All categories

3 years ago

Gina solved the inequality below. 16≥2y−4 What is the solution to Gina’s inequality? A 8≥y B y≤40 C y≤10 D 6≥y

Mathematics

1 answer:

Vinvika [58]3 years ago

4 0

Answer: y ≤ 10

Step-by-step explanation:

16 ≥ 2y -4

Add 4 to both sides

16 + 4 ≥ 2y - 4 + 4

20 ≥ 2y

divide through by 2

20/2 ≥ 2y / 2

10 ≥ y

Therefore :

y ≤ 10

You might be interested in

Write the decimal as a percent: 0.89

Paha777 [63]

Answer:

0.0089

Step-by-step explanation:

Because when we want to change decimal as percents we need to replace the period on the left side 2 times. But when we want to change percent as a decimal, then we need to move the period 2 times on the right.

ex:

.72 decimal= 0.0072

4 0

2 years ago

Read 2 more answers

Find two numbers which have a sum of 132 and that can be written as the simplified<br> ratio 3:8

Molodets [167]

Answer:

96 and 36

Step-by-step explanation:

3+8 = 11

132/11 = 12

so:

12 x 3 = 36

12 x 8 = 96

5 0

2 years ago

Y=|4x|-1 what is the absolute value

maks197457 [2]

Answer:

-1

Step-by-step explanation:

5 0

3 years ago

The lengths of two sides of a triangle are 10 and 24, and the third side is x. How many whole number values are possible for x.

solong [7]

Answer:

$20$ .

Step-by-step explanation:

In any triangle, the sum of the lengths of any two sides should be strictly greater than the length of the third side. For example, if the length of the three sides are $a$ , $b$ , and $c$ :

$a + b > c$ ,

$a + c > b$ , and

$b + c > a$ .

In this question, the length of the sides are $10$ , $24$ , and $x$ . The length of these sides should satisfty the following inequalities:

$10 + 24 > x$ ,

$10 + x > 24$ , and

$24 + x > 10$ .

Since $x > 0$ , the inequality $24 + x > 10$ is guarenteed to be satisfied.

Simplify $10 + 24 > x$ to obtain the inequality $x < 35$ .

Similarly, simplify $10 + x > 24$ to obtain the inequality $x > 14$ .

Since $x$ needs to be a whole number, the greatest $x$ that satisfies $x < 35$ would be $34$ . Similarly, the least $x\!$ that satisfies $x > 14$ would be $15$ . Thus, $x\!\!$ could be any whole number between $15\!$ and $34\!$ (inclusive.)

There are a total of $34 - 15 + 1 = 20$ distinct whole numbers between $15$ and $34$ (inclusive.) Thus, the number of possible whole number values for $x$ would be $20$ .