Ask question

Ask question

All categories

3 years ago

15

Arnold's entire workout consisted of 10 minutes of warm up exercise, 25 minutes of lifting weights, and 15 minutes on the treadm

ill. What was the ratio of the number of minutes he lifted weights to the total number of minutes of his entire workout?

2 answers:

lord [1]3 years ago

5 0

Add all the times together:

10 + 25 + 15 = 50 minutes total.

For the ratio divide the time for weights by total time:

25/50 which reduces to 1/2

The ratio is 1/2

aliina [53]3 years ago

3 0

Answer:

15:50 = 3:10

3:10 is the answer

You might be interested in

Which expressions are equivalent to (5⋅x)⋅3 ? Drag and drop the equivalent expressions into the box.

marysya [2.9K]

<h2>Greetings!</h2>

Answer:

3⋅(5⋅x)

5⋅(x⋅3)

15x

Step-by-step explanation:

As the values are inside the brackets, it does not matter what side the (x3) is on, so 3⋅(5⋅x) is equivalent.

Multiplying the contents of the brackets in the third one (x * 3) by 5 gives the same value as 3 * (x * 5) so 5⋅(x⋅3) is also equivalent.

On multiplying the brackets out:

5 * x = 5x

5x * 3 = 15x

So 15x is also equivalent.

<h2>Hope this helps!</h2>

3 0

3 years ago

A clothing salesman want to earn $6000 in March. He receives a base salary of $4000 per mont as well as a 10% commission for all

N76 [4]

He would have to sell $20,000 worth of clothes to get $2,000

6 0

2 years ago

Mason walked on Monday, Wensday, and Friday. Use these clues to find how far he walked each day. These distances were six-eights

krok68 [10]

Answer:

Monday he walked $\frac{1}{4}miles$

Wednesday he walked $\frac{6}{8}miles$

Friday he walked $\frac{1}{6}miles$

Step-by-step explanation:

Given Mason walked on Monday, Wednesday and Friday. These distances were six-eights mile, one-fourth mile, and one-sixth mile. He did not walk the farthest on Monday. He walked less on Friday than Monday. we have to find how far he walked each day.

distances are $\frac{6}{8}, \frac{1}{4}, \frac{1}{6}$ that are 0.75, 0.25 and 0.17 respectively.

Now, he didn't walk farthest on Monday and also walked less on Friday than Monday.

∴ Less distance travelled is $\frac{1}{6}$ which is on friday and then $\frac{1}{4}$ on monday.

Rest distance which is $\frac{6}{8}$ on wednesday.

Hence, Monday he walked $\frac{1}{4}miles$

Wednesday he walked $\frac{6}{8}miles$

Friday he walked $\frac{1}{6}miles$

4 0

3 years ago

Read 2 more answers

Hey there,so I wanted to help some people out and give this question 48 points. This is kinda easy,but i need a bit of help. Wha

madreJ [45]

Answer:

<em>All real solutions.</em>

Step-by-step explanation:

5(- 3x - 2) - (x - 3) = -4(4x + 5) + 13

-15x-10-x+3=-16x-20+13

-16x-7=-16x-7

All real solutions.

hope this halps... :D lol

4 0

3 years ago

PLEASE HELPPP!!!!!!!!!!!!!

Leto [7]

For this case we have:

Question 1:

According to Heron's formula, the area of a triangle is given by:

$A=\sqrt{(s(s-a)(s-b)(s-c))}$

Where a, b and c are the sides of the triangle and s is the semi-perimeter of the triangle given by:

$s=\frac{(a+b+c)}{2}$

Let:

$a = 20mm\\b = 23mm\\c = 27mm$

We have s:

$s=\frac{(20+23+27)}{2}\\s=\frac{70}{2}\\s=35$

Substituting in the formula of the area:

$A=\sqrt{(35(35-20)(35-23)(35-27))}\\A=\sqrt{(35(15)(12)(8))}\\A=\sqrt{(50400)}\\A=224.5mm^2$

Thus, the area of the triangle is $A = 224.5mm ^ 2$

Answer:

$A = 224.5mm ^ 2$

Question 2:

The area of a traingule can be expressed as:

$A=\frac{(b*h)}{2}$

Where:

b: Base of the triangle

h: Triangle height

Let:

$b = 20m\\h = 18m$

Substituting the values in the expression we have:

$A=\frac{(20*18)}{2}\\A=\frac{360}{2}\\A=180m^2$

Thus, the area of the triangle is $A = 180m ^ 2$

Answer:

$A = 180m ^ 2$

Question 3:

It is known that the area of a rectangle is given by:

$A = l * w$

Where:

l: It is the length of the rectangle

w: It is the width of the rectangle

So:

$l = 9 feet\\w = 8 feet$

Substituting:

$A = (9 * 8) feet ^ 2\\A = 72 feet ^ 2$

Thus, the area of Laura's carpet is given by $A = 72 feet ^ 2$

Answer:

$A = 72 feet ^ 2$

Question 4:

We must take out the area of the outer wall, given by a rectangle, then:

$A = l * a$

Where:

l: It is the length of the rectangle

a: It is the height of the rectangle

We have:

$l = 48 feet\\a = 9 feet$

Substituting in the formula we have:

$A = (48 * 9) feet ^ 2\\A = 432ft ^ 2$

Thus, the area of the exterior wall is given by: $A = 432 feet ^ 2$

$432ft ^ 2> 400ft ^ 2$

So, a can of paint is not enough to cover the exterior wall.

Answer:

A can of paint is not enough to cover the exterior wall.

Question 5:

A regular hexagon is formed by 6 equal triangles. The area of the hexagon is given by:

$A=\frac{(perimeter* apothem)}{2}$

Where the perimeter is given by the sum of the sides, that is:

$perimeter = (10 + 10 + 10 + 10 + 10 + 10) cm\\perimeter = 60cm$

And the apothem is the height of each of the triangles that make up the hexagon, that is:

$apothem = 5cm$

Substituting in the formula:

$A=\frac{(60*5)}{2}\\A=\frac{300}{2}\\A=150cm^2$

Thus, the area of the hexagon is $A = 150cm ^ 2$

Answer:

$A = 150cm ^ 2$

7 0

2 years ago