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Irina-Kira [14]
3 years ago
12

Pleases help (: Simplify the expression below √256

Mathematics
1 answer:
Ad libitum [116K]3 years ago
4 0

\sqrt{256} = \sqrt{2^8} = 2^4 = 16

Answer: 16

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Four less than the product of 7 and a number is equal to 8 Use the variable b for unknown number
Akimi4 [234]

Answer:12/7

Step-by-step explanation:

7b-4=8

7b-4+4=8+4

7b=12

b=12/7

3 0
3 years ago
If α and β are the zeros of the quadratic polynomial f(x) = 3x2–4x + 5, find a polynomialwhose zeros are 2α + 3β and 3α + 2β.
Lelechka [254]

Answer:

\boxed{\sf \ \ \ 3x^2-20x+37\ \ \ }

Step-by-step explanation:

Hello,

a and b are the zeros, we can say that

f(x)=3(x^2-\dfrac{4}{3}x+\dfrac{5}{3}) = 3(x-a)(x-b)=3(x-(a+b)x+ab)

So we can say that

a+b=\dfrac{4}{3}\\ab=\dfrac{5}{3}

Now, we are looking for a polynomial where zeros are 2a+3b and 3a+2b

for instance we can write

(x-2a-3b)(x-3a-2b)=x^2-(2a+3b+3a+2b)x+(2a+3b)(3a+2b)\\= x^2-5(a+b)x+6a^2+6b^2+9ab+4ab

and we can notice that

a^2+b^2=(a+b)^2-2ab so

(x-2a-3b)(x-3a-2b)=x^2-5(a+b)x+6[(a+b)2-2ab]+13ab\\= x^2-5(a+b)x+6(a+b)^2+ab

it comes

x^2-5*\dfrac{4}{3}x+6(\dfrac{4}{3})^2+\dfrac{5}{3}

multiply by 3

3x^2-20x+2*16+5=3x^2-20x+37

4 0
3 years ago
Find angle v and solve for x
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Okayskdkdkfnngng Kent djd djd rn rn rndnd dnd dnd Bf dndm kd djd s djd djd djd rn djd rn kd djd didnjendnd e
6 0
3 years ago
Solve for . Round to the nearest tenth, if necessary.​
Len [333]

Answer:

x ≈ 8.7

Step-by-step explanation:

Using the sine ratio in the right triangle

sin75° = \frac{opposite}{hypotenuse} = \frac{RS}{QS} = \frac{x}{9} ( multiply both sides by 9 )

9 × sin75° = x , then

x ≈ 8.7 ( to the nearest tenth )

7 0
2 years ago
mr. bird leaves for a walk at 7:50 a.m. she allows 25 minutes later what time does she arrive to work​
Tatiana [17]

Answer: 8:15

Step-by-step explanation:

If she leaves at 7:50 and arrives to work 25 minutes later, then she will get there at 8:15.

Something easy you could use:

7:50 25

+ 10 -10

____ ___

8:00 15

+ 15 -15

____ ____

8:15 0

5 0
2 years ago
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