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3 years ago

How do i solve step by step s+2 1/2s+20=48 ?

Mathematics

2 answers:

sammy [17]3 years ago

8 0

S+(2+1/2)s+20=48

2/2s+5/2s=48-20

7/2s=28

7s=28*2=2(20+8)=40+16

7s=56

s=56/7

s=8

Anuta_ua [19.1K]3 years ago

6 0

$s+2\frac{1}{2}s+20=48\\\\
3,5s+20=48\ \ \ \ | subtract\ 20\\\\
3,5s=28\ \ \ \ | divide\ by\ 3,5\\\\
s=8\\\\
Solution\ is\ s=8.$

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Mademuasel [1]

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Step-by-step explanation:

Using the rule of exponents

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What are the real and complex solutions of the polynomial equation? x^4-41x^2=-400. *work needed*

pickupchik [31]

<h3>Answer: x = -5, -4, 4 and 5.</h3><h3>All four zeros are real solutions.</h3>

Step-by-step explanation:

Given the polynomial equation $x^4-41x^2=-400$ .

Adding 400 on both sides to get rid 400 from right side and set 0 on right side, we get

$x^4-41x^2+400=-400+400$ .

$x^4-41x^2+400=0$ .

Factoring by product sum rule.

We need product of 400 and sum upto -41.

We can see that 400 = -25 × -16 = 400 and -25-16 = -41.

Therefore,

$x^4-25x^2-16x^2+400=0$

Making it into two groups, we get

$(x^4-25x^2)+(-16x^2+400)=0$

Factoring out GCF of each group, we get

$x^2(x^2-25)-16(x^2-25)=0$

$(x^2-25)(x^2-16) =0$

Factoring out $(x^2-25)$ and $(x^2-16)$ separately by difference of the squares identity $a^2-b^2=(a-b)(a+b)$ , we get

$(x^2-25) = x^2-5^2= (x-5)(x+5)$ and

$x^2-16 = x^2-4^2 =(x-4)(x+4)$ .

Therefore,

$(x-5)(x+5)(x-4)(x+4) =0$

Applying zero product rule,

x-5=0

x+5=0

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x+4=0.

Therefore,

<h3>x = -5, -4, 4 and 5.</h3><h3>All four zeros are real solutions.</h3>

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3 years ago