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Vera_Pavlovna [14]

3 years ago

15

a car travels 83 7/10 miles on 2 1/4 gallons of fuel. which is the best estimate of the car travels on one gallon of fuel

1 answer:

Komok [63]3 years ago

5 0

It will travel 39.11 miles on one gallon of fuel
hope it helps :)

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Subtract: (b^2 + 9 ab + 5a ) - ( 3b^2 -25ab + 1)

Brrunno [24]

Answer:

- 2b^2 + 34ab + 5a - 1

Step-by-step explanation:

<em>here's</em><em> your</em><em> solution</em>

<em>=</em><em>></em><em> </em><em>(</em><em>b^</em><em>2</em><em> </em><em>+</em><em> </em><em>9</em><em>a</em><em>b</em><em> </em><em>+</em><em> </em><em>5</em><em>a</em><em>)</em><em> </em><em>-</em><em> </em><em>(</em><em>3</em><em>b</em><em>^</em><em>2</em><em> </em><em>-</em><em> </em><em>2</em><em>5</em><em>a</em><em>b</em><em> </em><em>+</em><em> </em><em>1</em><em>)</em>

<em>=</em><em>></em><em> </em><em>b^</em><em>2</em><em> </em><em>+</em><em> </em><em>9</em><em>a</em><em>b</em><em> </em><em>+</em><em> </em><em>5</em><em>a</em><em> </em><em>-</em><em> </em><em>3</em><em>b</em><em>^</em><em>2</em><em> </em><em>+</em><em> </em><em>2</em><em>5</em><em>a</em><em>b</em><em> </em><em>-</em><em> </em><em>1</em>

<em>=</em><em>></em><em> </em><em>-</em><em> </em><em>2</em><em>b</em><em>^</em><em>2</em><em> </em><em>+</em><em> </em><em>3</em><em>4</em><em>a</em><em>b</em><em> </em><em>+</em><em> </em><em>5</em><em>a</em><em> </em><em>-</em><em> </em><em>1</em>

<em> </em><em> </em><em> </em><em> </em>

<em> </em><em> </em><em>hope</em><em> it</em><em> helps</em>

6 0

2 years ago

Plzz help Solve for x x ÷3 3/10 =2 2/5

Svet_ta [14]

Answer:

$\huge\boxed{x=7\dfrac{23}{25}}$

Step-by-step explanation:

$x\div3\dfrac{3}{10}=2\dfrac{2}{5}\\\\\text{convert the mixed number to the impropper fraction}\\\\3\dfrac{3}{10}=\dfrac{3\cdot10+3}{10}=\dfrac{33}{10}\\\\2\dfrac{2}{5}=\dfrac{2\cdot5+2}{5}=\dfrac{12}{5}\\\\x\div\dfrac{33}{10}=\dfrac{12}{5}\\\\x\times\dfrac{10}{33}=\dfrac{12}{5}\qquad\text{multiply both sides by}\ \dfrac{33}{10}\\\\x\times\dfrac{10\!\!\!\!\!\diagup}{33\!\!\!\!\!\diagup}\times\dfrac{33\!\!\!\!\!\diagup}{10\!\!\!\!\!\diagup}=\dfrac{12}{5}\times\dfrac{33}{10}\\\\x=\dfrac{396}{50}$

$x=\dfrac{198}{25}\\\\x=\dfrac{175+23}{25}\\\\x=\dfrac{175}{25}+\dfrac{23}{25}\\\\x=7\dfrac{23}{25}$

3 0

3 years ago

A serving of chickenpeas contain 1,750 milligrams of potassium. how many grams of potassium is that

Sidana [21]

1.75 Grams is what I got. Every 1 Milligram = 0.001 Gram.

Hope it helps :)

8 0

3 years ago

In Exercises 45–48, let f(x) = (x - 2)2 + 1. Match the<br> 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.

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

$y=g(3)=1$

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

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:

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

$g(0)=1$

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

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:

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

$g(2)=3$

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

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:

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

$g(2)=-2$

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

<u />

I hope it helps you!

<u />

8 0

3 years ago

Classify 4x^5+2x^4-5x^3+12 by number of terms

Semenov [28]

That is 4 terms long.

7 0

3 years ago

Read 2 more answers