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yarga [219]
3 years ago
8

Can you solve for x pls

Mathematics
1 answer:
Ber [7]3 years ago
6 0

Answer:

x=11

Step-by-step explanation:

9x+6= 8X+17

X= 11

if my answer helps please mark as brainliest.

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Ali serves 24 customers an hour

Alyssa severs 60 customer an hour

Alyssa severs more customers in one hour

Step-by-step explanation:

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Determine if x + 2 is a factor of p(x) = x ^4 + 3x^ 3 + 4x ^2 - 8 and explain why.
zlopas [31]
X^4+ 3x^3+4x^2-8 |   x+2   =  x^3+x^2+2x-4
 -x^4-2x^3
-----------------------     
         x^3+4x^2-8
         -x^3-2x^2
-------------------------
                 2x^2-8
                 -2x^2-4x
                --------------
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                    4x+8
----------------------------
                      / /

is a factor because (x+2)(x^3+x^2+2x-4)=x^4+3x^3+4x^2-8
4 0
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Use the quadratic formula to solve each equation.
Elden [556K]

Answer:

1) x_1 = \frac{2 - 2\sqrt{13}}{2}= 1-\sqrt{13}

x_2 = \frac{2 + 2\sqrt{13}}{2}= 1+\sqrt{13}

2) x_1 = \frac{6 - 2\sqrt{10}}{1}= 6-2\sqrt{10}

x_2 = \frac{6 + 2\sqrt{10}}{1}= 6+2\sqrt{10}

3) p_1 = \frac{-8 - 2\sqrt{30}}{4}= -2-\frac{1}{2}\sqrt{30}

p_2 = \frac{-8 + 2\sqrt{30}}{4}= -2+\frac{1}{2}\sqrt{30}

4) y_1 = \frac{-3 - 9}{4}=-3

y_2 = \frac{-3 + 9}{4}= \frac{3}{2}

Step-by-step explanation:

The quadratic formula is given by:

x = \frac{-b \pm \sqrt{b^2 -4ac}}{2a}

We can use this formula in order to solve the following equations:

1. x^2 − 2x = 12 → a = 1, b = −2, c = −12

For this case if we apply the quadratic formula we got:

x = \frac{-(-2) \pm \sqrt{(-2)^2 -4(1)(-12)}}{2(1)}

x_1 = \frac{2 - 2\sqrt{13}}{2}= 1-\sqrt{13}

x_2 = \frac{2 + 2\sqrt{13}}{2}= 1+\sqrt{13}

2. 1/2x^2 − 6x = 2 → a = 1 / 2, b = −6, c = −2

For this case if we apply the quadratic formula we got:

x = \frac{-(-6) \pm \sqrt{(-6)^2 -4(1/2)(-2)}}{2(1/2)}

x_1 = \frac{6 - 2\sqrt{10}}{1}= 6-2\sqrt{10}

x_2 = \frac{6 + 2\sqrt{10}}{1}= 6+2\sqrt{10}

3. 2p^2 + 8p = 7 → a = 2, b = 8, c = −7

For this case if we apply the quadratic formula we got:

p = \frac{-(8) \pm \sqrt{(8)^2 -4(2)(-7)}}{2(2)}

p_1 = \frac{-8 - 2\sqrt{30}}{4}= -2-\frac{1}{2}\sqrt{30}

p_2 = \frac{-8 + 2\sqrt{30}}{4}= -2+\frac{1}{2}\sqrt{30}

4. 2y^2 + 3y − 5 = 4 → a = 2, b = 3, c = −9

For this case if we apply the quadratic formula we got:

y = \frac{-(3) \pm \sqrt{(3)^2 -4(2)(-9)}}{2(2)}

y_1 = \frac{-3 - 9}{4}=-3

y_2 = \frac{-3 + 9}{4}= \frac{3}{2}

8 0
4 years ago
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