - y1 = m(x - x1)[/tex]

[tex]y - 1 = -\frac{1}{3} (x-0)[/tex]

[tex]y = -\frac{1}{3}x+1[/tex]

question (b)

coordinates:

(0,-2) , (4, -1 )

slope:

[tex]\frac{-1--2}{4-0}[/tex]

[tex]\frac{1}{4}[/tex]

equation:

[tex]y - y1 = m(x - x1)[/tex]

[tex]y + 1 = \frac{1}{4} (x - 4)[/tex]

[tex]y=\frac{1}{4} x -1-1[/tex]

[tex]y=\frac{1}{4} x -2[/tex]

Step-by-step explanation:

Let use have a interger,n

Thus three consectice intergers can be represented by

[tex]n[/tex]

[tex]n + 1[/tex]

[tex]n + 2[/tex]

Multiply, and we get

[tex]n(n + 1)(n + 2)[/tex]

[tex]n( {n}^{2} + 3n + 2)[/tex]

[tex]n {}^{3} + 3 {n}^{2} + 2n[/tex]

Then add the middle number, n+1.

[tex] {n}^{3} + 3n {}^{2} + 3n + 1[/tex]

We must prove that the middle number cubed = the equation.

Using the binomial theroem,

[tex](n + 1) {}^{3} = {n}^{3} + 3( {n}^{2} 1 {}^{1} ) + 3n( {1}^{2} ) + 1 {}^{3} [/tex]

[tex](n + 1 ){}^{3} = {n}^{3} + 3 {n}^{2} + 3n + 1[/tex]

So this is indeed true.

You can do the subsitue variables for n to prove it.