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3 years ago

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Mathematics

1 answer:

Anastasy [175]3 years ago

6 0

Y = 12x where x is the # of students and y is the # of adults

Or for every 12 students there's 1 adult

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2x + 2y - x - (8 + 1)?

kotegsom [21]

Answer:

x + 2 y − 9

Step-by-step explanation:

4 0

2 years ago

Read 2 more answers

Two numbers have an absolute value of 16 . Which of the two numbers is greater than 12. Can someone plz help me answer this ques

KATRIN_1 [288]

Answer:

Step-by-step explanation:

Absolute value means the modulus of a number.

-16 and 16 give an absolute value of 16

16 > 12

6 0

3 years ago

Given rectangle QUAD, If DU = 5x - 4 and QP = 2x + 7, then find the length of DU.

ira [324]

Answer:

DU = 86

Step-by-step explanation:

Properties of a rectangle,

- Opposite sides are equal in measure.

- Diagonals of a rectangle are equal.

- Diagonals bisect each other.

In the given rectangle QUAD,

Diagonals AQ and DU bisect each other at P.

AQ = DU

2(QP) = DU [Since, AQ = 2(PQ)]

2(2x + 7) = 5x - 4

4x + 14 = 5x - 4

5x - 4x = 14 + 4

x = 18

Since, DU = 5x - 4

DU = 5(18) - 4

= 90 - 4

= 86

Therefore, measure of DU = 86 units

3 0

2 years ago

PLEASE HELP WILL GIVE BRAINLIEST AND 20 POINTS

Georgia [21]

1250 lbs of sand.

So first let's find the volume of the sand in the sandbox. You have a square sandbox, with a side length of 5, meaning you have a length and width of 5. You also have half a foot of sand in the sandbox, meaning that you have a height of 0.5. To find the volume of the sand, multiply the length by the width by the height:

$5 * 5 * .5 = 12.5$

So you have 12.5 cubic feet of sand. If each cubic foot of sand is 100 lbs, then you just multiply the two:

$12.5 * 100 = 1250.0

$

6 0

3 years ago

Help please!! How do I solve this

egoroff_w [7]

This is an exponential equation. We will solve in the following way. I do not have special symbols, functions and factors, so I work in this way
2 on (2x) - 5 2 on x + 4=0 =>. (2 on x)2 - 5 2 on x + 4=0 We will replace expression ( 2 on x) with variable t => 2 on x=t =. t2-5t+4=0 => This is quadratic equation and I solve this in the folowing way => t2-4t-t+4=0 => t(t-4) - (t-4)=0 => (t-4) (t-1)=0 => we conclude t-4=0 or t-1=0 => t'=4 and t"=1 now we will return t' => 2 on x' = 4 => 2 on x' = 2 on 2 => x'=2 we do the same with t" => 2 on x" = 1 => 2 on x' = 2 on 0 => x" = 0 ( we know that every number on 0 gives 1). Check 1: 2 on (2*2)-5*2 on 2 +4=0 => 2 on 4 - 5 * 4+4=0 => 16-20+4=0 =. 0=0 Identity proving solution.
Check 2: 2 on (2*0) - 5* 2 on 0 + 4=0 => 2 on 0 - 5 * 1 + 4=0 =>
1-5+4=0 => 0=0 Identity provin solution.

5 0

3 years ago