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Yakvenalex [24]
3 years ago
11

Please please help marking brainlist but if you dont know than dont answer please thank you

Mathematics
1 answer:
zloy xaker [14]3 years ago
5 0

Answer:

1. z = 5,  2. w = 9

Step-by-step explanation:

1. 5z - 4 = 21

5z = 21 + 4

5z = 25

z = 25/5

z = 5

2. w/9 + 2 = 3

w/9 = 3 - 2

w/9 = 1

w = 1 x 9

w = 9

I hope this helps :)

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If you tell me the concepts you’re referring to I might be able to help
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Determine the rational zeros for the function f(x)=x^3+7x^2+7x-15
alexandr402 [8]

Answer:

1,-3,-5

Step-by-step explanation:

Given:

f(x)=x^3+7x^2+7x-15

Finding all the possible rational zeros of f(x)

p= ±1,±3,±5,±15 (factors of coefficient of last term)

q=±1(factors of coefficient of leading term)

p/q=±1,±3,±5,±15

Now finding the rational zeros using rational root theorem

f(p/q)

f(1)=1+7+7-15

   =0

f(-1)= -1 +7-7-15

     = -16

f(3)=27+7(9)+21-15

    =96

f(-3)= (-3)^3+7(-3)^2+7(-3)-15

     = 0

f(5)=5^3+7(5)^2+7(5)-15

    =320

f(-5)=(-5)^3+7(-5)^2+7(-5)-15

      =0    

f(15)=(15)^3+7(15)^2+7(15)-15

     =5040

f(-15)=(-15)^3+7(-15)^2+7(-15)-15

      =-1920

Hence the rational roots are 1,-3,-5 !

4 0
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Step-by-step explanation:

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3 years ago
Solve 2x^2 + x - 4 = 0 <br> X2 +
damaskus [11]

Answer:

\large \boxed{\sf \ \ x = -\dfrac{\sqrt{33}+1}{4} \ \ or \ \ x = \dfrac{\sqrt{33}-1}{4} \ \ }

Step-by-step explanation:

Hello, please find below my work.

2x^2+x-4=0\\\\\text{*** divide by 2 both sides ***}\\\\x^2+\dfrac{1}{2}x-2=0\\\\\text{*** complete the square ***}\\\\x^2+\dfrac{1}{2}x-2=(x+\dfrac{1}{4})^2-\dfrac{1^2}{4^2}-2=0\\\\\text{*** simplify ***}\\\\(x+\dfrac{1}{4})^2-\dfrac{1+16*2}{16}=(x+\dfrac{1}{4})^2-\dfrac{33}{16}=0

\text{*** add } \dfrac{33}{16} \text{ to both sides ***}\\\\(x+\dfrac{1}{4})^2=\dfrac{33}{16}\\\\\text{**** take the root ***}\\\\x+\dfrac{1}{4}=\pm \dfrac{\sqrt{33}}{4}\\\\\text{*** subtract } \dfrac{1}{4} \text{ from both sides ***}\\\\x = -\dfrac{1}{4} -\dfrac{\sqrt{33}}{4} \ \ or \ \ x = -\dfrac{1}{4} +\dfrac{\sqrt{33}}{4}

Hope this helps.

Do not hesitate if you need further explanation.

Thank you

4 0
3 years ago
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