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Softa [21]
3 years ago
7

You know that 5^2=25. How can you use this fact to evaluate 5^4

Mathematics
1 answer:
PIT_PIT [208]3 years ago
7 0

Answer:

5^4 = 625

Step-by-step explanation:

5^4 is just 25 times 25

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Classify the following triangle as acute
BaLLatris [955]
It is an acute triangle since all the sides are less than 90 degrees (it's also equilateral)
8 0
3 years ago
The ratio of boys to girls in Ms. smith’s class is 3 to 4. If there are 12 boys, how many girls are there?
Lorico [155]

✧・゚: *✧・゚:*    *:・゚✧*:・゚✧

                  Hello!

✧・゚: *✧・゚:*    *:・゚✧*:・゚✧

❖ There will be 16 girls.

The relationship is x4. If we multiply 3 x 4, we get 12. What we do to one number we do to the other so 4 x 4 = 16.

~ ʜᴏᴘᴇ ᴛʜɪꜱ ʜᴇʟᴘꜱ! :) ♡

~ ᴄʟᴏᴜᴛᴀɴꜱᴡᴇʀꜱ

4 0
3 years ago
Solve -44-7k=12+k answer quick please
nirvana33 [79]

Answer:

k = -7

Step-by-step explanation:

Solve for k:

-7 k - 44 = k + 12

Subtract k from both sides:

(-7 k - k) - 44 = (k - k) + 12

-7 k - k = -8 k:

-8 k - 44 = (k - k) + 12

k - k = 0:

-8 k - 44 = 12

Add 44 to both sides:

(44 - 44) - 8 k = 44 + 12

44 - 44 = 0:

-8 k = 12 + 44

12 + 44 = 56:

-8 k = 56

Divide both sides of -8 k = 56 by -8:

(-8 k)/(-8) = 56/(-8)

(-8)/(-8) = 1:

k = 56/(-8)

The gcd of 56 and -8 is 8, so 56/(-8) = (8×7)/(8 (-1)) = 8/8×7/(-1) = 7/(-1):

k = 7/(-1)

Multiply numerator and denominator of 7/(-1) by -1:

Answer: k = -7

7 0
3 years ago
Read 2 more answers
BRAINLIESTT ASAP! PLEASE HELP ME :)
taurus [48]

Answer:

The answer is they are congruent.

Step-by-step explanation:

They have two congruent sides and share one side, so they are congruent by SSS.

I hope this helps! :)

8 0
3 years ago
Read 2 more answers
Find the smallest 4 digit number such that when divided by 35, 42 or 63 remainder is always 5
alex41 [277]

The smallest such number is 1055.

We want to find x such that

\begin{cases}x\equiv5\pmod{35}\\x\equiv5\pmod{42}\\x\equiv5\pmod{63}\end{cases}

The moduli are not coprime, so we expand the system as follows in preparation for using the Chinese remainder theorem.

x\equiv5\pmod{35}\implies\begin{cases}x\equiv5\equiv0\pmod5\\x\equiv5\pmod7\end{cases}

x\equiv5\pmod{42}\implies\begin{cases}x\equiv5\equiv1\pmod2\\x\equiv5\equiv2\pmod3\\x\equiv5\pmod7\end{cases}

x\equiv5\pmod{63}\implies\begin{cases}x\equiv5\equiv2\pmod 3\\x\equiv5\pmod7\end{cases}

Taking everything together, we end up with the system

\begin{cases}x\equiv1\pmod2\\x\equiv2\pmod3\\x\equiv0\pmod5\\x\equiv5\pmod7\end{cases}

Now the moduli are coprime and we can apply the CRT.

We start with

x=3\cdot5\cdot7+2\cdot5\cdot7+2\cdot3\cdot7+2\cdot3\cdot5

Then taken modulo 2, 3, 5, and 7, all but the first, second, third, or last (respectively) terms will vanish.

Taken modulo 2, we end up with

x\equiv3\cdot5\cdot7\equiv105\equiv1\pmod2

which means the first term is fine and doesn't require adjustment.

Taken modulo 3, we have

x\equiv2\cdot5\cdot7\equiv70\equiv1\pmod3

We want a remainder of 2, so we just need to multiply the second term by 2.

Taken modulo 5, we have

x\equiv2\cdot3\cdot7\equiv42\equiv2\pmod5

We want a remainder of 0, so we can just multiply this term by 0.

Taken modulo 7, we have

x\equiv2\cdot3\cdot5\equiv30\equiv2\pmod7

We want a remainder of 5, so we multiply by the inverse of 2 modulo 7, then by 5. Since 2\cdot4\equiv8\equiv1\pmod7, the inverse of 2 is 4.

So, we have to adjust x to

x=3\cdot5\cdot7+2^2\cdot5\cdot7+0+2^3\cdot3\cdot5^2=845

and from the CRT we find

x\equiv845\pmod2\cdot3\cdot5\cdot7\implies x\equiv5\pmod{210}

so that the general solution x=210n+5 for all integers n.

We want a 4 digit solution, so we want

210n+5\ge1000\implies210n\ge995\implies n\ge\dfrac{995}{210}\approx4.7\implies n=5

which gives x=210\cdot5+5=1055.

5 0
2 years ago
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