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2 years ago

8

I need someone to help me with mý math homework please help!

1 answer:

Julli [10]2 years ago

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<img src="https://tex.z-dn.net/?f=%5Cfrac%7B12x%7D%7B2x-3x-10%7D%20-%20%5Cfrac%7Bx%7D%7B%28x-5%29%7D" id="TexFormula1" title="\f

uysha [10]

Answer:

Step-by-step explanation:

$\frac{12x}{2x-3x-10}-\frac{x}{x-5}=\frac{12x}{-x-10}-\frac{x}{x-5}\\\\=\frac{12x}{-(x+10)}-\frac{x}{x-5}\\\\=\frac{-12x}{x+10}-\frac{x}{x-5}\\\\=\frac{-12x*(x-5)}{(x+10)*(x-5)}-\frac{x*(x+10)}{(x-5(x+10)}\\\\=\frac{-12x*x-5*(-12x)}{(x+10)(x-5)}-\frac{x*x+x*10}{(x-5)(x+10)}\\\\=\frac{-12x^{2}+60x}{(x+10)(x-5)}-\frac{x^{2}+10x}{(x-5)(x+10)}\\\\=\frac{-12x^{2}+60x-(x^{2}+10x)}{(x+10)(x-5)}\\\\=\frac{-12x^{2} + 60x - x^{2}-10x}{x^{2}+5x-50}\\\\=\frac{-13x^{2}+50x}{x^{2}+5x-50}$

8 0

2 years ago

Read 2 more answers

35 POINTS ANSWER FAST PLZ!!!!!!!!!!!!!!!

Sladkaya [172]

The formula to solve for thee area of a square pyramid is [ V = 1/3 * b² * h ]

b² = base area (b = length of one side of the base)

h = height

The only option that as a base length of 12cm. and a height of 14cm. is the first option.

Solve for the volume if needed.

V = 1/3(12²)(14)

V = 1/3(144)(14)

V = 1/3(2,016)

V = 672cm²

Best of Luck!

8 0

2 years ago

Read 2 more answers

What is the distance formula?

alina1380 [7]

Answer:

A

Step-by-step explanation:

4 0

2 years ago

Solve the system of equations x + y = 0 and x + 9y= -48 by combing the eqatuons

lidiya [134]

Answer:

x = 6

y = -6

Step-by-step explanation:

x + y = 0 (Multiply by -1)

x + 9y= -48

-x - y = 0

x + 9y= -48

8y = -48

y = -6

x + y = 0

x - 6 = 0

x = 6

7 0

2 years ago

If A is nonsingular, then (AT )-1 =(A-1) T . (a) Verify this theorem for 2 × 2 matrices

kipiarov [429]

Answer:

Verified

Step-by-step explanation:

Let the 2x2 matrix A be in the form of:

$\left[\begin{array}{cc}a&b\\c&d\end{array}\right]$

Where det(A) = ad - bc # 0 so A is nonsingular:

Then the transposed version of A is

$A^T = \left[\begin{array}{cc}a&c\\b&d\end{array}\right]$

Then the inverted version of transposed A is

$(A^T)^{-1} = \frac{1}{ad - cb} \left[\begin{array}{cc}a&-c\\-b&d\end{array}\right]$

The inverted version of A is:

$A^{-1} = \frac{1}{ad - bc}\left[\begin{array}{cc}a&-b\\-c&d\end{array}\right]$

The transposed version of inverted A is:

$(A^{-1})^T = \frac{1}{ad - bc}\left[\begin{array}{cc}a&-c\\-b&d\end{array}\right]$

We can see that

$(A^T)^{-1} = (A^{-1})^T$

So this theorem is true for 2 x 2 matrices

3 0

2 years ago