3 years ago

What is one-ninth times negative one and three over one?

Mathematics

1 answer:

denis23 [38]3 years ago

5 0

Answer:

I think the answer is one over three

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Help me please !!!!!!!!!!!

Ymorist [56]

Answer:

Step-by-step explanation:

first choice

1/8 x +1/8 x =2/8=1/4 x

7 0

2 years ago

Kylie has $70 to spend on clothes. She wants to buy a pair of jeans for $30 and spend the rest on t-shirts. Each t-shirt costs $

NeX [460]

70-30=40

40 divided by 8 = 5

x ≤ 5

she can buy at the most 5 T-shirts.

3 0

3 years ago

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4 (x - 1) = }(12x – 3)

slamgirl [31]

Answer:

x= -1/8

Step-by-step explanation:

try using symbolab

4 0

2 years ago

For the following telescoping series, find a formula for the nth term of the sequence of partial sums

gtnhenbr [62]

I'm guessing the sum is supposed to be

$\displaystyle\sum_{k=1}^\infty\frac{10}{(5k-1)(5k+4)}$

Split the summand into partial fractions:

$\dfrac1{(5k-1)(5k+4)}=\dfrac a{5k-1}+\dfrac b{5k+4}$

$1=a(5k+4)+b(5k-1)$

If $k=-\frac45$ , then

$1=b(-4-1)\implies b=-\frac15$

If $k=\frac15$ , then

$1=a(1+4)\implies a=\frac15$

This means

$\dfrac{10}{(5k-1)(5k+4)}=\dfrac2{5k-1}-\dfrac2{5k+4}$

Consider the $n$ th partial sum of the series:

$S_n=2\left(\dfrac14-\dfrac19\right)+2\left(\dfrac19-\dfrac1{14}\right)+2\left(\dfrac1{14}-\dfrac1{19}\right)+\cdots+2\left(\dfrac1{5n-1}-\dfrac1{5n+4}\right)$

The sum telescopes so that

$S_n=\dfrac2{14}-\dfrac2{5n+4}$

and as $n\to\infty$ , the second term vanishes and leaves us with

$\displaystyle\sum_{k=1}^\infty\frac{10}{(5k-1)(5k+4)}=\lim_{n\to\infty}S_n=\frac17$

7 0

3 years ago

What value of x makes the equation 5(4x - 1)=4(2.5x+8) true?

monitta

Answer:

37/10

( I hope this was helpful) >;D

7 0

3 years ago

Read 2 more answers