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3 years ago

—бу – 3 = 3 – бу No solution All real numbers I need the answer quickly!!

Mathematics

1 answer:

topjm [15]3 years ago

5 0

Answer:

no solutions

Step-by-step explanation:

-6y-3=3-6y

1. add 6y to both sides:

-6y-3 +6y=3-6y +6y

-3=3

since -3 doesn't equal 3, there are no solutions.

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PLEASE help me with 26 and 27 asap

wariber [46]

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3 years ago

A jeweler had 38 inches of silver chain. She need five times that much to make some necklaces and 3 times that amount to make so

aivan3 [116]

Answer:

304 IN

Step-by-step explanation:

total inches needed for necklace : 38 x 5 = 190

total inches needed for necklace : 38 x 3 = 114

190 + 114 = 304 inches

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3 years ago

Answer the question in the photo pls!

galben [10]

Answer:

6x + 2y

Step-by-step explanation:

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2 years ago

Could someone please help me with this thank you very much

elena-s [515]

Answer:

Answer for #10:

$y = 2 \: and \: x = 1$

Answer for #1:

$x = 2 \: and \: y = - 1$

Step-by-step explanation:

In order to solve a simultaneous equation using the substitution method you have to:

Use one of the equation and substitute it into the other equation, in this case equation B was substituted into equation A.

 $equ. \: a = (2x - 3y = 7) \\ equ. \: b = (x = 1 - y)$ 

Therefore since x = 1 - y, equation A would be:

$2(1 - y) - 3y = 7 \\ (2 - 2y) - 3y = 7 \\ - 2y - 3y = 7 - 2 \\ - 5y = 5 \\ y = - 1 \: and \: x = 2$

8 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