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Naddika [18.5K]

4 years ago

12

A football team gained 35 yards on a first down, lost 15 yards on the second down and gained 12 yards on the third down. How man

y yards do they need to gain on the fourth down to have a ten yard gain from their starting position?

1 answer:

lidiya [134]4 years ago

5 0

13 yards

35-15=20
20+12=32
32-35=3
3+10=13

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Simplify (s)(-3st)(-1/3).

nlexa [21]

$(s)(-3st)\left(-\dfrac{1}{3}\right)=(-3)\left(-\dfrac{1}{3}\right)(s\cdot st)=(1)(s^2t)=\boxed{s^2t}$

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3 years ago

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There is only one positive integer which is exactly twice the sum of its digits. find this 2 digit number

ad-work [718]

There is only one positive integer which is exactly twice the sum of its digits. find this 2 digit number

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3 years ago

If a model scale is 1 foot to 5 feet, how long will the model be

dalvyx [7]

The model will be 5 times longer that stated.

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3 years ago

Evaluate the double integral. 2y2 dA, D is the triangular region with vertices (0, 1), (1, 2), (4, 1) D

zubka84 [21]

Answer:

$\mathbf{\iint _D y^2 dA= \dfrac{22}{3}}$

Step-by-step explanation:

From the image attached below;

We need to calculate the limits of x and y to find the double integral

We will notice that y varies from 1 to 2

The line equation for (0,1),(1,2) is:

$y-1 = \dfrac{2-1}{1-0}(x-0)$

y - 1 = x

The line equtaion for (1,2),(4,1) is:

$y-2 = \dfrac{1-2}{4-1}(x-1) \\ \\ y-2 = -\dfrac{1}{3}(x-1)$

-3(y-2) = (x -1)

-3y + 6 = x - 1

-x = 3y - 6 - 1

-x = 3y - 7

x = -3y + 7

This implies that x varies from y - 1 to -3y + 7

Now, the region D = {(x,y) | 1 ≤ y ≤ 2, y - 1 ≤ x ≤ -3y + 7}

The double integral can now be calculated as:

$\iint _D y^2 dA= \int ^2_1 \int ^{-3y +7}_{y-1} \ 2y ^2 \ dx \ dy$

$\iint _D y^2 dA= \int ^2_1 \bigg[ 2xy ^2 \bigg]^{-3y+7}_{y-1} \ dy$

$\iint _D y^2 dA= \int ^2_1 \bigg[2(-3y+7)y^2-2(y-1)y^2 \bigg ] \ dy$

$\iint _D y^2 dA= \int ^2_1 \bigg[-6y^3 +14y^2 -2y^3 +2y^2 \bigg ] \ dy$

$\iint _D y^2 dA= \int ^2_1 \bigg[-8y^3 +16y^2 \bigg ] \ dy$

$\iint _D y^2 dA= \bigg[-8(\dfrac{y^4}{4}) +16(\dfrac{y^3}{3})\bigg ] ^2_1$

$\iint _D y^2 dA= \bigg[-8(\dfrac{16}{4}-\dfrac{1}{4}) +16(\dfrac{8}{3}-\dfrac{1}{3})\bigg ]$

$\iint _D y^2 dA= \bigg[-8(\dfrac{15}{4}) +16(\dfrac{7}{3})\bigg ]$

$\iint _D y^2 dA= -30 + \dfrac{112}{3}$

$\iint _D y^2 dA= \dfrac{-90+112}{3}$

$\mathbf{\iint _D y^2 dA= \dfrac{22}{3}}$

4 0

3 years ago

12)What is the resistance of a 600-foot #12 (solid) aluminum conductor? Hint: The resistance of the conductor is 3.18 Ohms/1000

Butoxors [25]

Answer:

1.91 Ohms

Step-by-step explanation:

It is given that the resistance of the conductor is 3.18 Ohms/1000 feet.

It means that the resistance of the 1000-foot conductor is 3.18 Ohms.

So, the resistance of the 600-foot #12 (solid) aluminium conductor is $\frac{600}{1000} (3.18)$

= 6 × 0.318

= 1.908

= 1.91 Ohms (rounded to two decimal places)

6 0

3 years ago