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Ludmilka [50]
3 years ago
10

Can you solve the equation? Test your brain power. 8 x f =

Mathematics
2 answers:
Marrrta [24]3 years ago
6 0

Answer: Hmm, i dont know if this is right, but heres one anwser, 11.2, and the second one, 16.

Please tell me which is right.

jeyben [28]3 years ago
5 0

Answer:

f = 0

Step-by-step explanation:

8 x f = 0               multiply 8 by f

8f= 0                    divide both sides by 8 to get f

8f/8 = 0/8

f = 0

hope this helps! if you have any questions, pls let me know!

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2 years ago
Perform the indicated operations. Write the answer in standard form, a+bi.<br> (8 - i)^2 + (8 + i)^2
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Answer:

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3 years ago
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Step-by-step explanation:

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3 years ago
In 2018, the number of students at The Villages High School was 975 and is increasing at a rate of 2.5% per year. Write and use
avanturin [10]

Answer:

A(t)=975(1.025)^t

In 2025,the number of students at the villages high school=1159

Step-by-step explanation:

We are given that in 2018

Number of students at the villages high school=975

Increasing rate,r=2.5%=0.025

We have to write and use of exponential growth function to project the populating in 2025.

A_0=975,t=0

According to question

Number of students at the villages high School is given by

A(t)=A_0(1+r)^t

Substitute the values

A(t)=975(1+0.025)^t=975(1.025)^t

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Then, the number of students at the villages high school in 2025

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8 0
3 years ago
Read 2 more answers
Write an equation of the hyperbola given that the center is at (2, -3), the vertices are at (2, 3) and (2, - 9), and the foci ar
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Check the picture below.

so, the hyperbola looks like so, clearly a = 6 from the traverse axis, and the "c" distance from the center to a focus has to be from -3±c, as aforementioned above, the tell-tale is that part, therefore, we can see that c = 2√(10).

because the hyperbola opens vertically, the fraction with the positive sign will be the one with the "y" in it, like you see it in the picture, so without further adieu,

\bf \textit{hyperbolas, vertical traverse axis }&#10;\\\\&#10;\cfrac{(y- k)^2}{ a^2}-\cfrac{(x- h)^2}{ b^2}=1&#10;\qquad &#10;\begin{cases}&#10;center\ ( h, k)\\&#10;vertices\ ( h,  k\pm a)\\&#10;c=\textit{distance from}\\&#10;\qquad \textit{center to foci}\\&#10;\qquad \sqrt{ a ^2 + b ^2}\\&#10;asymptotes\quad  y= k\pm \cfrac{a}{b}(x- h)&#10;\end{cases}\\\\&#10;-------------------------------

\bf \begin{cases}&#10;h=2\\&#10;k=-3\\&#10;a=6\\&#10;c=2\sqrt{10}&#10;\end{cases}\implies \cfrac{[y- (-3)]^2}{ 6^2}-\cfrac{(x- 2)^2}{ b^2}=1&#10;\\\\\\&#10;\cfrac{(y+3)^2}{ 36}-\cfrac{(x- 2)^2}{ b^2}=1&#10;\\\\\\&#10;c^2=a^2+b^2\implies (2\sqrt{10})^2=6^2+b^2\implies 2^2(\sqrt{10})^2=36+b^2&#10;\\\\\\&#10;4(10)=36+b^2\implies 40=36+b^2\implies 4=b^2&#10;\\\\\\&#10;\sqrt{4}=b\implies 2=b\\\\&#10;-------------------------------\\\\&#10;\cfrac{(y+3)^2}{ 36}-\cfrac{(x- 2)^2}{ 2^2}=1\implies \cfrac{(y+3)^2}{ 36}-\cfrac{(x- 2)^2}{ 4}=1

3 0
3 years ago
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