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bulgar [2K]
3 years ago
11

HELP?!? idk what im doing and nobody has asked this. will give brainliest answrr or whatever its called.

Mathematics
1 answer:
Illusion [34]3 years ago
6 0

9514 1404 393

Answer:

  \textbf{A.   }x=\dfrac{-4\pm2\sqrt{13}}{3}

Step-by-step explanation:

First of all, subtract 12 from both sides of the equation to put it into standard form:

  3x² +8x -12 = 0

Then identify the a, b, c of this form:

  ax² +bx +c = 0

You see that a=3, b=8, c=-12.

Now use those values in the quadratic formula for the solutions:

  x=\dfrac{-b\pm\sqrt{b^2-4ac}}{2a}\\\\x=\dfrac{-8\pm\sqrt{8^2-4(3)(-12)}}{2(3)}\\\\x=\dfrac{-8\pm\sqrt{208}}{6}=\dfrac{-8\pm4\sqrt{13}}{6}\\\\\boxed{x=\dfrac{-4\pm2\sqrt{13}}{3}}\qquad\text{factor 2 from numerator and denominator}

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The graph shows Ms. Padilla's monthly cell phone cost, where x is the number of minutes she uses the phone during the month.
balu736 [363]

Answer: the X intercept mean this here

Ms.

Padilla is charged $0.04/min.

Ms. Padilla pays $16 for months she makes 0 min of calls

Step-by-step explanation:

7 0
1 year ago
Solve x2 = 12x - 15 by completing the square. Which is the solution set of the equation?
Novosadov [1.4K]

Answer:

(6-√21,  6+√21)

Step-by-step explanation:

x^2 - 12x = -15

(x - 6)^2 - 36 = -15

(x - 6)^2 = 21

x - 6 = ± √21

x = 6  ± √21

8 0
3 years ago
Pre-Calc: Find all the zeros of the function.
uysha [10]

The zeros of the polynomial function are y = 4/5, y = -4/5 and y = ±4/5√i and the polynomial as a product of the linear factors is f(y) = (5y - 4)(5y + 4)(25y^2 + 16)

<h3>What are polynomial expressions?</h3>

Polynomial expressions are mathematical statements that are represented by variables, coefficients and operators

<h3>How to determine the zeros of the polynomial?</h3>

The polynomial equation is given as

f(y) = 625y^4 - 256

Express the terms as an exponent of 4

So, we have

f(y) = (5y)^4 - 4^4

Express the terms as an exponent of 2

So, we have

f(y) = (25y^2)^2 - 16^2

Apply the difference of two squares

So, we have

f(y) = (25y^2 - 16)(25y^2 + 16)

Apply the difference of two squares

So, we have

f(y) = (5y - 4)(5y + 4)(25y^2 + 16)

Set the equation to 0

So, we have

(5y - 4)(5y + 4)(25y^2 + 16) = 0

Expand the equation

So, we have

5y - 4 = 0, 5y + 4 = 0 and 25y^2 + 16 = 0

This gives

5y = 4, 5y = -4 and 25y^2 = -16

Solve the factors of the equation

So, we have

y = 4/5, y = -4/5 and y = ±4/5√i

Hence, the zeros of the polynomial function are y = 4/5, y = -4/5 and y = ±4/5√i

How to write the polynomial as a product of the linear factors?

In (a), we have

The polynomial equation is given as

f(y) = 625y^4 - 256

Express the terms as an exponent of 4

So, we have

f(y) = (5y)^4 - 4^4

Express the terms as an exponent of 2

So, we have

f(y) = (25y^2)^2 - 16^2

Apply the difference of two squares

So, we have

f(y) = (25y^2 - 16)(25y^2 + 16)

Apply the difference of two squares

So, we have

f(y) = (5y - 4)(5y + 4)(25y^2 + 16)

Hence, the polynomial as a product of the linear factors is f(y) = (5y - 4)(5y + 4)(25y^2 + 16)

Read more about polynomial at

brainly.com/question/17517586

#SPJ1

5 0
1 year ago
I need helpabdjbajdbajdb
gizmo_the_mogwai [7]

Answer:

c=10

Step-by-step explanation:

pythagreom theroem

c (hypotenuse) = \sqrt{a^{2} +b^{2} }

plug in values for a and b

c=\sqrt{6^{2} +8^{2} }

c=\sqrt{36+64\\}

c=\sqrt{100\\}

c=10

7 0
2 years ago
Read 2 more answers
Four friends go to an amusement arcade.
galben [10]

<u><em>Answer:</em></u>

<u>The rule is: </u>

T(x) = 8x + 20

<em><u>Explanation:</u></em>

<u>We are given that:</u>

<u>The rule for the amount that one friend pays is:</u>

C(x) = 2x + 5

<u>Now, we know that:</u>

Each of the four friends will pay the entry fee which is $5 per person

The 4 friends will play the same number of games represented by x

<u>This means that:</u>

We can simply get the rule for the total amount to be paid by the four friends (T(x)) by multiplying the amount paid by each friend by 4

<u>This means that:</u>

T(x) = 4 * C(x)

T(x) = 4(2x + 5)

T(x) = 8x + 20

Hope this helps :)

4 0
3 years ago
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