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3 years ago

Simplify: (7^5)^6. Write your answer using an exponent.

Mathematics

1 answer:

lana66690 [7]3 years ago

4 0

Answer:

We know that,

$→ { ({a}^{m}) }^{n} = {a}^{(m \times n)}$

$→{( {7}^{5}) }^{6} = {7}^{(5 \times 6)} = \boxed{{7}^{30} }✓$

7³⁰ is the right answer.

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PLEASE HELP I NEED HELP ASAP TRANSVERSAL ANGLES A LOT OF POINTS ON THE LINE.

guapka [62]

Answer:

Step-by-step explanation:

so basically you have to look for the opposite side of what they give you. then just plot it in

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3 years ago

Solve the system of equations by elimination. 2x+8y=10 -4x-9y=-13 x = y =

Reptile [31]

So use cancelation
multiply first equation by 2
2x+8y=10
times 2
4x+16y=20
now add

4x+16y=20
-4x-9y=-13 +
0x+7y=7
7y=7
divide by 7
y=1

subsittue

2x+8y=10
2x+8(1)=10
2x+8=10
subtract 8 from both sides
2x=2
divide by 2
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x=1
y=1

8 0

3 years ago

What’s the correct answer for this?

grigory [225]

Answer:

The answer is 20

Step-by-step explanation:

Reasoning it is correct because for questions like this the bottom numbers for each shape are 10 and 8 they are very close to each other on the number line there fore since the number is two off you subtract 2 and then two again since it is 2 off to get 21 the closest answer is 20

It is a little confusing but since it is two places off you subtract two twice

8 0

3 years ago

Marty has 2 4/5 quarts of juice. he pours the same amount of juice into 2 bottles. how much does he pour into each bottle?

Ad libitum [116K]

2.8!!!!!! ;pppppppp-

8 0

3 years ago

Prove that if $w,z$ are complex numbers such that $|w|=|z|=1$ and $wz\ne -1$, then $\frac{w+z}{1+wz}$ is a real number.

mr_godi [17]

Answer:

See proof below

Step-by-step explanation:

Let $r=\frac{w+z}{1+wz}$ . If w=-z, then r=0 and r is real. Suppose that w≠-z, that is, r≠0.

Remember this useful identity: if x is a complex number then $x\bar{x}=|x|^2$ where $\bar{x}$ is the conjugate of x.

Now, using the properties of the conjugate (the conjugate of the sum(product) of two numbers is the sum(product) of the conjugates):

$\frac{r}{\bar{r}}=\frac{w+z}{1+wz} \left(\frac{1+\bar{w}\bar{z}}{\bar{w}+\bar{z}}{\right)$

= $\frac{(w+z)(1+\bar{w}\bar{z})}{(1+wz)(\bar{w}+\bar{z})}=\frac{w+z+w\bar{w}\bar{z}+z\bar{z}\bar{w}}{\bar{w}+\bar{z}+\bar{w}wz+\bar{z}zw}=\frac{w+z+w+|w|^2\bar{z}+|z|^2\bar{w}}{\bar{w}+\bar{z}+|w|^2z+|z|^2w}=\frac{w+z+\bar{z}+\bar{w}}{\bar{w}+\bar{z}+z+w}=1$

Thus $\frac{r}{\bar{r}}=1$ . From this, $r=\bar{r}$ . A complex number is real if and only if it is equal to its conjugate, therefore r is real.

3 0

3 years ago