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3 years ago

Steps by solving substitution -x+6y=10 and 3x-7y=-8

1 answer:

7 0

(2,2) would be your answer. solve for the first variable in one of the equations , then substitute the result into the other equation! i hope this helps :)

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building across the street casts a shadow that is 24 feet long. Your friend is 6 feet tall and casts a shadow that is 4 feet lon

atroni [7]

Answer:

6/4=b/24

1.5=b/24

24(1.5)=b

36=b

Building is 36 ft tall

Step-by-step explanation:

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3 years ago

In one year a company found 9 defective parts out of 4500 parts picked for testing

FromTheMoon [43]

Answer:

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3 years ago

Correct answer will get brainliest

Sophie [7]

Answer:

4.) Choice B

5.) Choice A

Step-by-step explanation:

Question 4:

First you have to solve for y to get slope intercept form.

$2x-5y=6\\5y=-2x+6\\y=-\frac{2}{5}x+\frac{6}{5}$

The slope is a fraction but you need it to convert it to a decimal so that'll be Choice B.

Question 5:

For this one I we will use this formula $m= \frac{y_{2}-y_{1}}{x_{2}-x_{1}}$

I will choose the easy pairs to plug it in $(-12,-2) (20,6)$

$\frac{6--2}{20--12}\\\\\\\frac{8}{32}=\frac{1}{4}$

The slope is a fraction again but you can convert it and that'll be Choice A.

5 0

3 years ago

I need help please can someone help

Andreyy89

Answer:

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Step-by-step explanation:

8 0

3 years ago

Read 2 more answers

The sum of the first 10 terms of an arithmetic sequence is 235 and the

Degger [83]

Answer:

$\bold{a_1 = -42.65}\\\\\bold{d=14.7}$

Step-by-step explanation:

The sum of the first 10 terms of an arithmetic sequence is:

$S_{10}=a_1+a_2+...+a_{10}=\dfrac{a_1+a_{10}}{2}\cdot10=\dfrac{2a_1+(10-1)d}{2}\cdot10$

$\dfrac{2a_1+(10-1)d}{2}\cdot10=235\\\\(2a_1+9d)\cdot5=235\\\\2a_2+9d=47$

the sum of the second 10 terms is: a₁₁ + a₁₂+...+ a₂₀

And the sum of the first 20 terms of an arithmetic sequence is:

$S_{20}=a_1+a_2+...+a_{10}+a_{11}+...+a_{20}=\dfrac{2a_1+(20-1)d}{2}\cdot10$

so the sum of the second 10 terms is:

$a_{11}+a_{12}+...+a_{20}=S_{20}-S_{10}$

Therefore we have:

$\dfrac{2a_1+(20-1)d}{2}\cdot10-\dfrac{2a_1+(10-1)d}{2}\cdot10=735\\\\(2a_1+19d)\cdot5-(2a_1+9d)\cdot5=735\\\\2a_1+19d-(2a_1+9d)=147\\\\10d=147\\\\d=14.7$

and:

$2a_1+9\cdot14.7=47\\\\2a_1+132.3=47\\\\2a_1=-85.3\\\\a_1=-42,65$

3 0

3 years ago