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2 years ago

After three plays in which a football team lost 7 yd, gained 3 yd, and

Mathematics

1 answer:

True [87]2 years ago

4 0

Answer:

25 Yard line was the previous location.

Step-by-step explanation:

-7+3-1= -8+3= -5

30-5=25.

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On a 15 hour trip, Rob drove 35% of the time and Mark drove the remainder of the time. How many hours did Mark drive? Explain 9.

Zigmanuir [339]

Answer:

9.75 hours

Step-by-step explanation:

If Rob drove 35% of the time and Mark drove the remainder of the time, the total is 100 %

100-35 = 65

so Mark drove 65% of the time

The total time driven was 15 hours, so we multiply the total time driven by the percentage of time that Mark drove to find the number of hours that Mark drove.

15 * 65%

15*.65

9.75 hours

7 0

3 years ago

How would you convert the repeating, non terminating decimal into a fraction? Explain the process as you solve the problem 0.151

lorasvet [3.4K]

Divide 1515 by 100 and you would get 15.15 the divide the .15 by 3 and and you will get 3.5 so the answer will be 15 3/5

5 0

2 years ago

Read 2 more answers

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aniked [119]

Answer:

the last one :)

Step-by-step explanation:

3 0

2 years ago

Find the area of the part of the plane 3x 2y z = 6 that lies in the first octant.

gavmur [86]

The area of the part of the plane 3x 2y z = 6 that lies in the first octant is mathematically given as

A=3 √(4) units ^2

<h3>What is the area of the part of the plane 3x 2y z = 6 that lies in the first octant.?</h3>

Generally, the equation for is mathematically given as

The Figure is the x-y plane triangle formed by the shading. The formula for the surface area of a z=f(x, y) surface is as follows:

$A=\iint_{R_{x y}} \sqrt{f_{x}^{2}+f_{y}^{2}+1} d x d y(1)$

The partial derivatives of a function are f x and f y.

$\begin{aligned}&Z=f(x)=6-3 x-2 y \\&=\frac{\partial f(x)}{\partial x}=-3 \\&=\frac{\partial f(y)}{\partial y}=-2\end{aligned}$

When these numbers are plugged into equation (1) and the integrals are given bounds, we get:

$&=\int_{0}^{2} \int_{0}^{3-\frac{3}{2} x} \sqrt{(-3)^{2}+(-2)^2+1dxdy} \\\\&=\int_{0}^{2} \int_{0}^{3-\frac{3}{2} x} \sqrt{14} d x d y \\\\&=\sqrt{14} \int_{0}^{2}[y]_{0}^{3-\frac{3}{2} x} d x d y \\\\&=\sqrt{14} \int_{0}^{2}\left[3-\frac{3}{2} x\right] d x \\\\$

$&=\sqrt{14}\left[3 x-\frac{3}{2} \cdot \frac{1}{2} \cdot x^{2}\right]_{0}^{2} \\\\&=\sqrt{14}\left[3-\frac{3}{2} \cdot \frac{1}{2} \cdot x^{2}\right]_{0}^{2} \\\\&=\sqrt{14}\left[3.2-\frac{3}{2} \cdot \frac{1}{2} \cdot 3^{2}\right] \\\\&=3 \sqrt{14} \text { units }{ }^{2}$

In conclusion, the area is

A=3 √4 units ^2