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SVETLANKA909090 [29]
3 years ago
6

Find the two positive numbers whose difference is 24 and whose product is 945

Mathematics
1 answer:
irina [24]3 years ago
5 0
X=45
Y=21
Take a look at the picture for the work

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Pork chops cost $1.39 per pound.
matrenka [14]

Answer: a) $4.17

b) $6.95

Step-by-step explanation:

3 * 1.39 = 4.17

5 * 1.39 = 6.95

5 0
3 years ago
Read 2 more answers
Which of the values satisfy the following inequality
Ira Lisetskai [31]

Answer:

  A:  x = 20

  C:  x = -9

Step-by-step explanation:

You can solve the inequality and compare that with the offered choices, or you can try the choices in the inequality to see if it is true. Either approach works, and they take about the same effort.

<u>Solving it</u>:

Unfold it ...

  -17 ≤ x -7.5 ≤ 17

Add 7.5 ...

  -9.5 ≤ x ≤ 24.5

The numbers 20 and -9 are in this range: answer choices A and C.

_____

<u>Trying the choices</u>:

A:  |20 -7.5| = 12.5 ≤ 17 . . . . this works

B:  |-10 -7.5| = 17.5 . . . doesn't work

C:  |-9 -7.5| = 16.5 ≤ 17 . . . . .this works

D:  |27-7.5| = 19.5 . . . doesn't work

The choices that work are answer choices A and C.

6 0
3 years ago
If u have 3/12 of an orange how many fourths do u have
sveta [45]
The answer would be 1/4 :)
5 0
4 years ago
Read 2 more answers
Explain how to write a quadratic equation given the following three points on the graph (5,31) (3,11) (0,11)
masha68 [24]

Given:

The graph of a quadratic function passes through the points (5,31) (3,11) (0,11).

To find:

The equation of the quadratic function.

Solution:

A quadratic function is defined as:

y=ax^2+bx+c            ...(i)

It is passes through the point (0,11). So, substitute x=0,y=11 in (i).

11=a(0)^2+b(0)+c

11=c

Putting c=11 in (i), we get

y=ax^2+bx+11               ...(ii)

The quadratic function passes through the point (5,31). So, substitute x=5,y=31 in (ii).

31=a(5)^2+b(5)+11

31-11=a(25)+5b

20=25a+5b

Divide both sides by 5.

4=5a+b                  ...(iii)

The quadratic function passes through the point (3,11). So, substitute x=3,y=11 in (ii).

11=a(3)^2+b(3)+11

11-11=a(9)+3b

0=9a+3b

Divide both sides by 3.

0=3a+b                 ...(iv)

Subtracting (iv) from (iii), we get

4-0=5a+b-3a-b

4=2a

\dfrac{4}{2}=a

2=a

Putting a=2 in (iv), we get

0=3(2)+b

0=6+b

-6=b

Putting a=2,b=-6 in (ii), we get

y=(2)x^2+(-6)x+11

y=2x^2-6x+11

Therefore, the required quadratic equation is y=2x^2-6x+11.

7 0
3 years ago
Make x the subject of the formulae: y=ax+b
Svet_ta [14]
I'm assuming this means solve for x in which case:
y=ax+b
y-b=ax
(y-b)/a = x
4 0
3 years ago
Read 2 more answers
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