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3 years ago

91 is the product of Greg’s age 7 .use variable g to represent Greg’s age

Mathematics

2 answers:

svetoff [14.1K]3 years ago

8 0

I believe your equation will be: 7g = 91

Ilya [14]3 years ago

6 0

The other number is 13. 13 x 7 = 91.

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2.A cell phone company charges $50 for the phone plus a monthly service charge of $30. The

Mama L [17]

Answer:

o find total, we'll need to add all the costs! There's the cost for the phone which is paid once, monthly fee for all the months you use it.

total = 50 + (30 × months )

In this case, the question wants the equation in x and y, so we can plug those in:

y = 50 + 30x

Step-by-step explanation:

8 0

3 years ago

How do you find the x

lions [1.4K]

Because those are straight lines, the 46 degrees must be equal to the (2x - 4).
2x -4 = 46.
2x = 46 + 4
2x = 50
x = 50/2
x = 25.

7 0

3 years ago

Trace metals in drinking water affect the flavor and an unusually high concentration can pose a health hazard. We obtain the fol

Vitek1552 [10]

Answer:

Please see attachment

Step-by-step explanation:

Please see attachment

5 0

3 years ago

If I have 5 Orange Cats And 14 Black cats How many Cats Will there be left altogether and in colours if 2 Orange Cats get lost a

eimsori [14]

So, we will start by subtracting the orange cats

5-2=3 orange cats left

Then the black cats

14-6=8

Now we will add them back together:

8+3=11 Cats remaining

7 0

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:

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

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