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2 years ago

You ran 3 1/2 mi in 3/4 hr your friend ran 1 2/5 mi in 1/3hr which of you ran faster

1 answer:

6 0

"You" ran faster. 3 1/2 mi. turn into 3 5/10 mile. 1 2/5 mi. turn into 1 4/10 mile. 3/4 of an hour is 45 min. 1/3 of an hour is 20 min. Therefore you multiply 1 4/10 times 2 to get 2 8/10 mi. in 40 min. Since your friend needs 5 more min. to be at the same time as you, you divide 20 (1/3 of an hour) by 4. Which gives you 5 minutes. Now, take the 1 2/5 of the mile. and divide that by 4. Which according to Symbolab.com gives you 1/10. Now go back to the beginning, you ran 3 5/10 mi. in 45 minutes. Your friend ran 2 8/10 in 40 min. Now that we know 5 min. equals 1/10 miles, you can add that to your friend. which means she/he ran 2 9/10 mi. in 45 min. In conclusion, you ran farther that your friend in 45 min. "You" ran faster. (:

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MariettaO [177]

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3 years ago

What is the area of the figure? A. 66 B. 84 C. 60 D. 72

Lerok [7]

Answer:

C.60

Step-by-step explanation:

4×12=48

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3 0

3 years ago

Read 2 more answers

How many solutions does 9(z+8)=-9z-72 have

zloy xaker [14]

Answer:

1 solution.

General Formulas and Concepts:

Pre-Algebra

Equality Properties

Step-by-step explanation:

Step 1: Define equation

9(z + 8) = -9z - 72

Step 2: Solve for x

Distribute 9: 9x + 72 = -9z - 72
Add 9z to both sides: 18z + 72 = -72
Subtract 72 on both sides: 18z = -144
Divide 18 on both sides: z = -8

Here we see that we will get only 1 solution for z.

4 0

2 years ago

You know the measure of the exterior angle which forms a linear pair with the vertex angle. Describe two ways you can find measu

aev [14]

Answer:

Step-by-step explanation:

A linear pair are two given angles whose sum is equal to $180^{o}$ .

Let the exterior angle be represented by $x^{o}$ , and the three interior angles of the triangle be represented by $a^{o}$ , $b^{o}$ and $c^{o}$ .

Assume that the vertex angle $c^{o}$ is the linear pair to the exterior angle $x^{o}$ .

i.e $x^{o}$ + $c^{o}$ = $180^{o}$ (sum of angles on a straight line)

Thus,

i. $a^{o}$ + $b^{o}$ = $x^{o}$ (sum of two opposite interior angles is equal to an exterior angle)

⇒ $a^{o}$ = $x^{o}$ - $b^{o}$

Also,

$b^{o}$ = $x^{o}$ - $a^{o}$

But,

$a^{o}$ + $b^{o}$ + $c^{o}$ = $180^{o}$ (sum of angles in a triangle)

⇒ $c^{o}$ = $180^{o}$ - ( $a^{o}$ + $b^{o}$ )

and

$a^{o}$ + $b^{o}$ = $180^{o}$ - $c^{o}$

ii. $a^{o}$ + $b^{o}$ + $c^{o}$ = $180^{o}$ (sum of angles in a triangle)

$a^{o}$ = $180^{o}$ - ( $b^{o}$ + $c^{o}$ )

Also,

$a^{o}$ + $b^{o}$ = $x^{o}$

⇒ $b^{o}$ = $x^{o}$ - $a^{o}$

$c^{o}$ = $180^{o}$ - $x^{o}$

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The caterpillar ate the salt

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