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zavuch27 [327]
3 years ago
10

36m+16 (no spaces in your answer Factor and Expand

Mathematics
1 answer:
N76 [4]3 years ago
4 0

uh..? im dead lol PLS

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What is the sum of the arithmetic series: s10 = -20, d = 4, a1 =? ​
Otrada [13]
<h2>Step-by-step explanation:</h2><h3>\green\star Given</h3>

\blue\stars10 = -20

\blue\starn = 10

\blue\star an = -20

\green\star solution

\green\stars10 = n\2 [a+(n-1)d]

\green\star-20 = 10a2[a+(10-1)4]

\green\star-4 = a +36

\green\stara1 = -40

\small\boxed{a = -40}

Hope it helps.

Hope it helps

8 0
3 years ago
Read 2 more answers
The mean of 100 observation is found to be 40. If at the time of compution two items were wrongly taken as 30 and 27 instead of
Colt1911 [192]

Answer:

40.18

Step-by-step explanation:

According to this question, the mean of 100 observation is found to be 40. This means that the number of values in the set of data is 100. However, at the time of computation two items were wrongly taken as 30 and 27 instead of 3 and 72.

The mean is derived by dividing the sum of values in the data (Σx) by the number of observations (n) i.e.

Mean = Σx/n

For the computation error:

Mean = 40

Let x represent the sun of other 98 values except 30 and 27

n = 100

40 = x + 30 + 27/100

40 = 57 + x/100

Cross multiply

4000 = 57 + x

x = 4000 - 57

x = 3943.

Sum of other 98 values = 3943.

Now, let's find the correct mean with correct values of 3 and 72;

Mean = Σx/n

Mean = 3943 + 3 + 72/100

Mean = 4018/100

Mean = 40.18

7 0
2 years ago
I'm confused on how to solve a right triangle given only one side measurement and one interior angle.
SVEN [57.7K]

Answer:

You use trigonometry.

Step-by-step explanation:

You can use soh cah toa. Basically you pic one side and if the interior angle is opposite or next to it, you you soh, or cah, or toa, it's based on where the angle is.

8 0
2 years ago
Solve the inequality <br><br><img src="https://tex.z-dn.net/?f=%20%5Cfrac%7B2x%7D%7Bx%20%2B%203%7D%20%20%5Cleqslant%203" id="Tex
Fofino [41]

Domain:x\neq-3\\\\\dfrac{2x}{x+3}\leq3\qquad\text{subtract 3 from both sides}\\\\\dfrac{2x}{x+3}-3\leq0\\\\\dfrac{2x}{x+3}-\dfrac{3(x+3)}{x+3}\leq0\\\\\dfrac{2x}{x+3}-\dfrac{3x+9}{x+3}\leq0\\\\\dfrac{2x-(3x+9)}{x+3}\leq0\\\\\dfrac{2x-3x-9}{x+3}\leq0\\\\\dfrac{-x-9}{x+3}\leq0\Rightarrow(x+3)(-x-9)\leq0\\\\x=-3,\ x=-9\\\\x\in(-\infty,\ -9]\ \cup\ (-3,\ \infty)

3 0
3 years ago
Can anyone help me with this question?
mina [271]

Answer:

download photomath

Step-by-step explanation:

8 0
3 years ago
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