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gayaneshka [121]
3 years ago
7

Which twos ratios represent quantities that are proportional?

Mathematics
1 answer:
Inessa05 [86]3 years ago
8 0

Answer: it’s 10/15 and 14/21 ;)

Step-by-step explanation:

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Divide:(10x²-3x+4)÷(2x-5)
Helen [10]

Answer:

\frac{10x^2-3x+4}{2x-5}=5x+11+\frac{59}{2x-5}

Quotient: 5x+11

Remainder: 59

Step-by-step explanation:

I'm going to do long division.

The bottom goes on the outside and the top goes in the inside.

  Setup:

        ---------------------------------

2x-5 |  10x^2    -3x      +4

  Starting the problem from the setup:

            5x     +11                        (I put 5x on top because 5x(2x)=10x^2)

        ---------------------------------   (We are going to distribute 5x to the divisor)

2x-5 |  10x^2    -3x      +4

        -(10x^2  -25x)                  (We are now going to subtract to see what's left.)

      -----------------------------------

                      22x      +4          (I know 2x goes into 22x, 11 times.)

                                                ( I have put +11 on top as a result.)

                    -(22x     -55)        (I distribute 11 to the divisor.)

                 -----------------------

                                  59          (We are done since the divisor is higher degree.)

The quotient is 5x+11.

The remainder is 59.

The result of the division is equal to:

5x+11+\frac{59}{2x-5}.

We can actually use synthetic division as well since the denominator is linear.

Let's solve 2x-5=0 to find what to put on the outside of the synthetic division setup:

2x-5=0

Add 5 on both:

2x=5

Divide both sides by 2:

x=5/2

Or realize that 2x-5 is the same as 2(x-(5/2)) which you will have to do anyways if you choose this route:

So 5/2 will go on the outside:

5/2  |    10          -3            4

      |                  25         55

         ------------------------------

           10          22        59

So we have:

\frac{10x^2-3x+4}{2x-5}

=\frac{10x^2-3x+4}{2(x-\frac{5}{2})}=\frac{1}{2} \cdot \frac{10x^2-3x+4}{x-\frac{5}{2}}=\frac{1}{2}(10x+22+\frac{59}{x-\frac{5}{2}})

Distribute the 1/2 back:

\frac{10x^2-3x+4}{2x-5}=\frac{10x+22}{2}+\frac{59}{2(x-\frac{5}{2})}

\frac{10x^2-3x+4}{2x-5}=5x+11+\frac{59}{2x-5}

3 0
4 years ago
Helppp!!<br><br> Identify the slope of the following equation:<br> -4x + 3y = 3
AlexFokin [52]

Answer:

the answer is -4

Step-by-step explanation:

7 0
3 years ago
Read 2 more answers
PLEASE HELP ITS FOR A MAJOR TEST
GuDViN [60]

Answer:

57

Step-by-step explanation:

the sum of those two interior angles is equal to the exterior (116), so y = 15 and the measure of angle A is 57 if you plug it in

6 0
3 years ago
Read 2 more answers
Which measures are accurate regarding triangle JKL? Check all that apply.
mixas84 [53]

Answer:

Option (1), (3) and (6) is correct.

m∠K = 84° , k ≈ 3.7 units and KL ≈ 3.2 units.

Step-by-step explanation:

  Given : ∠J = 58° , ∠L = 38°  and length of side JK = 2.3 units.

We need to check all the options and choose that follows.

First we find the measure of ∠K

Using angle sum property , Sum of angles of a triangle  is 180°

⇒ ∠J + ∠k + ∠L = 180°

⇒ 58° + ∠k + 38° = 180°

⇒  ∠k + 96° = 180°

⇒  ∠k   = 180° - 96°

⇒  ∠k   = 84°

Also using sine rule on ΔJKL , we get,

\frac{KL}{\sin J}=\frac{JL}{\sin K}=\frac{JK}{\sin L}

Substitute the values, we get,

\frac{KL}{\sin 58^{\circ}}=\frac{k}{\sin 84^{\circ}}=\frac{2.3}{\sin 38^{\circ}}

Consider the last two ratios, we have,

\frac{k}{\sin 84^{\circ}}=\frac{2.3}{\sin 38^{\circ}}

{k}=\frac{2.3}{\sin 38^{\circ}}\times {\sin 84^{\circ}}

On solving we get,

{k}=3.71

Also, now consider the first and last ratio, we get,

\frac{KL}{\sin 58^{\circ}}=\frac{2.3}{\sin 38^{\circ}}

{KL}=\frac{2.3}{\sin 38^{\circ}}\times {\sin 58^{\circ}

{KL}=3.16

Thus, k ≈ 3.7 units and KL ≈ 3.2 units.

Option (1), (3) and (6) is correct.

5 0
4 years ago
Read 2 more answers
HELP!!!
Bogdan [553]
The answer is One solution because the two lines are crossing each other and has one point in common
4 0
3 years ago
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