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mart [117]
3 years ago
12

If a man weighs 185 pounds, and 60% of his body weight is water, how many pounds of water does he have?

Mathematics
1 answer:
Kitty [74]3 years ago
6 0

Answer:

111 lbs

Step-by-step explanation:

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Lisa's mom is making cookies to sell at the bake sale. She spent 12 minutes mixing the cookie batter. It took 18 minutes to bake
Llana [10]

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38 min

Step-by-step explanation:

cuz 12+8+18= 38

pls mark brainliest

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3 years ago
On a coordinate plane, a line goes through (negative 4, 0) and (0, 2).
Kazeer [188]

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A C D

Step-by-step explanation:

3 0
3 years ago
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The length of a rectangle is increasing at a rate of 6 cm/s and its width is increasing at a rate of 5 cm/s. When
atroni [7]

Answer:

The area of the rectangle is increasing at a rate of 84 square centimeters per second.

Step-by-step explanation:

The area for a rectangle is given by the formula:

A=w\ell

Where <em>w</em> is the width and <em>l</em> is the length.

We are given that the length of the rectangle is increasing at a rate of 6 cm/s and that the width is increasing at a rate of 5 cm/s. In other words, dl/dt = 6 and dw/dt = 5.

First, differentiate the equation with respect to <em>t</em>, where <em>w</em> and <em>l</em> are both functions of <em>t: </em>

\displaystyle \frac{dA}{dt}=\frac{d}{dt}\left[w\ell]

By the Product Rule:

\displaystyle \frac{dA}{dt}=\frac{dw}{dt}\ell +\frac{d\ell}{dt}w

Since we know that dl/dt = 6 and that dw/dt = 5:

\displaystyle \frac{dA}{dt}=5\ell + 6w

We want to find the rate at which the area is increasing when the length is 12 cm and the width is 4 cm. Substitute:

\displaystyle \frac{dA}{dt}=5(12)+6(4)=84\text{ cm}^2\text{/s}

The area of the rectangle is increasing at a rate of 84 square centimeters per second.

8 0
3 years ago
1/3x +5<br><br> x=-6<br><br> pls help!!!
lara [203]
Answer:

x = - 9/8

Explanation:
3 0
3 years ago
What is the measure of ZA?<br> 100"<br> A. 30<br> B. 50°<br> C. 70°<br> D. 130°
dmitriy555 [2]
Answe is B: 50
It needs to equal 180 so
100+30=130
180-130=50
8 0
3 years ago
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