What's another way to write (6+7)x?

Mathematics

2 answers:

Akimi4 [234]2 years ago

7 0

You can write it as 13x. I hope this helps!

6+7=13 (13)x=13x

Annette [7]2 years ago

6 0

I think 6x+7x because i think u distribute

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VladimirAG [237]

Answer:

(-2,0)

Step-by-step explanation:

i did it in my head im not so sure

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2 years ago

Select the correct answer from each drop-down menu. The general form of the equation of a circle is 7x2 + 7y2 − 28x + 42y − 35 =

nignag [31]

The general form of the equation of a circle is,

x² + y² + Dx + Ey + F = 0, where D, E, F are constants.

The given equation is 7x² + 7y² − 28x + 42y − 35 = 0. So, to convert this equation into a general form we just need to get rid of the leading coefficient 7.

Hence, divide both sides of the equation by 7. So,

$\frac{7x^2+ 7y^2-28x + 42y-35}{7} =0$

x² + y² − 4x + 6y − 5=0.

So, the general form of the equation is x² + y² − 4x + 6y − 5=0.

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2 years ago

Due to inflation, a fixed quantity of money can buy 3% less every year. If Pam has enough in her wallet to buy 47 movie tickets

Nitella [24]

3% x 10 years is 30%. 47 - 30% is 14.1 or 14

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2 years ago

Sara is working on a Geometry problem in her Algebra class. The problem requires Sara to use the two quadrilaterals below to ans

zloy xaker [14]

Part A:

Given a square with sides 6 and x + 4. Also, given a rectangle with sides 2 and 3x + 4

The perimeter of the square is given by 4(x + 4) = 4x + 16

The area of the rectangle is given by 2(2) + 2(3x + 4) = 4 + 6x + 8 = 6x + 12

For the perimeters to be the same

4x + 16 = 6x + 12
4x - 6x = 12 - 16
-2x = -4
x = -4 / -2 = 2

The value of x that makes the perimeters of the quadrilaterals the same is 2.

Part B:

The area of the square is given by

$Area=(x+4)^2=x^2+8x+16$

The area of the rectangle is given by 2(3x + 4) = 6x + 8

For the areas to be the same

$x^2+8x+16=6x+8 \\ \\ \Rightarrow x^2+8x-6x+16-8=0 \\ \\ \Rightarrow x^2+2x+8=0 \\ \\ \Rightarrow x= \frac{-2\pm\sqrt{2^2-4(8)}}{2} \\ \\ = \frac{-2\pm\sqrt{4-32}}{2} = \frac{-2\pm\sqrt{-28}}{2} \\ \\ = \frac{-2\pm2i\sqrt{7}}{2} =-1\pm i\sqrt{7}$

Thus, there is no real value of x for which the area of the quadrilaterals will be the same.

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2 years ago

Which of the following are solutions to the equation sinx cosx = 1/4? Check all that apply.

N76 [4]

The solution is B. π/12+nπ

proof
sinx cosx = 1/4 is equivalent to 2 sinx cosx = 1/2 or sin2x =1/2
so 2x = arcsin(1/2) = π/6 + 2nπ, so x = π/12+nπ

8 0

3 years ago

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