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3 years ago

Find The value of each variable

Mathematics

1 answer:

AnnyKZ [126]3 years ago

7 0

So what’s the problem

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If 10a+10b=35, what is the average (arithmetic mean) of a and b?

worty [1.4K]

Answer:

1.75

Step-by-step explanation:

If a and b are two numbers, then their arithmetic mean is

$\dfrac{a+b}{2}$

Given:

$10a+10b=35$

Divide this equation by 10:

$a+b=3.5$

Now, divide it by 2:

$\dfrac{a+b}{2}=\dfrac{3.5}{2}\\ \\\dfrac{a+b}{2}=1.75$

6 0

3 years ago

Find the slope of the given line on the graph. REMEMBER: Rise over run! <br> Pls show work!

Leviafan [203]

Answer:

the slope is $-\frac{1}{3}$

Step-by-step explanation:

Notice that you are given two points to use for the slope calculation:

$(x_1,y_1)=(-4,4) \,\,and\,\, (x_2,y_2)=(5,1)$

So we use the formula for the slope of a line given two points:

$slope\,=\,\frac{y_2-y_1}{x_2-x_1} \\slope\,=\,\frac{1-4}{5-(-4)} \\slope\,=\,\frac{-3}{9} \\slope\,=\,-\frac{1}{3} \\$

3 0

2 years ago

1. (x + 2) < −3(x + 4)<br><br> 2. −3x + 3 < 6

Leno4ka [110]

number 1:

$x>\frac{-7}{2}$

number 2:

$x>-1$

7 0

3 years ago

What’s the pattern of -1, -8, -27, and -64

AURORKA [14]

Hello from MrBillDoesMath!

Answer:

a(n) = (-n)^3 where n = 1,2,3,...

Discussion:

The pattern 1,8,27, 64... is immediately recognizable as the the cube of the positive integers. But this question has a minus sign appearing before each entry, suggesting we try this:

- 1 = (-1)^3

-8 = (-2)^3

-27 = (-3)^3

-64 = (-4)^3

That's what the problem statement asked for . The answer is equivalently

-1 * (n^3)

Thank you,

MrB

6 0

3 years ago

Solve using the method of elimination and determine if the system has 1 solution, no solutions, or infinite solutions

Minchanka [31]

$\begin{array}{rrrrr} 10x&-&18y&=&2\\ -5x&+&9y&=&-1 \end{array}~\hfill \implies ~\hfill \stackrel{\textit{second equation }\times 2}{ \begin{array}{rrrrr} 10x&-&18y&=&2\\ 2(-5x&+&9y&)=&2(-1) \end{array}} \\\\[-0.35em] ~\dotfill\\\\ \begin{array}{rrrrr} 10x&-&18y&=&2\\ -10x&+&18y&=&-2\\\cline{1-5} 0&+&0&=&0 \end{array}\qquad \impliedby \textit{another way of saying \underline{infinite solutions}}$

if we were to solve both equations for "y", we'd get

$10x-18y=2\implies 10x-2=18y\implies \cfrac{10x-2}{18}=y\implies \cfrac{5}{9}x-\cfrac{1}{9}=y \\\\\\ -5x+9y=-1\implies 9y=5x-1\implies y=\cfrac{5x-1}{9}\implies y = \cfrac{5}{9}x-\cfrac{1}{9}$

notice, the 1st equation is really the 2nd in disguise, since both lines are just pancaked on top of each other, every point in the lines is a solution or an intersection, and since both go to infinity, well, there you have it.

8 0

1 year ago