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Greeley [361]
3 years ago
11

2x-3y=133x+y=3can you please explain the steps on how to get the answer?

Mathematics
1 answer:
Rufina [12.5K]3 years ago
5 0
Lets do this the elimination way. to do this, you multiply the bottom equation by three, and now you have
2x-3y=13
9x+3y=9

lets add these two equations together. you get
11x=22
x=2
now substitute 2 for x in the first equation.
2(2)-3y=13
4-3y=13
-3y=9
y=-3


y=-3, x=2

You might be interested in
Determine the sample size needed to construct a 99​% confidence interval to estimate the average GPA for the student population
Volgvan

Answer: n = 14

Step-by-step explanation: margin of error = critical value × σ/√n

Where σ = population standard deviation = 1

n = sample size = ?

We are to construct a 99% confidence interval, hence the level of significance is 1%.

The critical value for 2 tailed test at 1% level of significance is gotten from a standard normal distribution table which is 2.58

Margin of error = 0.7

0.7 = 2.58×1/√n

0.7 = 2.58/√n

By cross multipying

0.7×√n = 2.58

By squaring both sides

0.7^2 × n = 2.58^2

0.49 × n = 6.6564

n = 6.6564/0.49

n = 14

3 0
3 years ago
For the function f(x)=(x+6)^3, rind f^-1(x)
qaws [65]

Answer:

f^{-1}(x)=\sqrt[3]{x}-6

Step-by-step explanation:

f(x)=(x+6)^3

y=(x+6)^3

x=(y+6)^3

\sqrt[3]{x}=y+6

\sqrt[3]{x}-6=y

5 0
3 years ago
A fraction becomes 4÷5 if 1 is added to both numerator and denominator. If, however, 5 is subtracted from both numerator and den
tia_tia [17]

Answer:

\dfrac{7}{9}

Step-by-step explanation:

\dfrac{x+1}{y+1}=\dfrac{4}{5}\\\Rightarrow 5x-4y=-1

\dfrac{x-5}{y-5}=\dfrac{1}{2}\\\Rightarrow 2x-y=5

Putting it in matrix form

\begin{bmatrix}a_{1}&b_{1}\\a_{2}&b_{2}\end{bmatrix}{\begin{bmatrix}x\\y\end{bmatrix}=\begin{bmatrix}{c_{1}}\\{c_{2}}\end{bmatrix}\\\Rightarrow\begin{bmatrix}5 & -4\\2 & -1\end{bmatrix}\begin{bmatrix}x\\ y\end{bmatrix}=\begin{bmatrix}-1\\ 5\end{bmatrix}

From Cramer's rule we have

x=\dfrac{\begin{vmatrix}c_1 &b_1 \\ c_2 & b_2\end{vmatrix}}{\begin{vmatrix}a_1 &b_1 \\ a_2& b_2\end{vmatrix}}\\\Rightarrow x=\dfrac{\begin{vmatrix}-1 &-4 \\ 5 & -1\end{vmatrix}}{\begin{vmatrix}5 &-4 \\ 2& -1\end{vmatrix}}\\\Rightarrow x=\dfrac{1+20}{-5+8}\\\Rightarrow x=7

y=\dfrac{\begin{vmatrix}a_1 &c_1 \\ a_2 & c_1\end{vmatrix}}{\begin{vmatrix}a_1 &b_1 \\ a_2& b_2\end{vmatrix}}\\\Rightarrow y=\dfrac{\begin{vmatrix}5 &-1 \\ 2 & 5\end{vmatrix}}{\begin{vmatrix}5 &-4 \\ 2& -1 \end{vmatrix}}\\\Rightarrow y=\dfrac{25+2}{-5+8}\\\Rightarrow y=9

Verifying the results

\dfrac{7+1}{9+1}=\dfrac{8}{10}=\dfrac{4}{5}

\dfrac{7-5}{9-5}=\dfrac{2}{4}=\dfrac{1}{2}

Hence, the fraction is \dfrac{7}{9}.

6 0
3 years ago
Which of the following are prime factorizations of the number 70? Check all<br> that apply.
Shtirlitz [24]

The prime factorization of 70 is 2*5*7

Further explanation:

First of all let us define prime factorization

Prime factorization consists of all the prime multiples of a number i.e. the factors should also be prime numbers

Given number is 70

The multiples of 70 are

70 = 2*35 => 35 is composite number

70 = 5*14 => 14 is a composite number

70 = 7*10 => 10 is a composite number

70 = 2*5*7 => All factors are prime so

The prime factorization of 70 is 2*5*7

Keywords: Factorization, Prime factors

Learn more about prime factorization at:

  • brainly.com/question/12938965
  • brainly.com/question/13018049

#LearnwithBrainly

4 0
3 years ago
Can u answer these for me with the work shown
babymother [125]

Answer:

\frac{x^3 + 2x^2 -9x-18}{x^3-x^2-6x}= \frac{(x+3)}{x}

\frac{3x^2 - 5x - 2}{x^3 - 2x^2} = \frac{3x + 1}{x^2}

\frac{6 - 2x}{x^2 - 9} * \frac{15 + 5x}{4x}=-\frac{5}{2x}

\frac{x^2 -6x + 9}{5x - 15} / \frac{5}{3-x} = \frac{-(x-3)^2}{25}

\frac{x^3 - x^2 -x + 1}{x^2 - 2x+1}= x +1

\frac{9x^2 + 3x}{6x^2} = \frac{3x + 1}{2x}

\frac{x^2-3x+2}{4x} * \frac{12x^2}{x^2 - 2x} / \frac{x - 1}{x} = 3x

Step-by-step explanation:

Required

Simplify

Solving (1):

\frac{x^3 + 2x^2 -9x-18}{x^3-x^2-6x}

Factorize the numerator and the denominator

\frac{x^2(x + 2) -9(x+2)}{x(x^2-x-6)}

Factor out x+2 at the numerator

\frac{(x^2 -9)(x+2)}{x(x^2-x-6)}

Express x^2 - 9 as difference of two squares

\frac{(x^2 -3^2)(x+2)}{x(x^2-x-6)}

\frac{(x -3)(x+3)(x+2)}{x(x^2-x-6)}

Expand the denominator

\frac{(x -3)(x+3)(x+2)}{x(x^2-3x+2x-6)}

Factorize

\frac{(x -3)(x+3)(x+2)}{x(x(x-3)+2(x-3))}

\frac{(x -3)(x+3)(x+2)}{x(x+2)(x-3)}

Cancel out same factors

\frac{(x+3)}{x}

Hence:

\frac{x^3 + 2x^2 -9x-18}{x^3-x^2-6x}= \frac{(x+3)}{x}

Solving (2):

\frac{3x^2 - 5x - 2}{x^3 - 2x^2}

Expand the numerator and factorize the denominator

\frac{3x^2 - 6x + x - 2}{x^2(x- 2)}

Factorize the numerator

\frac{3x(x - 2) + 1(x - 2)}{x^2(x- 2)}

Factor out x - 2

\frac{(3x + 1)(x - 2)}{x^2(x- 2)}

Cancel out x - 2

\frac{3x + 1}{x^2}

Hence:

\frac{3x^2 - 5x - 2}{x^3 - 2x^2} = \frac{3x + 1}{x^2}

Solving (3):

\frac{6 - 2x}{x^2 - 9} * \frac{15 + 5x}{4x}

Express x^2 - 9 as difference of two squares

\frac{6 - 2x}{x^2 - 3^2} * \frac{15 + 5x}{4x}

Factorize all:

\frac{2(3 - x)}{(x- 3)(x+3)} * \frac{5(3 + x)}{2(2x)}

Cancel out x + 3 and 3 + x

\frac{2(3 - x)}{(x- 3)} * \frac{5}{2(2x)}

\frac{3 - x}{x- 3} * \frac{5}{2x}

Express 3 - x as -(x - 3)

\frac{-(x-3)}{x- 3} * \frac{5}{2x}\\

-1 * \frac{5}{2x}

-\frac{5}{2x}

Hence:

\frac{6 - 2x}{x^2 - 9} * \frac{15 + 5x}{4x}=-\frac{5}{2x}

Solving (4):

\frac{x^2 -6x + 9}{5x - 15} / \frac{5}{3-x}

Expand x^2 - 6x + 9 and factorize 5x - 15

\frac{x^2 -3x -3x+ 9}{5(x - 3)} / \frac{5}{3-x}

Factorize

\frac{x(x -3) -3(x-3)}{5(x - 3)} / \frac{5}{3-x}

\frac{(x -3)(x-3)}{5(x - 3)} / \frac{5}{3-x}

Cancel out x - 3

\frac{(x -3)}{5} / \frac{5}{3-x}

Change / to *

\frac{(x -3)}{5} * \frac{3-x}{5}

Express 3 - x as -(x - 3)

\frac{(x -3)}{5} * \frac{-(x-3)}{5}

\frac{-(x-3)(x -3)}{5*5}

\frac{-(x-3)^2}{25}

Hence:

\frac{x^2 -6x + 9}{5x - 15} / \frac{5}{3-x} = \frac{-(x-3)^2}{25}

Solving (5):

\frac{x^3 - x^2 -x + 1}{x^2 - 2x+1}

Factorize the numerator and expand the denominator

\frac{x^2(x - 1) -1(x - 1)}{x^2 - x-x+1}

Factor out x - 1 at the numerator and factorize the denominator

\frac{(x^2 - 1)(x - 1)}{x(x -1)- 1(x-1)}

Express x^2 - 1 as difference of two squares and factor out x - 1 at the denominator

\frac{(x +1)(x-1)(x - 1)}{(x -1)(x-1)}

x +1

Hence:

\frac{x^3 - x^2 -x + 1}{x^2 - 2x+1}= x +1

Solving (6):

\frac{9x^2 + 3x}{6x^2}

Factorize:

\frac{3x(3x + 1)}{3x(2x)}

Divide by 3x

\frac{3x + 1}{2x}

Hence:

\frac{9x^2 + 3x}{6x^2} = \frac{3x + 1}{2x}

Solving (7):

\frac{x^2-3x+2}{4x} * \frac{12x^2}{x^2 - 2x} / \frac{x - 1}{x}

Change / to *

\frac{x^2-3x+2}{4x} * \frac{12x^2}{x^2 - 2x} * \frac{x}{x-1}

Expand

\frac{x^2-2x-x+2}{4x} * \frac{12x^2}{x^2 - 2x} * \frac{x}{x-1}

Factorize

\frac{x(x-2)-1(x-2)}{4x} * \frac{12x^2}{x(x - 2)} * \frac{x}{x-1}

\frac{(x-1)(x-2)}{4x} * \frac{12x^2}{x(x - 2)} * \frac{x}{x-1}

Cancel out x - 2 and x - 1

\frac{1}{4x} * \frac{12x^2}{x} * \frac{x}{1}

Cancel out x

\frac{1}{4x} * \frac{12x^2}{1} * \frac{1}{1}

\frac{12x^2}{4x}

3x

Hence:

\frac{x^2-3x+2}{4x} * \frac{12x^2}{x^2 - 2x} / \frac{x - 1}{x} = 3x

8 0
3 years ago
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