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nikdorinn [45]
2 years ago
11

I need help to this it would be awesome if you could help.

Mathematics
1 answer:
mrs_skeptik [129]2 years ago
3 0

In the given diagram, the value of JI is 6

<h3>Solving equations </h3>

From the question, we are to determine JI

From the given diagram, we can write that

JI + IH = x + 22

Also,

JI = 2x + 28

and

IH = 5

Then, we can write that

2x + 28 + 5 = x + 22

Collect like terms

2x - x = 22 -28 -5

x = -11

Now, substitute the value of x into the equation

JI = 2x + 28

JI = 2(-11) + 28

JI = -22 +28

JI = 6

Hence, the value of JI is 6

Learn more on Solving equations here: brainly.com/question/3926291

#SPJ1

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11r+7s what is the answer please help!
Tasya [4]
That is the answer. you can't simplify it.
7 0
4 years ago
Loan A (bank) requires a total payback of $54,289.44 while Loan B (dealership) requires a total payback of $55,098.60.
dedylja [7]

Answer: Loan B requires more interest by $809.16.

Step-by-step explanation:

Subtract the total pay back of loan B from loan A

= $55,098.60 - $54,289.44

= $809.16

I hope this helps, please mark as brainliest answer.

8 0
3 years ago
Determine the solution type for each equation if one solution determine the value for x
sdas [7]

Step-by-step explanation:

Vamos resolver sua equação passo a passo.

10-7c=10+2(3-5c)

Etapa 1: simplifique os dois lados da equação.

10-7c=10+2(3-5c)

10+-7c=10+(2)(3)+(2)(-5c)(Distribuir)

10+-7c=10+6+-10c

-7c+10=(-10c)+(10+6)(Combine os termos semelhantes)

-7c+10=-10c+16

-7c+10=-10c+16

Etapa 2: adicione 10c a ambos os lados.

-7c+10+10c=-10c+16+10c

3c+10=16

Etapa 3: subtraia 10 de ambos os lados.

3c+10-10=16-10

3c=6

Etapa 4: divida os dois lados por três.

3c

3

=

6

3

c=2

4 0
3 years ago
Type the correct answer in each box. If necessary, use / for the fraction bar(s).
Eduardwww [97]

Answer:

x=5/6 and y=8/5

(5/6 , 8/5)

Step-by-step explanation:

You can use the elimination method! The point of this method is to add or subtract one equation from the other, multiplying them by constants if necessary, to cancel out one variable so you can solve for the other. Then, you can plug the value you got for your solution into either equation to solve for the other. I'll demonstrate.

Look at the two equations and the coefficients of both variables. You have 12x and -6x -- 6*2=12, so this is perfect. (You could also eliminate the y because 5*3=15, but I'll show you by eliminating the x instead.)

Here's what that looks like:

(-6x+1y=3)2

-12x+10y=6

So we just multiplied the second equation by 2 on both sides. Let's see how that helps us.

12x+15y=34

-12x+10y=6

If we add the two equations now, x will be canceled out and we can solve for y.

   12x+15y=34

+(-12x+10y=6)

___________

     0x+25y=40

           25y=40

0x=0, so we can get rid of the x. Now, we need to solve for y.

25y=40

y=40/25

You probably know how to simplify fractions, so divide both the numerator and the denominator by 5 to get y=8/5.

Now you can use this value in either equation and solve for x. I'll use the first. (This is called substitution.)

12x+15(8/5)=34

12x+3(8)=34 <-- What I did here is cancel out the 5 in the denominator with 15 to leave 3, because 5*3=15.

12x+24=34

12x=10

x=10/12

x=5/6 (Divide the numerator and denominator by 2.)

You can write your answer as a point, too. (5/6 , 8/5)

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Simora [160]

\bf ~~~~~~\textit{vertical parabola vertex form} \\\\ y=a(x- h)^2+ k\qquad \begin{cases} \stackrel{vertex}{(h,k)}\\\\ \stackrel{"a"~is~negative}{op ens~\cap}\qquad \stackrel{"a"~is~positive}{op ens~\cup} \end{cases} \\\\[-0.35em] ~\dotfill\\\\ y=-2(x-\stackrel{h}{3})^2+\stackrel{k}{4}\qquad\qquad \stackrel{vertex}{(\underline{3},4)}\qquad \qquad \stackrel{\textit{axis of symmetry}}{x=\underline{3}}

6 0
3 years ago
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