3^x = 27^a+b and a^2-b^2/(a-b)=5 What is x?

Drupady [299]

Again, I've attached an image showing the 30-60-90 relationship for triangles. Hopefully this should help as we work through this problem.

For the side adjacent to the 30 degree angle, we know that the side length is . Let's take a look at the given options and see which ones fit.

The options that fit are $x\sqrt{3}$ and $7\sqrt{3}$ .

For the side adjacent to the 60 degree angle, we know that the side length is just x. Let's take a look at the given options and see which ones fit.

The options that fit are x and 7.

Finally, the hypotenuse is the longest side of the triangle and its length is equal to 2x.

So, the options that fit are 2x and 14.

Hope this helps! :)

4 0

3 years ago

Semenov [28]

Answer:

Yes, because m<UVW is congruent to m<XVY and m<VUW is congruent to m<VXY

Step-by-step explanation:

we know that

If two figures are similar, then the ratio of its corresponding sides is equal and its corresponding angles are congruent

so

In this problem

we know that

$m< VUW=55\°$

the measure of angle VXY is equal to

$m$ -----> by supplementary angles

$m$

therefore

$m< VUW=m$

Remember that

m<UVW=m<XVY -----> is the same vertex

therefore

Triangles VUW and VXY are similar by AAA Similarity Theorem ( the three angles are congruent)

3 0

3 years ago

write two additional sentences where the sum is negative. in each sentence one added should be positive and the other

Goshia [24]

10 + (-5) = 10 - 5 = 5

-11 + 20 = 20 + (-11) = 20 - 11 = 9

102 + (-1) = 102 - 1 = 101

<span>-20 + 60 = 60 + (-20) = 60 - 20 = 40 </span>

6 0

3 years ago

The freezing point of oxygen is –218.790C and hydrogen is –252.80C. A lab is lowering the temperature inside a fridge. Which fre

alexandr402 [8]

Answer:

Oxygen; 34.01°C

Step-by-step explanation:

The oxygen freezes first as it has a higher (warmer) freezing point. For both to freeze, the temperature must drop a further 34.01°C as:

-218.79 - (-252.8)

= 252.8 - 218.79

= 34.01

3 0

3 years ago

Use induction to prove: For every integer n > 1, the number n5 - n is a multiple of 5.

nignag [31]

Answer:

we need to prove : for every integer n>1, the number $n^{5}-n$ is a multiple of 5.

1) check divisibility for n=1, $f(1)=(1)^{5}-1=0$ (divisible)

2) Assume that $f(k)$ is divisible by 5, $f(k)=(k)^{5}-k$

3) Induction,

$f(k+1)=(k+1)^{5}-(k+1)$

$=(k^{5}+5k^{4}+10k^{3}+10k^{2}+5k+1)-k-1$

$=k^{5}+5k^{4}+10k^{3}+10k^{2}+4k$

Now, $f(k+1)-f(k)$

$f(k+1)-f(k)=k^{5}+5k^{4}+10k^{3}+10k^{2}+4k-(k^{5}-k)$

$f(k+1)-f(k)=k^{5}+5k^{4}+10k^{3}+10k^{2}+4k-k^{5}+k$

$f(k+1)-f(k)=5k^{4}+10k^{3}+10k^{2}+5k$

Take out the common factor,

$f(k+1)-f(k)=5(k^{4}+2k^{3}+2k^{2}+k)$ (divisible by 5)

add both the sides by f(k)

$f(k+1)=f(k)+5(k^{4}+2k^{3}+2k^{2}+k)$

We have proved that difference between $f(k+1)$ and $f(k)$ is divisible by 5.

so, our assumption in step 2 is correct.

Since $f(k)$ is divisible by 5, then $f(k+1)$ must be divisible by 5 since we are taking the sum of 2 terms that are divisible by 5.

Therefore, for every integer n>1, the number $n^{5}-n$ is a multiple of 5.

3 0

3 years ago