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Tanya [424]
3 years ago
5

7/5 squared times blank equals 1

Mathematics
1 answer:
Pepsi [2]3 years ago
5 0

Answer: 5/7 squared


Step-by-step explanation:

To change 7/5 to 1, multiply by 5/7.

To change (7/5)^2 to 1, multiply by (5/7)^2.


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Ty, who weighs 120 lbs, sits 5 ft to the left side on a seesaw. In order to balance the seesaw, Charley has to sit 6 ft to the r
BabaBlast [244]

Ans: 144

Step-by-step explanation:

24 pounds per foot.

120+24=144

pls mark brainliest

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2 years ago
5x-3=11-2x. I gotta
ryzh [129]

Answer:

x = 2

Step-by-step explanation:

Add 2x from each side giving you

7x - 3 = 11

Add 3 to each side

7x = 14

Divide by 7

x = 2

6 0
2 years ago
Read 2 more answers
Work out, giving your answer in its simplest form:<br> 4/7 divided by 2 1/3
kirill [66]

Answer:

12/49  

Step-by-step explanation:

First write 2 1/3 as a improper fraction, to do that, first multiply the 2 by the denominator, then add it with the 1, which will result in 7/3

when doing division of fraction you have to remember this, KFG

KFG: Keep Flip Change

you keep the first number or fraction, which is 4/7

then you flip the sign, so you change division in to multiplication

after that, you change or write the reciprocal of the second number or fraction, so 7/3 will be 3/7

Now the equation is

4/7 times 3/7, which is 12/49  

4 0
2 years ago
WHOEVER ANSWERS FIRST ILL MARK BRAINLIEST!!!
maksim [4K]

Answer:

17

Step-by-step explanation:

6 0
3 years ago
Let n be any natural number greater than 1. Explain why the numbers n! 2, n! 3, n! 4, ..., n! n must all be composite. (This exe
bagirrra123 [75]

Answer:

Because each term of the sequence generates numbers with more than 1 and itself as dividers

Step-by-step explanation:

Just for the sake of correction.

1. Explain\: why\: the\: numbers\: n! +2, n!+ 3, n!+ 4, ..., n! \\n \:must\: all\: be\: composite.

1) Let's consider that

n! =n(n-1)(n-2)(n-3)...

2)And examine some numbers of that sequence above:

n!+2

Every Natural number plugged in n, and added by two will a be an even number not only divisible by two, but in some cases by other numbers for example,n=4, then 4!+2=26 which has four dividers.

3) Similarly, the same happens to

n!+3 and n!+4

Where we can find many dividers.

There's an example of a sequence, let's start with a prime number greater than 1

Let n=11

\left \{ n!+2,n!+3,n!+4,n!+5,n!+6,...n!+n. \right \}\\\left \{ 11!+2,11!+3,11!+4,11!+5,11!+6,11!+7,11!+8,...11!+11 \right \}\\\\

That's a long sequence of consecutive composite numbers, n=11.

\left \{39916802, 39916803,39916804,39916805,39916806,39916807,...,39916811,39916812 \right \}

4 0
3 years ago
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