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Yuri [45]
3 years ago
14

I need help! If you could explain that would help alot

Mathematics
1 answer:
Shkiper50 [21]3 years ago
8 0

Answer:

x = 3. (Refresh if you cannot see the explanation.)

Step-by-step explanation:

\frac{4x+6}{3} = \frac{x+9}{2}\\\text{Cross multiply.}\\2(4x+6)=3(x+9)\\\text{Distribute.}\\8x+12=3x+27\\\text{Subtract 3x from both sides.}\\5x+12=27\\\text{Subtract 12 from both sides.}\\5x=15\\\text{Divide both sides by 15.}\\x=3\\\\\text{Proof:}\\\\\frac{4x+6}{3}=\frac{x+9}{2}\\\text{Substitute variable.}\\\frac{4(3)+6}{3}=\frac{3+9}{2}\\\text{Simplify numerators.}\\\frac{18}{3}=\frac{12}{2}\\\text{Divide both sides.}\\6=6

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What are the slopes of the asymptotes of a hyperbola with equation (y-2)^2/1^2 - (x-4)^2/3^2 = 1?
vladimir1956 [14]

Answer: 1/3 and -1/3

Step-by-step explanation:

8 0
3 years ago
Which equation involves a prime quadratic and cannot be solved by factoring? A. x2 + 5x − 4 = 0 B. x2 − x − 6 = 0 C. x2 + 3x − 4
Damm [24]

Answer:

Option A

Step-by-step explanation:

We can try to factorize all the options one by one.

For A:

x^{2} +5x-4\\=> x^{2}+4x+x-4\\=>x(x+4)+1(x-4)

We can see that the quadratic expression cannot be solved by factorization as the factors at the end of factorization are not equal in both brackets. So Option A is the correct answer for the given question.

Moreover we can also note that all the other quadratic expressions can be factorized ..

8 0
3 years ago
Need help ASAP! Anyone know?
VARVARA [1.3K]

Answer: x= 29°

Step-by-step explanation: Add angles D and C. Subtract from 180.

58/2 = 29

6 0
3 years ago
Meredith solved a quadratic equation. Her work is shown below.
Damm [24]

Answer:

Step 2

Step-by-step explanation:

Meredith made a fatal mistake or error in step 2;

Any quadratic problem will always have two solutions. Therefore since, we only have one solution here, it is wrong. The mistake was incurred in step 2;

  From the given problem ;

    <u> Wrong                                                  right</u>

    2(x + 4)² = 242

    (x + 4)² = 121     Step 1  

     x+4=11              Step 2                  x + 4  = ± 11

     x  = 7                Step 3                  x + 4  = 11 or x + 4  = -11

                                                           x  = 11- 4     or    x  = -11 - 4

                                                           x  = 7 or -15

3 0
3 years ago
A quadrilateral has vertices at A(-5,5), B(1,8), C(4,2), and D(-2,-2). Use slope to determine if the quadrilateral is a rectangl
AysviL [449]

Answer:

The quadrilateral is not a rectangle

Step-by-step explanation:

we know that

If a quadrilateral ABCD is a rectangle

then

Opposite sides are congruent and parallel and adjacent sides are perpendicular

Remember that

If two lines are parallel, then their slopes are the same

If two lines are perpendicular, then their slopes are opposite reciprocal (the product of their slopes is equal to -1)

The formula to calculate the slope between two points is equal to

m=\frac{y2-y1}{x2-x1}

we have

A(-5,5), B(1,8), C(4,2), and D(-2,-2)

Plot the figure to better understand the problem

see the attached figure

Find the slope of the four sides and then compare

step 1

Find slope AB

A(-5,5), B(1,8)

substitute in the formula

m=\frac{8-5}{1+5}

m=\frac{3}{6}

m_A_B=\frac{1}{2}

step 2

Find slope BC

B(1,8), C(4,2)

substitute in the formula

m=\frac{2-8}{4-1}

m=\frac{-6}{3}

m_B_C=-2

step 3

Find slope CD

C(4,2), and D(-2,-2)

substitute in the formula

m=\frac{-2-2}{-2-4}

m=\frac{-4}{-6}

m_C_D=\frac{2}{3}

step 4

Find slope AD

A(-5,5), D(-2,-2)

substitute in the formula

m=\frac{-2-5}{-2+5}

m=\frac{-7}{3}

m_A_D=-\frac{7}{3}

step 5

Verify if the opposites are parallel

Remember that

If two lines are parallel, then their slopes are the same

The opposite sides are

AB and CD

BC and AD

we have

m_A_B=\frac{1}{2}

m_C_D=\frac{2}{3}

so

m_A_B \neq m_C_D

It is not necessary to continue verifying, because two of the opposite sides are not parallel

therefore

The quadrilateral is not a rectangle

4 0
3 years ago
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