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3 years ago

7

What is the length, in feet, of the hypotenuse of a right triangle with legs that are 3 feet long and 4 feet long? Question 10 o

ptions: 25 5 7 7√

1 answer:

wel3 years ago

7 0

A^2+b^2=c^2 so 4(4) +3(3) is 16+9= 25 and the square root of that is 5.

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Identify the end behavior brain list for whoever gets it right

morpeh [17]

Answer:

Number 4, the last one

Step-by-step explanation:

8 0

3 years ago

Read 2 more answers

The formula for finding the perimeter of a rectangle is P = 2L + 2W. If a rectangle has a perimeter of 68 inches and the length

Artemon [7]

Answer:

Width = 10 inches

Step-by-step explanation:

Given the perimeter of a rectangle of 68 inches, and a length that is 14 inches longer than its width.

We can establish the following values to help us solve for the width of a rectangle:

Perimeter (P) = 68 inches

Length (L) = 14 + W inches

Width (W) = unknown

<h3 /><h3><u>Solve for the Width (W)</u></h3>

P = 2(L + W) ⇒ This is the same as P = 2L + 2W, except that 2 is factored out from the right-hand side.

Divide both sides by 2:

$\displaystyle\mathsf{\frac{P}{2}\:=\:\frac{2(L\:+\:W)}{2}}$

$\displaystyle\mathsf{\frac{P}{2}\:=L\:+\:W}$

Substitute the value of the Perimeter and the length (L) into the formula:

$\displaystyle\mathsf{\frac{68}{2}\:=14\:+W\:+\:W}$

Combine like terms on the right-hand side, and simplify the left-hand side of the equation:

$\displaystyle\mathsf{34\:=14\:+2W}$

Subtract 14 from both sides:

34 - 14 = 14 - 14 + 2W

20 = 2W

Divide both sides by 2 to solve for the width (W):

$\displaystyle\mathsf{\frac{20}{2}\:=\:\frac{2W}{2}}$

W = 10 inches

Therefore, the width of the rectangle is 10 inches.

<h3 /><h3><u>Double-check:</u></h3>

Verify whether the derived value for the width is correct:

P = 2L + 2W

68 = 2(14 + 10) + 2(10)

68 = 2(34) + 20

68 = 48 + 20

68 = 68 (True statement).

Thus, the length of the rectangle is 34 inches, and the width is 10 inches.

4 0

3 years ago

In physics, if a moving object has a starting position at so, an initial velocity of vo, and a constant acceleration a, the

emmasim [6.3K]

Answer:

$a=\frac{2S -2v_ot-2s_o}{t^2}$

Step-by-step explanation:

We have the equation of the position of the object

$S = \frac{1}{2}at ^2 + v_ot+s_o$

We need to solve the equation for the variable a

$S = \frac{1}{2}at ^2 + v_ot+s_o$

Subtract $s_0$ and $v_0t$ on both sides of the equality

$S -v_ot-s_o = \frac{1}{2}at ^2 + v_ot+s_o - v_ot- s_o$

$S -v_ot-s_o = \frac{1}{2}at ^2$

multiply by 2 on both sides of equality

$2S -2v_ot-2s_o = 2*\frac{1}{2}at ^2$

$2S -2v_ot-2s_o =at ^2$

Divide between $t ^ 2$ on both sides of the equation

$\frac{2S -2v_ot-2s_o}{t^2} =a\frac{t^2}{t^2}$

Finally

$a=\frac{2S -2v_ot-2s_o}{t^2}$

5 0

3 years ago

Last year Billy weighed 75kg. This year he weighed 90kg. With what percentage did his weight increase. SHOW ALL WORK!!!

k0ka [10]

Answer:

20%

Step-by-step explanation:

90 - 75 = 15

15 ÷ 75 = 0.2

0.2 x 100% = 20%

8 0

3 years ago

On a trip, a motorist drove 150 miles in the morning and 50 miles in the afternoon. His average rate in the morning was twice hi

Lina20 [59]

Answer:

Morning's average rate = 50 mph, and Afternoon's average rate = 25 mph.

Step-by-step explanation:

Suppose he drove 150 miles for X hours, then his average rate in the morning was (150/X) mph.

Given that he spent 5 hours in driving.

And he drove 50 miles for (5-X) hours, then his average rate in the afternoon was 50/(5-X) mph.

Given that his average rate in the morning was twice his average rate in the afternoon.

(150/x) = 2 * 50/(5-x)

150/x = 100/(5-x)

Cross multiplying terms, we get:-

150*(5-x) = 100*x

750 - 150x = 100x

750 = 100x + 150x

750 = 250x

x = 750/250 = 3.

It means he spent 3 hours in the morning and 2 hours in the afternoon.

So morning's average rate = 150/3 = 50 mph.

and afternoon's average rate = 50/(5-3) = 25 mph.

8 0

3 years ago