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choli [55]
3 years ago
14

Choose the missing digit of these equivalent fractions.

Mathematics
2 answers:
jok3333 [9.3K]3 years ago
6 0

Answer:

where is the question

kindly attach it

Katarina [22]3 years ago
5 0

Answer:

4

Step-by-step explanation:

\frac{2}{5} \frac{x}{10}

Look at the denominators. 5 x 2 is 10. So, multiply the numerator (2) by 2 as well. Your answer is 4. This makes the numerator of the other fraction 4.

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Can someone help please?
natima [27]
Question 4 is C and im not sure about the other questions
8 0
3 years ago
Can someone please help? :((
natulia [17]

Just find the vertex and then compare with graph

#a

  • y=2x²-8
  • y=2(x-0)²-8

Parabola opening upwards

  • Vertex at (0,-8)

Graph 3

#2

  • y=(x+3)²+0

Vertex at (-3,0)

Graph IV

#3

  • y=-2(x-4)²+8

Parabola opening downwards as a is -ve

Graph I

#4

One graph is left

  • Graph Ii
8 0
2 years ago
Read 2 more answers
C. The quotient of 3 and a number x.​
Yakvenalex [24]

The answer you are looking for is 3÷x. Well, the better way to write this would be 3 over x, or 3/x, which is the fractional form. Since "quotient" modifies division, you would divide 3 by x.

So, therefore, the best final answer to this problem is 3/x.

Hope this helped answer your question for you. Good day to you.

4 0
3 years ago
Read 2 more answers
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
A population of rabbits doubles evey 4 months. if the population starts out with 8 rabbits how many rabbits will there be in 1 y
Airida [17]
If it doubles in every 4 months, then in 1 year it will double by 12/4 = 3 times.

So, the expression would be: n³
n = number of rabbit at starting point = 8

So, after 1 year, it would be = 8³ = 512

So, your final answer is 512

Hope this helps!
4 0
3 years ago
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