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2 years ago

Find the solution to the equation:

Mathematics

1 answer:

Volgvan2 years ago

3 0

Answer:

x = 1

Step-by-step explanation:

To find the solution to the equation given, we will follow the steps below:

2(3x-6)= -12+6

Open the bracket

6x - 12 = -12 + 6

6x - 12 =-6

Add 12 to both-side of the equation

6x -12 + 12 = -6 + 12

6x =6

Divide both-side of the equation by 6

6x/6 = 6/6

At the left-hand side of the equation, 6 will cancel-out 6 while at the right-hand side of the equation, 6 will be divided by 6

x = 1

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(please zoom / zoom out if needed to see question. (: )

lisov135 [29]

You essentially have four triangles. The area of a triangle is: A=bh/2 (b=base length and h=height)

So the area is simply the sum of the four triangle areas...

A=(3*2)/2+(3*2)/2+(7*2)/2+(7*2)/2

A=20 ft^2

7 0

2 years ago

Read 2 more answers

Find the cosine of angle B.

Alexandra [31]

$\boxed { \text {Cosine Rule : } a^2 = b^2 + c^2 - 2bc \cos(A) }$

9² = 12² + 15² - 2 (12) (15) cos (B)

81 = 144 + 225 - 360 cos(B)

81 = 369 - 360 cos (B)

360 cos (B) = 369 - 81

360 cos (B) = 288

cos (B) = 0.8

Answer: Cosine B = 0.8

$\boxed { \text {Cosine Rule : } a^2 = b^2 + c^2 - 2bc \cos(A) }$

12² = 15² + 9² - 2 (15)(9) cos (A)

144 = 225 + 81 - 270 Cos A

144 = 306 - 270 Cos A

270 Cos A = 162

Cos A = 3/5 or 0.6

Answer: Cosine Angle A = 3/5

4 0

2 years ago

Consider the following function. f(x) = 9 − x2/3 Find f(−27) and f(27). f(−27) = f(27) = Find all values c in (−27, 27) such tha

oksano4ka [1.4K]

I guess the function is $f(x)=9-x^{2/3}$ . Then $f(-27)=0$ and $f(27)=0$ .

The derivative is $f'(x)=-\dfrac23 x^{-1/3}$ , but there is no $c$ such that

$-\dfrac23c^{-1/3}=0$

This doesn't contradict Rolle's theorem because $f'(0)$ does not exists. In other words, $f$ is not differentiable on (-27, 27), so the conditions of Rolle's theorem are not met. (Looks like that would be the last option, or the second to last option if the last one is "Nothing can be concluded")