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3 years ago

Can someone help me with this question?

Mathematics

1 answer:

soldier1979 [14.2K]3 years ago

4 0

Answer:

C. Graph Y

Step-by-step explanation:

I graphed the equation below and found that it was the same as graph Y.

If this answer is correct, please make me Brainliest!

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Use the graph to write a linear function that relates y to x<br> NO LINKS THANK YOU

kipiarov [429]

Answer:

I uploaded a file hosting link for your question.

See below for more details.

Step-by-step explanation:

$m=\frac{change(y)}{change(x)}$

$=\frac{10-2}{6-0}$

$=\frac{8}{6}$

$m=\frac{4}{3}$

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3 years ago

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EXPRESS 56 AS THE PRODUCT OF PRIME NUMBERS USING PRIME FACTORIZATION METHOD

Degger [83]

Answer:

Prime factorization: 56 = 2 x 2 x 2 x 7, which can also be written 56 = 2³ x 7

Step-by-step explanation:

56 = 2³ x 7

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3 years ago

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What is the slope of the line graphed on the coordinate plane?

lina2011 [118]

Answer:

18*

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4 years ago

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4x + 1 = 2x – 5 can someone help me. Whoever gets it right gets it right gets brainliest

castortr0y [4]

Answer:

x=-3

Step-by-step explanation:

subtract "2x" to "4x + 1" and that'll leave -5 by itself and leave 2x+1 on the left and -5 on the right then subtract -1 to -5 and it'll add up to -6 then after that you'd divide 2x to -6 and get -3 and it'll leave 'x' by itself so therefore X = -3

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3 years ago

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Perform the indicated operations. Write the answer in standard form, a+bi.<br> 5-3i / -2-9i

Vsevolod [243]

$\huge \boxed{\mathfrak{Answer} \downarrow}$

$\large \bf\frac { 5 - 3 i } { - 2 - 9 i } \\$

Multiply both numerator and denominator of $\sf \frac{5-3i}{-2-9i} \\$ by the complex conjugate of the denominator, -2+9i.

$\large \bf \: Re(\frac{\left(5-3i\right)\left(-2+9i\right)}{\left(-2-9i\right)\left(-2+9i\right)}) \\$

Multiplication can be transformed into difference of squares using the rule: $\sf\left(a-b\right)\left(a+b\right)=a^{2}-b^{2}$ .

$\large \bf \: Re(\frac{\left(5-3i\right)\left(-2+9i\right)}{\left(-2\right)^{2}-9^{2}i^{2}}) \\$

By definition, i² is -1. Calculate the denominator.

$\large \bf \: Re(\frac{\left(5-3i\right)\left(-2+9i\right)}{85}) \\$

Multiply complex numbers 5-3i and -2+9i in the same way as you multiply binomials.

$\large \bf \: Re(\frac{5\left(-2\right)+5\times \left(9i\right)-3i\left(-2\right)-3\times 9i^{2}}{85}) \\$

Do the multiplications in $\sf5\left(-2\right)+5\times \left(9i\right)-3i\left(-2\right)-3\times 9\left(-1\right)$ .

$\large \bf \: Re(\frac{-10+45i+6i+27}{85}) \\$

Combine the real and imaginary parts in -10+45i+6i+27.

$\large \bf \: Re(\frac{-10+27+\left(45+6\right)i}{85}) \\$

Do the additions in $\sf-10+27+\left(45+6\right)i$ .

$\large \bf Re(\frac{17+51i}{85}) \\$

Divide 17+51i by 85 to get $\sf\frac{1}{5}+\frac{3}{5}i \\$ .

$\large \bf \: Re(\frac{1}{5}+\frac{3}{5}i) \\$

The real part of $\sf \frac{1}{5}+\frac{3}{5}i \\$ is $\sf \frac{1}{5} \\$ .

$\large \boxed{\bf\frac{1}{5} = 0.2} \\$

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3 years ago