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3 years ago

5

On the first play, the football team lost 6 yards. On the second play, the team lost 5yards. What was their total change in yard

s?

1 answer:

WINSTONCH [101]3 years ago

6 0

The total change in yards the team lost was 11

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Find the perimeter of the quadrilateral whose vertices are as follows: (-1,1), (-2,4), (-4,2),(-3,-1).

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Answer:

Transcribed Image Text from this Question. 13. Find the perimeter of the quadrilateral whose vertices are as follows. +F (-1,1).(-2,4).(-4,2),(-3,-1) ...

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3 years ago

Someone help me solve this please, I'd appreciate it if you can walk me through the steps. If not, just answering it is okay :)

Marat540 [252]

Answer:

c. -8

Step-by-step explanation:

Given:

$\frac{4b \div c}{1 - a}$

a = 5, b = 8, and c = 1

To evaluate, plug in the values into the expression given.

Thus:

$\frac{4(8) \div 1}{1 - 5}$

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3 years ago

According to of Plymouth plantation what was the pilgrim’s opinion of Cape Cod in the first weeks after landing

Makovka662 [10]

They thought it was depleted, fruitless and unfertile.

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4 years ago

Which of the following is a quadratic function?

serious [3.7K]

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3 years ago

Write an equation for an ellipse centered at the origin, which has foci at (\pm\sqrt{12},0)(± 12 ,0)left parenthesis, plus min

lora16 [44]

Answer:

$\mathbf{\dfrac{x^2}{49^2} +\dfrac{y^2}{37^2} =1}$

Step-by-step explanation:

Given that :

the foci of the ellipse is (±√12,0) and C0-vertices are (0,±√37)

The foci are (-C,0) and (C ,0)

the focus has x-coordinates so the focus is lie on x- axis.

The major axis also lie on x-axis

The minor axis lies on y-axis so C0-vertices are (0,±√37)

The given focus C = ae = √12

Given co-vertices ( minor axis) (0,±b) = (0,±√37)

b= √37

We can therefore express the relation between the focus and semi major axes and semi minor axes as:

$\mathbf{c^2 = a^2 - b^2 } \\ \\ \mathbf{a^2 = c^2 + b^2 } \\ \\ \mathbf{c^2 = ( \sqrt12)^2 - (\sqrt 37)^2 } \\ \\ \mathbf{c^2 = 49 } \\ \\ \mathbf{c = \sqrt{49 }}$

The equation of ellipse formula is:

$\dfrac{x^2}{a^2} +\dfrac{y^2}{b^2} =1$

and we know that $\mathbf{a=\sqrt{49} \ \ and \ \ b=\sqrt{37}}$

Thus ; the equation of the ellipse at the origin is

$\mathbf{\dfrac{x^2}{49^2} +\dfrac{y^2}{37^2} =1}$

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4 years ago