Write the equation of the line that passes through the points (-2,3) and (-2,5)

Will mark u BRAINLIEST! Please help this is due ASAP

serious [3.7K]

Answer + Explanation:

let x be the measure of angle x

let y be the measure of angle y

$\left \{ {{x + y = 180} \atop {x=24+y}} \right.$

Now, solve:

plug x = 24 + y into x + y = 180

24 + y + y = 180

24 + 2y = 180

2y = 156

y = 78

Now, plug this back into the equation x = 24 + 78

x = 102

8 0

3 years ago

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Help this is due tomorrow!!!

r-ruslan [8.4K]

Answer:

V=4/3*3.14*r^3

V=4/3*3.14*3^3

V=113.1

Volume of Himiphere= Volume of sphere/2

113.1/2

V=56.6

Step-by-step explanation:

The answer is 56.6

4 0

2 years ago

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The length of a rectangle is six times it's width. If the area is 216 inches squared, find its perimeter

Lana71 [14]

I had a problem like this but it was different but what i divided was 216 divided by 12 and got 18 and 216 divided by 6 and got 36

8 0

3 years ago

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30 points!!<br> What is the sum of the first six terms of the series?<br> 48 - 12 + 3 - 0.75 +...

Lunna [17]

Answer:

The sum of the first six terms is 38.39

Step-by-step explanation:

This is a geometric sequence since the common difference between each term is $-\frac{1}{4}$

Thus, $r=-\frac{1}{4}$

To find the sum of first six terms, we need to find the fifth and sixth term of the sequence.

To find the fifth term:

The general form of geometric sequence is $a_{n}=a_{1} \cdot r^{n-1}$

To find the fifth term, substitute $n=5$ in $a_{n}=a_{1} \cdot r^{n-1}$

$\begin{aligned}a_{5} &=(48) \cdot\left(-\frac{1}{4}\right)^{5-1} \\&=(48) \cdot\left(-\frac{1}{4}\right)^{4} \\&=(48)\left(\frac{1}{256}\right) \\a_{5} &=0.1875\end{aligned}$

To find the sixth term, substitute $n=6$ in $a_{n}=a_{1} \cdot r^{n-1}$

$\begin{aligned}a_{6} &=(48) \cdot\left(-\frac{1}{4}\right)^{6-1} \\&=(48) \cdot\left(-\frac{1}{4}\right)^{5} \\&=(48)\left(-\frac{1}{1024}\right) \\a_{5} &=-0.046875\end{aligned}$

To find the sum of the first six terms:

The general formula to find Sn for $|r|$ is $S_{n}=\frac{a\left(1-r^{n}\right)}{1-r}$

$\begin{aligned}S_{6} &=\frac{48\left(1-\left(-\frac{1}{4}\right)^{6}\right)}{1-\left(-\frac{1}{4}\right)} \\&=\frac{48\left(1-\frac{1}{4096}\right)}{1+\frac{1}{4096}} \\&=\frac{48(0.95)}{5} \\&=\frac{48(0.9998)}{5} \\&=\frac{48(0.9998)}{5} \\&=\frac{47.9904}{5} \\&=38.39\end{aligned}$