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mamaluj [8]
1 year ago
7

For the triangles to be congruent by hl, what must be the value of x? 8 9 17 34

Mathematics
1 answer:
lora16 [44]1 year ago
8 0

Using equations we know that the value of x needs to be (E) 4 to make HL congruent to AC.

<h3>What are equations?</h3>

A mathematical equation is a formula that uses the equals sign to represent the equality of two expressions.

The point-slope form, standard form, and slope-intercept form are the three main types of linear equations.

So, HL is a hypotenuse that will be congruent to the hypotenuse AC.

We know that HL is 3x + 3.

Ac is 15.

Then the equation will be:

3x + 3 = 15

Now, solve the equation to get x as follows:

3x + 3 = 15

3x = 15 - 3

3x = 12

x = 12/3

x = 4

Therefore, the value of x needs to be (E) 4 to make HL congruent to AC.

Know more about equations here:

brainly.com/question/28937794

#SPJ4

Correct question:
For the triangles to be congruent by hl, what must be the value of x?

a. 8

b. 9

c. 17

d. 3

e. 4

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Whixh list of numbers is ordered from least to greatest?
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B is the solution

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-6.53 is the furthest from 0

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Please help me on this problem
Anna71 [15]

Answer:

The pairs of integer having two real solution forax^{2} -6x+c = 0 are

  1. a = -4, c = 5
  2. a = 1, c = 6
  3. a = 2, c = 3
  4. a = 3, c = 3

Step-by-step explanation:

Given

ax^{2} -6x+c = 0

Now we will solve the equation by putting all the 6 pairs so we get the  following

-3x^{2} -6x-5 = 0 for a = -3 , c=-5

-4x^{2} -6x+5 = 0 for a = -4 , c=5

1x^{2} -6x+6 = 0 for a = 1 , c=6

2x^{2} -6x+3 = 0 for a = 2 , c=3

3x^{2} -6x+3 = 0 for a = 3 , c=3

5x^{2} -6x+4 = 0 for a = 5 , c=4

The above  all are Quadratic equations inn general form ax^{2} +bx+c=0

where we have a,b and c constant values

So for a real Solution we must have

Disciminant , b^{2} -4\timesa\timesc \geq 0

for a = -3 , c=-5 we have

Discriminant =-24 which is less than 0 ∴ not a real solution.

for a = -4 , c=5 we have

Discriminant = 116 which is greater than 0 ∴ a real solution.

for a = 1 , c=6 we have

Discriminant =12 which is greater than 0 ∴ a real solution.

for a = 2 , c=3 we have

Discriminant =12 which is greater than 0 ∴ a real solution.

for a = 3 , c=3 we have

Discriminant =0 which is equal to 0 ∴ a real solution.

for a = 5 , c=4 we have

Discriminant =-44 which is less than 0 ∴ not a real solution.

7 0
3 years ago
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