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2 years ago

8

An office upgrades its internet speed from 10^4 bits per second to 10^8 bits per second. How many times greater is the new inter

net speed?

1 answer:

Kamila [148]2 years ago

8 0

10000 times greater
10^4 is 10000
And 10^8 is 100000000

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PLEASE HELP ILL GIVE BRAINLIESTTTT :)

Taya2010 [7]

Answer:

Yes, these triangles are congruent

∠E≅∠M

Step-by-step explanation:

Yes, these triangles are similar.

<em>Proof:</em>

Each side in LNM is $1\frac{2}{3}$ times longer than the sides in GFE

$30/18 = 1\frac{2}{3}$

$25/15=1\frac{2}{3}$

$20/12 = 1\frac{2}{3}$

According CPCTC (corresponding parts of congruent triangles are congruent), angle E is similar to angle M.

Hope this helps :)

Have a good day!

4 0

2 years ago

What is the value of x?<br> X= 10^2 + 8^2

Luden [163]

$\: \: \: \: \: \: \: \: \: x = {10}^{2} + {8}^{2} \\ x = 164$

Hope it helps

7 0

2 years ago

Read 2 more answers

Will give brainliest Question on the picture

raketka [301]

Answer:

12

Step-by-step explanation:

since i helped can i have brainlist please :D

3 0

2 years ago

Please help with this I am completely stuck on it

vaieri [72.5K]

Answer:

$f(x)=\sqrt[3]{x-4} , g(x)=6x^{2}\textrm{ or }f(x)=\sqrt[3]{x},g(x)=6x^{2} -4$

Step-by-step explanation:

Given:

The function, $H(x)=\sqrt[3]{6x^{2}-4}$

Solution 1:

Let $f(x)=\sqrt[3]{x}$

If $f(g(x))=H(x)=\sqrt[3]{6x^{2}-4}$ , then,

$\sqrt[3]{g(x)} =\sqrt[3]{6x^{2}-4}\\g(x)=6x^{2}-4$

Solution 2:

Let $f(x)=\sqrt[3]{x-4}$ . Then,

$f(g(x))=H(x)=\sqrt[3]{6x^{2}-4}\\\sqrt[3]{g(x)-4}=\sqrt[3]{6x^{2}-4} \\g(x)-4=6x^{2}-4\\g(x)=6x^{2}$

Similarly, there can be many solutions.

7 0

3 years ago

Each day her goal is to make 4 more baskets than she made thr day before. If she makes 10 baskets the first day and meets her go

irina [24]

Answer:

Janelle will make <u>62 baskets</u> on 14th day.

Step-by-step explanation:

Given,

Number of goals on 1st day = 10

We need to find the number of goals on 14th day.

Solution,

Since Janelle made 10 goals on 1st day and want to make 4 more basket from the day before.

So we can say that;

$a_1=10$ and $d=4$

And also $n=14$

here $a_1$ = first term

d = common difference

n = number of terms

Now we apply the formula of A.P.

$T_n=a_1+(n-1)d$

now substituting the given values, we get;

$T_{14}=10+(14-1)4\\\\T_{14}=10+13\times 4\\\\T_{14}=10+52 =62$

Hence Janelle will make <u>62 basket</u> on 14th day.

7 0

2 years ago