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EastWind [94]
4 years ago
9

Write an equation for the problem.Then Solve. What number is 8% of 50?

Mathematics
1 answer:
Maslowich4 years ago
8 0

\frac{50}{100}  \times 8 = x \\ .5 \times 8 = x \\ 4 = x \\ 4
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Can someone please help me
Natalija [7]
B. 

Using this will give you proper representations in similar triangles. 
3 0
3 years ago
Please help me solve this​
Viktor [21]

You can factor the 32 out of the sum:

\displaystyle \sum_{m=1}^\infty 32 \cdot \left(\dfrac{1}{2}\right)^{m-1} = 32\sum_{m=1}^\infty \left(\dfrac{1}{2}\right)^{m-1}

We can also change the index as follows

\displaystyle 32\sum_{m=1}^\infty \left(\dfrac{1}{2}\right)^{m-1} = 32\sum_{m=0}^\infty \left(\dfrac{1}{2}\right)^{m}

Now, we have a theorem that states that the series

\displaystyle \sum_{m=1}^\infty a^m

converges if and only if |a|, and in this case we have

\displaystyle \sum_{m=1}^\infty a^m = \dfrac{1}{1-a}

This is your case, because you have

|a|=\dfrac{1}{2}

which implies that your series converges, and the value is

\displaystyle 32\sum_{m=0}^\infty \left(\dfrac{1}{2}\right)^{m} = 32 \cdot \dfrac{1}{1-\frac{1}{2}} = 32\cdot 2 = 64

3 0
3 years ago
0=2x+3(3x-4)-(-x+14)<br><br> Fraction answer need help!!!
blondinia [14]
<span>0=2x+3(3x-4)-(-x+14) 
Follow BEDMAS
</span><span>0=2x+9x-12-(-x+14) 
</span>0=2x+9x-12+x-14
Add like terms
0=11x-12+x-14
0=12x-12-14
0=12x-26
Flip the eqaution
12x-26=0
Move -26 across the equal sign
12x=0+26
12x=26
x=26/12
x=13/6

Hope this helps! A thanks/brainiest answer would be appreciated :)
5 0
4 years ago
PLEASE HELP ME !!!!​
Natalka [10]

Answer:

B

Step-by-step explanation:

There are no repeating values for x, for different values plugged in for y.

3 0
3 years ago
Prove that an = 4^n + 2(-1)^nis the solution to
olga nikolaevna [1]

Answer:

See proof below

Step-by-step explanation:

We have to verify that if we substitute a_n=4^n+2(-1)^n in the equation a_n=3a_{n-1}+4a_{n-2} the equality is true.

Let's substitute first in the right hand side:

3a_{n-1}+4a_{n-2}=3(4^{n-1}+2(-1)^{n-1})+4(4^{n-2}+2(-1)^{n-2})

Now we use the distributive laws. Also, note that (-1)^{n-1}=\frac{1}{-1}(-1)^n=(-1)(-1)^{n} (this also works when the power is n-2).

=3(4^{n-1})+6(-1)^{n-1}+4(4^{n-2})+8(-1)^{n-2}

=3(4^{n-1})+(-1)(6)(-1)^{n}+4^{n-1}+(-1)^2(8)(-1)^{n}

=4(4^{n-1})-6(-1)^{n}+8(-1)^{n}=4^n+2(-1)^n=a_n

then the sequence solves the recurrence relation.

4 0
3 years ago
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