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4 years ago

12

Pete drove 224 miles in 4 hours on Saturday if he drives at the same rate on Sunday,how long would it take him to drive 490 mile

s

2 answers:

grin007 [14]4 years ago

8 0

Answer:

Should be 8.75 hours

Step-by-step explanation:

There are 56 miles per an hour.

224 / 4 = 56

So divide 490 by 56

490 / 56 = 8.75

julia-pushkina [17]4 years ago

5 0

Answer: 8.9 hours

Step-by-step explanation: 244 in 4 hours is 55 mph

so I divide 490 by 55 and got 8.9 hours

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Please answer correctly! I will mark you Brainliest!

vovikov84 [41]

Answer:

D. 480

Step-by-step explanation:

V= <u>LWH / 3 </u>

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V = <u>12x12x10 / 3</u>

<u>V= 480</u>

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3 years ago

Read 2 more answers

A circle whose area is 4 has a radius of x find the area of a circle whose radius is 3x

____ [38]

The first one is 1.12867 because you need to set it up like this.
4=3.14(x)squared
4/3.14=x(squared)
x=1.12867

The second one is done simply by multiplying it out.
A=3.14(3x)squared
A=3.14(9x[squared])
A=28.26x[squared] -The squared is referring to the x.

7 0

3 years ago

Read 2 more answers

Set up an integral for the volume of the solid obtained by rotating the region bounded by the given curves about the specified l

Sloan [31]

Answer:

The integral of the volume is:

$V = 32\pi\int\limits^1_0 {\sqrt{(1 - y^2)} \, dy$

The result is: $V = 78.97731$

Step-by-step explanation:

Given

Curve: $x^2 + 4y^2 = 4$

About line $x = 2$ --- Missing information

Required

Set up an integral for the volume

$x^2 + 4y^2 = 4$

Make x^2 the subject

$x^2 = 4 - 4y^2$

Square both sides

$x = \sqrt{(4 - 4y^2)$

Factor out 4

$x = \sqrt{4(1 - y^2)$

Split

$x = \sqrt{4} * \sqrt{(1 - y^2)$

$x = \±2 * \sqrt{(1 - y^2)$

$x = \±2 \sqrt{(1 - y^2)$

Split

$x_1 = -2 \sqrt{(1 - y^2)}\ and\ x_2 = 2 \sqrt{(1 - y^2)}$

Rotate about x = 2 implies that:

$r = 2 - x$

So:

$r_1 = 2 - (-2 \sqrt{(1 - y^2)})$

$r_1 = 2 +2 \sqrt{(1 - y^2)}$

$r_2 = 2 - 2 \sqrt{(1 - y^2)}$

Using washer method along the y-axis i.e. integral from 0 to 1.

We have:

$V = 2\pi\int\limits^1_0 {(r_1^2 - r_2^2)} \, dy$

Substitute values for r1 and r2

$V = 2\pi\int\limits^1_0 {(( 2 +2 \sqrt{(1 - y^2)})^2 - ( 2 -2 \sqrt{(1 - y^2)})^2)} \, dy$

Evaluate the squares

$V = 2\pi\int\limits^1_0 {(4 +8 \sqrt{(1 - y^2)} + 4(1 - y^2)) - (4 -8 \sqrt{(1 - y^2)} + 4(1 - y^2))} \, dy$

Remove brackets and collect like terms

$V = 2\pi\int\limits^1_0 {4 - 4 + 8\sqrt{(1 - y^2)} +8 \sqrt{(1 - y^2)}+ 4(1 - y^2) - 4(1 - y^2)} \, dy$

$V = 2\pi\int\limits^1_0 { 16\sqrt{(1 - y^2)} \, dy$

Rewrite as:

$V = 16* 2\pi\int\limits^1_0 {\sqrt{(1 - y^2)} \, dy$

$V = 32\pi\int\limits^1_0 {\sqrt{(1 - y^2)} \, dy$

Using the calculator:

$\int\limits^1_0 {\sqrt{(1 - y^2)} \, dy = \frac{\pi}{4}$

So:

$V = 32\pi\int\limits^1_0 {\sqrt{(1 - y^2)} \, dy$

$V = 32\pi * \frac{\pi}{4}$

$V =\frac{32\pi^2}{4}$

$V =8\pi^2$

Take:

$\pi = 3.142$

$V = 8* 3.142^2$

$V = 78.97731$ --- approximated

3 0

3 years ago

Find factors of g(x)=2x^3+3x^2-23x-12 when (2x+1) is a factor

Sliva [168]

Answer:

Step-by-step explanation:

STEP

1

:

Equation at the end of step 1

(((2 • (x3)) - 3x2) - 23x) + 12

STEP

2

:

Equation at the end of step

2

:

((2x3 - 3x2) - 23x) + 12

STEP

3

:

Checking for a perfect cube

3.1 2x3-3x2-23x+12 is not a perfect cube

Trying to factor by pulling out :

3.2 Factoring: 2x3-3x2-23x+12

Thoughtfully split the expression at hand into groups, each group having two terms :

Group 1: -23x+12

Group 2: 2x3-3x2

Pull out from each group separately :

Group 1: (-23x+12) • (1) = (23x-12) • (-1)

Group 2: (2x-3) • (x2)

Bad news !! Factoring by pulling out fails :

The groups have no common factor and can not be added up to form a multiplication.

8 0

3 years ago

1 matrix question help (with picture)

Dominik [7]

Answer:

a = 9/2

Step-by-step explanation:

In order for the equation to be true, the value of the scalar multiplier must be ...

15/3 = -10/-2 = 35/7 = 5

So, you're solving ...

(2/3)a + 2 = 5

(2/3)a = 3 . . . . . . subtract 2

a = 9/2 . . . . . . . . multiply by 3/2

3 0

4 years ago