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erastova [34]
3 years ago
9

Which statement correctly explains how Mari could find the solution to the following system of linear equations using

Mathematics
1 answer:
just olya [345]3 years ago
7 0

Multiply the first equation by 7 and the second equation by 2, and then add.

Step-by-step explanation:

2f - 5g=-9

-7f + 3g=4

To eliminate any variable , we need to make the coefficients same with different sign and add it

the coefficient of 'f' in first equation is 2

the coefficient of 'f' in second equation is -7

We already have different sign so we make the coefficients same

To make the coefficient same , we multiply the first equation by 7  and second equation by 2

So equation becomes

14f - 35g = -63

-14f + 6g = 8

Now when we add both equations, 'f' gets cancelled

-29g = -55

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Makovka662 [10]

Answer:

28/q

Step-by-step explanation:

4 times 7 equals 28 and then divide 28 by q

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What is the slope of the line that contains
emmainna [20.7K]

Answer:

-4/5 or -0.8

Step-by-step explanation:

Slope: (y2 - y1)/(x2 - x1)

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4 0
3 years ago
Remember to show work and explain. Use the math font.
MrMuchimi

Answer:

\large\boxed{1.\ f^{-1}(x)=4\log(x\sqrt[4]2)}\\\\\boxed{2.\ f^{-1}(x)=\log(x^5+5)}\\\\\boxed{3.\ f^{-1}(x)=\sqrt{4^{x-1}}}

Step-by-step explanation:

\log_ab=c\iff a^c=b\\\\n\log_ab=\log_ab^n\\\\a^{\log_ab}=b\\\\\log_aa^n=n\\\\\log_{10}a=\log a\\=============================

1.\\y=\left(\dfrac{5^x}{2}\right)^\frac{1}{4}\\\\\text{Exchange x and y. Solve for y:}\\\\\left(\dfrac{5^y}{2}\right)^\frac{1}{4}=x\qquad\text{use}\ \left(\dfrac{a}{b}\right)^n=\dfrac{a^n}{b^n}\\\\\dfrac{(5^y)^\frac{1}{4}}{2^\frac{1}{4}}=x\qquad\text{multiply both sides by }\ 2^\frac{1}{4}\\\\\left(5^y\right)^\frac{1}{4}=2^\frac{1}{4}x\qquad\text{use}\ (a^n)^m=a^{nm}\\\\5^{\frac{1}{4}y}=2^\frac{1}{4}x\qquad\log_5\ \text{of both sides}

\log_55^{\frac{1}{4}y}=\log_5\left(2^\frac{1}{4}x\right)\qquad\text{use}\ a^\frac{1}{n}=\sqrt[n]{a}\\\\\dfrac{1}{4}y=\log(x\sqrt[4]2)\qquad\text{multiply both sides by 4}\\\\y=4\log(x\sqrt[4]2)

--------------------------\\2.\\y=(10^x-5)^\frac{1}{5}\\\\\text{Exchange x and y. Solve for y:}\\\\(10^y-5)^\frac{1}{5}=x\qquad\text{5 power of both sides}\\\\\bigg[(10^y-5)^\frac{1}{5}\bigg]^5=x^5\qquad\text{use}\ (a^n)^m=a^{nm}\\\\(10^y-5)^{\frac{1}{5}\cdot5}=x^5\\\\10^y-5=x^5\qquad\text{add 5 to both sides}\\\\10^y=x^5+5\qquad\log\ \text{of both sides}\\\\\log10^y=\log(x^5+5)\Rightarrow y=\log(x^5+5)

--------------------------\\3.\\y=\log_4(4x^2)\\\\\text{Exchange x and y. Solve for y:}\\\\\log_4(4y^2)=x\Rightarrow4^{\log_4(4y^2)}=4^x\\\\4y^2=4^x\qquad\text{divide both sides by 4}\\\\y^2=\dfrac{4^x}{4}\qquad\text{use}\ \dfrac{a^n}{a^m}=a^{n-m}\\\\y^2=4^{x-1}\Rightarrow y=\sqrt{4^{x-1}}

6 0
3 years ago
5x+10=5(x+2) what is the answer I’ll give u 85 points
asambeis [7]

Answer:

All real numbers of x.

Step-by-step explanation:

5x+10=5(x+2) have the exact same equations on both sides.

5x+10=5x+10

If both are graphed, they will intersect each other for every value of x.

5 0
2 years ago
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The zeroes of f(x) = x3 + 3x2 + 2x are <br> .
Vaselesa [24]

The zeros of the function f(x) = x^3 + 3x^2 + 2x are x = 0, x = -1 and x = -2

<h3>How to determine the zeros of the function?</h3>

The function is given as:

f(x) = x^3 + 3x^2 + 2x

Factor out x in the above function

f(x) = x(x^2 + 3x + 2)

Set the function to 0

x(x^2 + 3x + 2) = 0

Factorize the expression in the bracket

x(x + 1)(x + 2) = 0

Split the expression

x = 0, x + 1 = 0 and x + 2 = 0

Solve for x

x = 0, x = -1 and x = -2

Hence, the zeros of the function f(x) = x^3 + 3x^2 + 2x are x = 0, x = -1 and x = -2

Read more about zeros of function at

brainly.com/question/20896994

#SPJ1

7 0
1 year ago
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