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3 years ago

Does the unit rate always have to be set up as the denominator?

2 answers:

5 0

Yes, as if you said your car was going 50 miles for 2 hours, and did this:
2 ÷ 50
50 ÷ 50,
it would equal 0.04 miles per hour. That is clearly incorrect. But, if you did:
50 ÷ 2
2 ÷ 2,
it would equal 25 miles an hour. Conclusion, you must have the unit rate set up as the denominator

Sveta_85 [38]3 years ago

3 0

Yes. It always has to be /1.

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The rainfall amount last year was 6 inches below normal. The rainfall this year was 13 inches more than last year. What is the d

Lostsunrise [7]

Answer:

Step-by-step explanation:

The term difference means to subtract. Therefore, 13in - 6in, is equal to 7 inches.

13-6=7

Hope this Helps!!!

6 0

3 years ago

Find the value of x

Svet_ta [14]

Answer:

$150^o$

Step-by-step explanation:

Step 1: Distribute the plus

$(x + 15)^o + (135 - x)^o$

$x + 15 + 135 - x$

Step 2: Combine like terms

$x + 15 + 135 - x$

$(x - x)^o + (15 + 135)^o$

$0 + 150^o$

$150^o$

Answer: $150^o$

7 0

3 years ago

Read 2 more answers

Write the equation of the circle centered at (3, 5) and tangent to x = -1. I need help pleasse

Annette [7]

Answer:

(x - 3)² + (y - 5)² = 16

Step-by-step explanation:

the equation of a circle in standard form is

(x - h)² + (y - k)² = r²

where (h, k ) are the coordinates of the centre and r the radius

given (h, k ) = (3, 5 ) we require to find r

r is the distance from the centre to a point on the circle

given x = - 1 is a tangent the r is the distance between the x- coordinates

r = | 3 - (- 1) | = | 3 + 1 | = | 4 | = 4

then equation of circle is

(x - 3)² + (y - 5)² = 4² , that is

(x - 3)² + (y - 5)² = 16

7 0

2 years ago

In Exercises 45–48, let f(x) = (x - 2)2 + 1. Match the function with its graph

MA_775_DIABLO [31]

Answer:

45) The function corresponds to graph A

46) The function corresponds to graph C

47) The function corresponds to graph B

48) The function corresponds to graph D

Step-by-step explanation:

We know that the function f(x) is:

$f(x)=(x-2)^{2}+1$

45)

The function g(x) is given by:

$g(x)=f(x-1)$

using f(x) we can find f(x-1)

$g(x)=((x-1)-2)^{2}+1=(x-3)^{2}+1$

If we take the derivative and equal to zero we will find the minimum value of the parabolla (x,y) and then find the correct graph.

$g(x)'=2(x-3)$

$2(x-3)=0$

$x=3$

Puting it on g(x) we will get y value.

$y=g(3)=(3-3)^{2}+1$

$y=g(3)=1$

Then, the minimum point of this function is (3,1) and it corresponds to (A)

46)

Let's use the same method here.

$g(x)=f(x+2)$

$g(x)=((x+2)-2)^{2}+1$

$g(x)=(x)^{2}+1$

Let's find the first derivative and equal to zero to find x and y minimum value.

$g'(x)=2x$

$0=2x$

$x=0$

Evaluatinf g(x) at this value of x we have:

$g(0)=(x)^{2}+1$

$g(0)=1$

Then, the minimum point of this function is (0,1) and it corresponds to (C)

47)

Let's use the same method here.

$g(x)=f(x)+2$

$g(x)=(x-2)^{2}+1+2$

$g(x)=(x-2)^{2}+3$

Let's find the first derivative and equal to zero to find x and y minimum value.

$g'(x)=2(x-2)$

$0=2(x-2)$

$x=2$

Evaluatinf g(x) at this value of x we have:

$g(2)=(2-2)^{2}+3$

$g(2)=3$

Then, the minimum point of this function is (2,3) and it corresponds to (B)

48)

Let's use the same method here.

$g(x)=f(x)-3$

$g(x)=(x-2)^{2}+1-3$

$g(x)=(x-2)^{2}-2$

Let's find the first derivative and equal to zero to find x and y minimum value.

$g'(x)=2(x-2)$

$0=2(x-2)$

$x=2$

Evaluatinf g(x) at this value of x we have:

$g(2)=(2-2)^{2}-2$

$g(2)=-2$

Then, the minimum point of this function is (2,-2) and it corresponds to (D)

I hope it helps you!

8 0

2 years ago

Boat travel 27 miles in two hours how many miles travel in 1/2 hours

Sloan [31]

Answer:

6.75

Step-by-step explanation:

8 0

2 years ago