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2 years ago

Pls tell/show me how u did itAlso the answer that there is worng

Mathematics

2 answers:

bekas [8.4K]2 years ago

7 0

Answer:

y=4/3x+2

Step-by-step explanation:

y=mx+b

b is y intercept, and in this case, 2

m is the slope, and in this case, 4/3

goblinko [34]2 years ago

3 0

Answer:

y=4/3x+2

Step-by-step explanation:

y=mx+b

b is y intercept, and in this case, 2

m is the slope, and in this case, 4/3

Hope this answers your question.

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For the data shown in the stem-and-leaf plot, what is the effect on the mean and median of adding the value 10 to the data set?

alekssr [168]

The correct answer for the question that is being presented above is this one: "<span> C. The mean is decreased by about 1.2 and the median is decreased by 2." </span>For the data shown in the stem-and-leaf plot, the effect on the mean and median of adding the value 10 to the data set is that t<span>he mean is decreased by about 1.2 and the median is decreased by 2.</span>

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4 years ago

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A house has increased in value by 37% since it was purchased. If the original value was $720, 000, what is the value after the i

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Step-by-step explanation:

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2 years ago

Expand. If necessary, combine like terms. (x+3)(x-3)=

Ne4ueva [31]

Answer:

${x}^{2} - 9$

Step-by-step explanation:

$(x−3)(x+3)$

$x(x+3)−3(x+3)$

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$x \times x−3 \times 3$

${x}^{2} - 9$

<h3>Hope it is helpful....</h3>

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3 years ago

Please help with these partial fractions!!!

VARVARA [1.3K]

a. Factorize the denominator:

$\dfrac{x+14}{x^2-2x-8}=\dfrac{x+14}{(x-4)(x+2)}$

Then we're looking for $a,b$ such that

$\dfrac{x+14}{x^2-2x-8}=\dfrac a{x-4}+\dfrac b{x+2}$

$\implies x+14=a(x+2)+b(x-4)$

If $x=4$ , then $18=6a\implies a=3$ ; if $x=-2$ , then $12=-6b\implies b=-2$ . So we have

$\dfrac{x+14}{x^2-2x-8}=\dfrac3{x-4}-\dfrac2{x+2}$

as required.

b. Same setup as in (a):

$\dfrac{-3x^2+5x+6}{x^3+x^2}=\dfrac{-3x^2+5x+6}{x^2(x+1)}$

We want to find $a,b,c$ such that

$\dfrac{-3x^2+5x+6}{x^2(x+1)}=\dfrac ax+\dfrac b{x^2}+\dfrac c{x+1}$

Quick aside: for the second term, since the denominator has degree 2, we should be looking for another constant $b'$ such that the numerator of the second term is $b'x+b$ . We always want the polynomial in the numerator to have degree 1 less than the degree of the denominator. But we would end up determining $b'=0$ anyway.