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3 years ago

Hey someone please please help me with this please i am really bad at math

Mathematics

1 answer:

Viefleur [7K]3 years ago

3 0

Answer:

x-y

Step-by-step explanation:

when you subtract (-2x-8y)-(-3x-7y)

you have to distribute that negative onto -3x-7y which turns it positive

which results in -2x-8y+3x+7y

simplified to x-y

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In the figure, PQ is parallel to RS. The length of RP is 5cm; the length of PT is 30cm; the length of QT is 60cm. What is the le

rjkz [21]

The triangles PTQ and RTS are similar; so the ratios of corresponding sides are equal.

Let us denote the unknown length SQ as x.

We could make the following similarity relation;

$\begin{gathered} \frac{30}{30+5}=\frac{60}{60+x} \\ \end{gathered}$

Cross multiply to obtain;

$\begin{gathered} 30(60+x)=60(35) \\ 1800+30x=2100 \\ \text{collect like terms} \\ 30x=2100-1800 \\ 30x=300 \\ \text{divide both sides by 30} \\ \frac{30x}{30}=\frac{300}{30} \end{gathered}$

Therefore;

$x=10\operatorname{cm}$

Therefore, the answer is option A

3 0

1 year ago

Is this equation infinity many solution 5c+9=5c-12

MatroZZZ [7]

Answer:

The answer is no solution.

Step-by-step explanation: First, you would add + 12 to both sides of the equation so that you would have 5c + 21 = 5c. Then, you would subtract 5c from both sides of the equation. This would leave you with 21 = 0. Since this is not true, the equation would have no solution.

3 0

3 years ago

Read 2 more answers

What is 11/12 - 2/3 <br> Subtracting Proper Fractions

alexgriva [62]

Answer:

0.25

Step-by-step explanation:

11/12 = 0.917

2/3 = 0.667

0.917 - 0.667 = 0.25

4 0

2 years ago

Read 2 more answers

Let X and Y be discrete random variables. Let E[X] and var[X] be the expected value and variance, respectively, of a random vari

Ulleksa [173]

Answer:

(a) $E[X+Y]=E[X]+E[Y]$

(b) $Var(X+Y)=Var(X)+Var(Y)$

Step-by-step explanation:

Let X and Y be discrete random variables and E(X) and Var(X) are the Expected Values and Variance of X respectively.

(a)We want to show that E[X + Y ] = E[X] + E[Y ].

When we have two random variables instead of one, we consider their joint distribution function.

For a function f(X,Y) of discrete variables X and Y, we can define

$E[f(X,Y)]=\sum_{x,y}f(x,y)\cdot P(X=x, Y=y).$

Since f(X,Y)=X+Y

$E[X+Y]=\sum_{x,y}(x+y)P(X=x,Y=y)\\=\sum_{x,y}xP(X=x,Y=y)+\sum_{x,y}yP(X=x,Y=y).$

Let us look at the first of these sums.

$\sum_{x,y}xP(X=x,Y=y)\\=\sum_{x}x\sum_{y}P(X=x,Y=y)\\\text{Taking Marginal distribution of x}\\=\sum_{x}xP(X=x)=E[X].$

Similarly,

$\sum_{x,y}yP(X=x,Y=y)\\=\sum_{y}y\sum_{x}P(X=x,Y=y)\\\text{Taking Marginal distribution of y}\\=\sum_{y}yP(Y=y)=E[Y].$

Combining these two gives the formula:

$\sum_{x,y}xP(X=x,Y=y)+\sum_{x,y}yP(X=x,Y=y) =E(X)+E(Y)$

Therefore:

$E[X+Y]=E[X]+E[Y] \text{ as required.}$

(b)We want to show that if X and Y are independent random variables, then:

$Var(X+Y)=Var(X)+Var(Y)$

By definition of Variance, we have that:

$Var(X+Y)=E(X+Y-E[X+Y]^2)$

$=E[(X-\mu_X +Y- \mu_Y)^2]\\=E[(X-\mu_X)^2 +(Y- \mu_Y)^2+2(X-\mu_X)(Y- \mu_Y)]\\$Since we have shown that expectation is linear$\\=E(X-\mu_X)^2 +E(Y- \mu_Y)^2+2E(X-\mu_X)(Y- \mu_Y)]\\=E[(X-E(X)]^2 +E[Y- E(Y)]^2+2Cov (X,Y)$

Since X and Y are independent, Cov(X,Y)=0

$=Var(X)+Var(Y)$

Therefore as required:

$Var(X+Y)=Var(X)+Var(Y)$

7 0

3 years ago

63.92 divided by 4.7

FinnZ [79.3K]

The answer would be 13.6

7 0

3 years ago