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2 years ago

How do I solve these problems?

Mathematics

1 answer:

topjm [15]2 years ago

8 0

6) 18,26,34
Add 8 each time
… 42,50,58,66,74,82,90,98,106,114,122,130
The 15th term in the sequence would be 130

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The equation of a curve is y=f(x). the curve goes through the points (1,3) and (3,7). f'(x) = 4x+p, where p is a number.

Nimfa-mama [501]

$f'(x)=4x+p\\
f(x)=2x^2+px+C\\\\
3=2\cdot1^2+p\cdot1+C\\
7=2\cdot3^2+p\cdot3+C\\\\
3=2+p+C\\
7=18+3p+C\\\\
C=1-p\\
3p=-11-C\\\\
3p=-11-(1-p)\\
3p=-11-1+p\\
2p=-12\\
p=-6\\\\
C=1-(-6)\\
C=7\\\\
y=2x^2-6x+7
$

5 0

3 years ago

Solve <img src="https://tex.z-dn.net/?f=%5Csf%20%5Cdfrac%7B1%7D%7Bp%7D%20%2B%20%5Cdfrac%7B1%7D%7Bq%7D%20%2B%20%5Cdfrac%7B1%7D

Nostrana [21]

Answer:

$\displaystyle \begin{cases} \displaystyle {x} _{1} = - p \\ \displaystyle x _{2} = - q \end{cases}$

Step-by-step explanation:

we would like to solve the following equation for x:

$\displaystyle \frac{1}{p} + \frac{1}{q} + \frac{1}{x} = \frac{1}{p + q + x}$

to do so isolate $\frac{1}{x}$ to right hand side and change its sign which yields:

$\displaystyle \frac{1}{p} + \frac{1}{q} = \frac{1}{p + q + x} - \frac{1}{x}$

simplify Substraction:

$\displaystyle \frac{1}{p} + \frac{1}{q} = \frac{x - (q + p + x)}{x(p + q + x)}$

get rid of only x:

$\displaystyle \frac{1}{p} + \frac{1}{q} = \frac{ - (q + p )}{x(p + q + x)}$

simplify addition of the left hand side:

$\displaystyle \frac{q + p}{pq} = \frac{ - (q + p )}{x(p + q + x)}$

divide both sides by q+p Which yields:

$\displaystyle \frac{1}{pq} = \frac{ -1}{x(p + q + x)}$

cross multiplication:

$\displaystyle x(p + q + x) = - pq$

distribute:

$\displaystyle xp + xq + {x}^{2} = - pq$

isolate -pq to the left hand side and change its sign:

$\displaystyle xp + xq + {x}^{2} + pq = 0$

rearrange it to standard form:

$\displaystyle {x}^{2} + xp + xq + pq = 0$

now notice we end up with a quadratic equation therefore to solve so we can consider factoring method to use so

factor out x:

$\displaystyle x( {x}^{} + p ) + xq + pq = 0$

factor out q:

$\displaystyle x( {x}^{} + p ) +q (x + p)= 0$

group:

$\displaystyle ( {x}^{} + p ) (x + q)= 0$

by Zero product property we obtain:

$\displaystyle \begin{cases} \displaystyle {x}^{} + p = 0 \\ \displaystyle x + q= 0 \end{cases}$

cancel out p from the first equation and q from the second equation which yields:

$\displaystyle \begin{cases} \displaystyle {x}^{} = - p \\ \displaystyle x = - q \end{cases}$

and we are done!

3 0

3 years ago

Which expressions are equivalent to x + 2y + x + 2?

Pani-rosa [81]

Answer:

none of the above im pretty sure

Step-by-step explanation:

6 0

2 years ago

The answer bNamamakakKkKkkKk

skad [1K]

That is really easy how don't you know

4 0

3 years ago

Please answerrrrrrrrrrr

saul85 [17]

Answer:

a c e

Step-by-step explanation:

7 0

3 years ago