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2 years ago

Marcel made an error when he used the distributive property to write an equivalent

Mathematics

1 answer:

natima [27]2 years ago

5 0

Answer: Upper left corner

Explanation:

The error happens in step 2. Notice that x is <u>not</u> part of the second term 16, so there's no way to factor x from it unless you want a fraction to result.

The second step should be 8(4x-2)

Another way to factor is to say 16(2x-1)

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____& of 200 miles is 150 miles

juin [17]

Answer:

75%

Step-by-step explanation:

150/200 = .75

Convert to a percentage by multiplying by 100

.75 x 100 = 75%

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2 years ago

F(x)=2x+2 font the value of y when x=2

Marrrta [24]

I think the answer is f(2)=6 , but I'm not too sure. Hope this helped somehow!

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3 years ago

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Type a simplified fraction as an answer. PLEASE ANSWER I AM BEGGING!!!!!!

lana [24]

50/36 just keep dividing the numbers by two

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3 years ago

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Calculating percentages. no links

Alenkasestr [34]

Answer:

35%

Step-by-step explanation:

Its 7/20 so 0.35 and then times 100

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2 years ago

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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

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

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3 years ago