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Anuta_ua [19.1K]
3 years ago
12

8x+8(c+8)-6 show work its for my homework i get it but im just making sure i did it right:)

Mathematics
2 answers:
ioda3 years ago
7 0
8x+8c+164-6
=8x+8c+158
katovenus [111]3 years ago
4 0
8x+8c+64-6 *distribute*

8x+8c+58 *add like terms*
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What is the solution for the equation 6x - 8 = 4x? x =
GarryVolchara [31]

Answer:

x = 4

Step-by-step explanation:

6x - 8 = 4x

<em>subtract 4x from both sides of the equation</em>

6x - 4x -8 = 4x - 4x

2x - 8 = 0

<em>add 8 to both sides of the equation</em>

2x - 8 + 8 = 0 + 8

2x = 8

<em>divide both sides of the equation by 2</em>

2x/2 = 8/2

x = 4

4 0
3 years ago
Read 2 more answers
Math question picture attached
babunello [35]

To find the slope, you use the slope formula (m):

m=\frac{y_2-y_1}{x_2-x_1}    and plug in the two points

(-3, 4) = (x₁, y₁)

(2, -1) = (x₂, y₂)

m=\frac{y_2-y_1}{x_2-x_1}

m=\frac{-1-4}{2-(-3)}       [two negatives cancel each other out and become positive]

m=\frac{-1-4}{2+3}

m=\frac{-5}{5}

m = -1        The slope is -1

4 0
3 years ago
Read 2 more answers
Sin theta+costheta/sintheta -costheta+sintheta-costheta/sintheta+costheta=2sec2/tan2 theta -1
sleet_krkn [62]

\dfrac{sin\theta + cos\theta}{sin\theta-cos\theta}+\dfrac{sin\theta-cos\theta}{sin\theta+cos\theta}=\dfrac{2sec^2\theta}{tan^2\theta-1}

From Left side:

\dfrac{sin\theta + cos\theta}{sin\theta-cos\theta}\bigg(\dfrac{sin\theta+cos\theta}{sin\theta+cos\theta}\bigg)+\dfrac{sin\theta-cos\theta}{sin\theta+cos\theta}\bigg(\dfrac{sin\theta-cos\theta}{sin\theta-cos\theta}\bigg)

\dfrac{sin^2\theta+2cos\thetasin\theta+cos^2\theta}{sin^2\theta-cos^2\theta}+\dfrac{sin^2\theta-2cos\thetasin\theta+cos^2\theta}{sin^2\theta-cos^2\theta}

NOTE: sin²θ + cos²θ = 1

\dfrac{1 + 2cos\theta sin\theta}{sin^2\theta-cos^2\theta}+\dfrac{1-2cos\theta sin\theta}{sin^2\theta-cos^2\theta}

\dfrac{1 + 2cos\theta sin\theta+1-2cos\theta sin\theta}{sin^2\theta-cos^2\theta}

\dfrac{2}{sin^2\theta-cos^2\theta}

\dfrac{2}{\bigg(sin^2\theta-cos^2\theta\bigg)\bigg(\dfrac{cos^2\theta}{cos^2\theta}\bigg)}

\dfrac{2sec^2\theta}{\dfrac{sin^2\theta}{cos^2\theta}-\dfrac{cos^2\theta}{cos^2\theta}}

\dfrac{2sec^2\theta}{tan^2\theta-1}

Left side = Right side <em>so proof is complete</em>

8 0
3 years ago
Read 2 more answers
Determine the period.
soldier1979 [14.2K]

Answer:

Step-by-step explanation:

we are being told that each square is one unit.  so,  look at peak to peak and count the units.  

1,2,3,4,5,6,7

From peak to peak is 7 units, so the period is 7

8 0
2 years ago
Select the correct answer.
sp2606 [1]

The perimeter of the pentagon with apothem of 5cm is 28.87cm

<h3>How to calculate the perimeter of a pentagon</h3>

The sum of all the external sides of the pentagon is its perimeter.

First we need to calculate the half of one of its sides using the SOH CAH TOA

tan 30 = x/5

x = 5 tan 30

x = 2.89

One of the sides = 2(2.89) = 5.77 cm

Perimeter of th pentagon  = 5(5.77) = 28.87

Hence the perimeter of the pentagon with apothem of 5cm is 28.87cm

Learn more on perimeter of pentagon here: brainly.com/question/11484726

6 0
2 years ago
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