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3 years ago

Rewrite the function by completing the square 4x^2-16x +7

Mathematics

1 answer:

tia_tia [17]3 years ago

6 0

Answer:

$(2x-4)^2-9 => 4(x-2)^2-9$

Step-by-step explanation:

$=> 4x^2-16x+7\\ =>(2x)^2-2*(2x)*(4)+7\\ =>(2x)^2-2*(2x)*(4)+(4)^2-(4)^2+7\\ =>(2x-4)^2-9\\ =>4(x-2)^2-9$

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Separate the number 41 and the two parts so that the first number is eight more than twice the second number what are the two nu

vesna_86 [32]

The two numbers are 30 and 11

Solution:

Given that we have to separate the number 41 into two parts

Let the second number be "x"

Given that first number is eight more than twice the second number

first number = eight more than twice the second number

first number = 8 + twice the "x"

first number = 8 + 2x

So we can say first number added with second number ends up in 41

first number + second number = 41

8 + 2x + x = 41

8 + 3x = 41

3x = 41 - 8

3x = 33

x = 11

first number = 8 + 2x = 8 + 2(11) = 8 + 22 = 30

Thus the two numbers are 30 and 11

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3 years ago

#8 can anyone help? I need help solving this and showing the work

tamaranim1 [39]

Use distance formula
=√(3-3)^2 +(-4-4)^2
=√64
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3 years ago

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Erik buys 2.5 pounds of cashews.If each pound of cashews $7.70,how much will he pay for the cashews???I put Erik will pay $49250

Fittoniya [83]

The answer should be $ 19.25. All you have to do is $7.70 * 2.5 which = $19.25

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3 years ago

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Haddie flipped a fair coin 12 times. The coin landed on heads 11 times and tails 1 time. She flips the coin a 13th time. How lik

Firdavs [7]

Answer:

1 out of 2

Step-by-step explanation:

even though the coin was landing on heads more times than tails it still remains a fifty-fifty chance.

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2 years ago

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1. The position of a particle moving along a coordinate axis is given by: s(t) = t^2 - 5t + 1. a) Find the speed of the particle

zimovet [89]

Answer: $\left | 2t-5\right |,\ 2,\ 2t-5$

Step-by-step explanation:

Given

Position of the particle moving along the coordinate axis is given by

$s(t)=t^2-5t+1$

Speed of the particle is given by

$\Rightarrow v=\dfrac{ds}{dt}\\\\\Rightarrow v=\dfrac{d(t^2-5t+1)}{dt}\\\\\Rightarrow v=\left | 2t-5\right |$

Acceleration of the particle is

$\Rightarrow a=\dfrac{dv}{dt}\\\\\Rightarrow a=2$

velocity can be negative, but speed cannot

$\Rightarrow v=\dfrac{ds}{dt}\\\\\Rightarrow v=\dfrac{d(t^2-5t+1)}{dt}\\\\\Rightarrow v=2t-5$

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3 years ago