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3 years ago

The vertex form of the equation of

Mathematics

1 answer:

Phoenix [80]3 years ago

6 0

Answer:

The standard form of the equation is y = 2x^2 + 16x + 25

Step-by-step explanation:

To find the standard form, first square the parenthesis.

y = 2(x + 4)^2 - 7

y = 2(x^2 + -8x + 16) - 7

Now distribute the 2

y = 2(x^2 + 8x + 16) - 7

y = 2x^2 + 16x + 32 - 7

Now combine like terms

y = 2x^2 + 16x + 32 - 7

y = 2x^2 + 16x + 25

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The area of a rectangular wall of a barn is 192 sq. Ft. It's length is 8 ft. Longer than twice it's width. Find the length and w

Novay_Z [31]

Answer: length = 24 ft

Width = 8 ft

Step-by-step explanation:

Let L represent the length of the barn.

Let W represent the width of the barn.

The area of a rectangular wall of a barn is 192 sq. Ft. This means that

LW = 192- - - - - - - - - - - - - - -1

It's length is 8 ft longer than twice it's width. This means that

L = 2W + 8

Substituting L = 2W + 8 into equation 1, it becomes

W(2W + 8) = 192

2W² + 8W = 192

2W² + 8W - 192 = 0

Dividing through by 2, it becomes

W² + 4W - 96 = 0

W² + 12W - 8W - 96 = 0

W(W + 12) - 8(W + 12) = 0

(W - 8)(W + 12) = 0

W = 8 or W = - 12

Since the width cannot be negative, then W = 8

L = 192/W = 192/8

L = 24

6 0

3 years ago

Which of the following is true? Tangent is positive in Quadrant I. Sine is negative in Quadrant II. Cosine is positive in Quadra

DiKsa [7]

A) Tangent is positive in Quadrant I.

Since sine and cosine are both positive in Quadrant I and tangent is the ratio of sine to cosine, tangent is positive in Quadrant I

7 0

3 years ago

I need help with questions #7 and #8 plz

katen-ka-za [31]

Answer:

7. A = 40.8 deg; B = 60.6 deg; C = 78.6 deg

8. A = 20.7 deg; B = 127.2 deg; C = 32.1 deg

Step-by-step explanation:

Law of Cosines

$c^2 = a^2 + b^2 - 2ab \cos C$

You know the lengths of the sides, so you know a, b, and c. You can use the law of cosines to find C, the measure of angle C.

Then you can use the law of cosines again for each of the other angles. An easier way to solve for angles A and B is, after solving for C with the law of cosines, solve for either A or B with the law of sines and solve for the last angle by the fact that the sum of the measures of the angles of a triangle is 180 deg.

We use the law of cosines to find C.

$18^2 = 12^2 + 16^2 - 2(12)(16) \cos C$

$324 = 144 + 256 - 384 \cos C$

$-384 \cos C = -76$

$\cos C = 0.2$

$C = \cos^{-1} 0.2$

$C = 78.6^\circ$

Now we use the law of sines to find angle A.

Law of Sines

$\dfrac{a}{\sin A} = \dfrac{b}{\sin B} = \dfrac{c}{\sin C}$

We know c and C. We can solve for a.

$\dfrac{a}{\sin A} = \dfrac{c}{\sin C}$

$\dfrac{12}{\sin A} = \dfrac{18}{\sin 78.6^\circ}$

Cross multiply.

$18 \sin A = 12 \sin 78.6^\circ$

$\sin A = \dfrac{12 \sin 78.6^\circ}{18}$

$\sin A = 0.6535$

$A = \sin^{-1} 0.6535$

$A = 40.8^\circ$

To find B, we use

m<A + m<B + m<C = 180

40.8 + m<B + 78.6 = 180

m<B = 60.6 deg

I'll use the law of cosines 3 times here to solve for all the angles.

Law of Cosines

$a^2 = b^2 + c^2 - 2bc \cos A$

$b^2 = a^2 + c^2 - 2ac \cos B$

$c^2 = a^2 + b^2 - 2ab \cos C$

Find angle A:

$a^2 = b^2 + c^2 - 2bc \cos A$

$8^2 = 18^2 + 12^2 - 2(18)(12) \cos A$

$64 = 468 - 432 \cos A$

$\cos A = 0.9352$

$A = 20.7^\circ$

Find angle B:

$b^2 = a^2 + c^2 - 2ac \cos B$

$18^2 = 8^2 + 12^2 - 2(8)(12) \cos B$

$324 = 208 - 192 \cos A$

$\cos B = -0.6042$

$B = 127.2^\circ$

Find angle C:

$c^2 = a^2 + b^2 - 2ab \cos C$

$12^2 = 8^2 + 18^2 - 2(8)(18) \cos B$

$144 = 388 - 288 \cos A$

$\cos C = 0.8472$

$C = 32.1^\circ$

8 0

3 years ago

Find the slope of the line

Sergio039 [100]

Answer:

A) 2

Step-by-step explanation:

4 0

3 years ago

Read 2 more answers

If the function f (x) has a domain of (a,b] and a range of [c,d), then what is the domain and range of g (x) = m × f (x) + n?

xz_007 [3.2K]

<h3>Answer: Choice A</h3>

Domain = (a,b]

Range = [mc + n,md + n)

==============================================

Explanation:

The domain stays the same because we still have to go through f(x) as our first hurdle in order to get g(x).

Think of it like having 2 doors. The first door is f(x) and the second is g(x). The fact g(x) is dependent on f(x) means that whatever input restrictions are on f, also apply on g as well. So going back to the "2 doors" example, we could have a problem like trying to move a piece of furniture through them and we'd have to be concerned about the f(x) door.

-------------------

The range will be different however. The smallest value in the range of f(x) is y = c as it is the left endpoint. So the smallest f(x) can be is c. This means the smallest g(x) can be is...

g(x) = m*f(x) + n

g(x) = m*c + n

All we're doing is replacing f with c.

So that means mc+n is the starting point of the range for g(x).

The ending point of the range is md+n for similar reasons. Instead of 'c', we're dealing with 'd' this time. The curved parenthesis says we don't actually include this value in the range. A square bracket means include that value.

6 0

3 years ago