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Phantasy [73]
3 years ago
10

Without using a calculator, fill in the blanks with two consecutive integers to complete the following inequality. ____ < squ

are root of 181 < _____
Mathematics
1 answer:
Natasha2012 [34]3 years ago
3 0

Answer:

13.

Step-by-step explanation:

We have been given an inequality ______. We are asked to fill in the blank with two consecutive integers to complete the given inequality.

We know that square root of 181 is not an integer. We can see that 181 is greater than square root 169 and and square root of 196.

\sqrt{169}

We know that \sqrt{169}=13 and \sqrt{196}=14.

Therefore, our required inequality would be 13.

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3 years ago
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Which ordered pair is a solution to the system of equations?<br> (2x - y=-2<br> (5x – 2y = 7
mrs_skeptik [129]

Answer:

21

Step-by-step explanation:

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3 years ago
Help help help help help
SIZIF [17.4K]

Answer:

1) 2b + 6

2) 8w - 12

3) 2b + 6 + 6x + 20

4) 9w^2 + 27 + 8 - 27

5) Sorry.

Step-by-step explanation:

8 0
3 years ago
1. Find four consecutive even integers such that the sum of the first and third is 6 less than the largest
Sophie [7]

These are indeed quite a lot of exercises, but a lot of them are almost identical - only small computations will change. So, I'm glad to help you with one exercise from every cathegory, but I encourage you to solve the others on your own.

<h2>Exercise 1</h2>

You can call four consecutive integers as

x,\ x+1,\ x+2,\ x+3

So, the sum of the first and third is x+(x+2) = 2x+2

We want this quantity to be six less than the largest, i.e. (x+3)-6 = x-3

So, the equality is

2x+2 = x-3

Subtract x from both sides:

x+2 = -3

Subtract 2 from both sides:

x = -5

So, the consecutive integers are

-5,\ -4,\ -3,\ -2

In fact, the sum of the first and third is -5-3 = -8, which is indeed six less than the largest: -8 = -2-6

<h2>Exercise 2</h2>

If you call the first odd number x, the next consecutive odd numbers will be x,\ x+2,\ x+4,\ x+6

In fact, we have to count skipping two's, because we only want odd integers. From here, you go on like exercise 1: you write the largest (which is x+6), and set it to be two more than the sum of the other three (x, x+2 and x+4)

<h2>Exercise 3</h2>

By the same logic of exercise 2, two consecutive even integers are x,\ x+2, assuming that x is even.

So, you set the equation as usual: the smaller (which is x) is 26 less than three times the larger (which means 3(x+2)-26)

<h2>Exercise 4 to 8</h2>

These are all pretty identical to exercise 1: you start by listing three or four consecutive integers:

x,\ x+1,\ x+2\quad\text{or}\quad  x,\ x+1,\ x+2\ x+3

and then you translate the request of each exercise accordingly. Remember that expressions like "three times the second number" means that you have to multiply: 3(x+1), while expression like "six more than the first" or "thirteen less than the first" imply adding/subtracting: x+6 or x-13.

<h2>Exercise 9</h2>

A multiple of 5 can be written as 5k, for some integer k.

So, three consecutive multiples of 5 are

5k, 5(k+1), 5(k+2) = 5k, 5k+5, 5k+10

We want these three numbers to have a sum of 75. So, we have

5k, 5k+5, 5k+10 = 75 \iff 15k+15 = 75 \iff 15k = 60 \iff k = 4

So, the three numbers are

5k, 5(k+1), 5(k+2) = 5\cdot 4, 5\cdot 5, 5\cdot 6 = 20, 25, 30

3 0
3 years ago
Simplify.<br> <img src="https://tex.z-dn.net/?f=%5Cfrac%7B8%5Csqrt%7B6mn%7D%20%2B%206%5Csqrt%7B8mn%7D%20%7D%7B2%5Csqrt%7B2mn%7D%
nata0808 [166]
It’s the first answer
I used Math-way :)

3 0
3 years ago
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