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2 years ago

Asap! please help i’ll give brainliest

Mathematics

1 answer:

Aleonysh [2.5K]2 years ago

6 0

Answer: 2nd one

Step-by-step explanation: Look at the options. In all of the options, the number of 0's is 3 so you don't have to look for the zero's. Next look at the one's. The first 2 have 2 ones and the next 2 have 3. The number of ones in the data is 2 so you can eliminate the last 2. Then look at the 2's between the top 2. In the first one, there is 2 and in the second one there is only one. In the data table, there is one 2 so the answer has to be the 2nd one. Hope this helps :)

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9x-2y=-21 2x+3y=16 Substitution

hjlf

$9x-2y=-21 \\\\ 9x=-21+2y \\\\ => \boxed{x=\frac{2y-21}{9}} \\\\ 2x+3y=16 \\\\ 2*\frac{2y-21}{9}+3y^{(9}=16^{(9} \\\\ 2(2y-21)+27y=144 \\\\ 4y-42+27y=144 \\\\ 31y=144+42 \\\\ 31y=186 \\\\ \boxed{y=\frac{186}{31}=6} \\\\ x=\frac{2*6-21}{9} \\\\ x=\frac{12-21}{9} \\\\ x=\frac{-9}{9} \\\\ \boxed{x=-1}$

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3 years ago

A-2b, If a = - 2.5, b = - 3.3

lorasvet [3.4K]

Answer:

-2.5 - 2(-3.3)

-2.5 + 6.6 = 4.1

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2 years ago

Nadia spent 1/4 of her money on a shirt and 2/5 of her money on new shoes. What fraction of Nadias money was spent?

Ulleksa [173]

Answer:

13/20 of her money was spent.

Step-by-step explanation:

To find the answer you have to add these two fractions together, and to do that you must find a common denominator.

For this problem I chose to use 20 for the denominator since it is the smallest number that both 4 and 5 have in common.

Keep in mind that, whatever you do to the denominator, you must also do to the numerator, so if you multiply 5 by 4, you must also multiply 2 by 4.

5/20 + 8/20 = 13/20

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3 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:

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