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2 years ago

What is the solution to the equation 1/square root of 8 = 4^(m + 2)?

2 answers:

8 0

Take the log (base 4) of both sides of the equation.
$\log_{4}(\frac{1}{\sqrt{8}}) = m + 2$
$\frac{-1}{2} \log_{4}(4^{\frac{3}{2}})-2 = m$
$-2 \frac{3}{4}= m$

The appropriate choice is ...
m = -11/4

salantis [7]2 years ago

6 0

I agree with the other person, the answer is -11/4

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If a person studies for 8hrs a week. How long to they work each day

Andrew [12]

Answer:

about and hour and 14 minutes each day

Step-by-step explanation:

just divide 8 by 7 and round to the nearest hundredths place

7 0

2 years ago

Anyone mind helping!?

Ksju [112]

Answer:

B) $3\sqrt{3}$

Step-by-step explanation:

First simplify the radical on the right.

$\sqrt{75}$ is equal to $\sqrt{25} *\sqrt{3}$ , because 25 * 3 is 75, so we can separate the square roots into the product of these two square roots.

Then we know $\sqrt{25}$ is just equal to 25, which means the simplified version of $\sqrt{75}$ is $5\sqrt{3}$ .

Then we can use this new representation to finish solving.

$8\sqrt{3}-5\sqrt{3} =3\sqrt{3}$ , so the answer would be B

6 0

2 years ago

Read 2 more answers

i bought 3 small tapas dishes and 2 large ones on a visit to café ba-ba-reeba and. spent $14. I went back another day and got 4

satela [25.4K]

The small dish costs $2 each and the large dish costs $4 each,

2(3)+4(2)=6+8=14

2(4)+4(1)=8+4=12

5 0

3 years ago

Jacob is asked to solve the follow problem: "Kayla is 9 years older than twice Hannah’s age. If the difference in their ages is

tatiyna

Answer:

Step-by-step explanation:

k = 2h+9

h+16=k

h+16=2h+9

-h = -7

h = 7

hannah is 7 and kayla is 23

jacob is wrong because if u do 16*2+7 it’s 39 and 39-16 does not equal 16

6 0

2 years ago

Find x Thanks for the help in advance!

Marina86 [1]

I'm not sure if this is the easiest way of doing this, but it surely work.

Let the base of the triangle be AB, and let CH be the height. Just for reference, we have

$AH=2,\quad HB=6,\quad AC=x$

Moreover, let CH=y and BC=z

Now, AHC, CHB and ABC are all right triangles. If we write the pythagorean theorem for each of them, we have the following system

$\begin{cases}4+y^2=x^2\\36+y^2=z^2\\x^2+z^2=64\end{cases}$

If we solve the first two equations for y squared, we have

$y^2=x^2-4\\y^2=z^2-36$

And we can deduce

$z^2 = x^2+32$

So that the third equation becomes

$x^2+x^2+32=64 \iff 2x^2 = 32 \iff x^2=16 \iff x=4$

(we can't accept the negative root because negative lengths make no sense)

7 0

3 years ago