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goblinko [34]
3 years ago
5

HELPP PLEASE

Mathematics
2 answers:
Aleksandr [31]3 years ago
6 0
152.6315 is because If you dived 870 by 5.7 it will be that.
jok3333 [9.3K]3 years ago
4 0

Answer:

Answer is f(t)= 870(1.0582)t

5.82% increased per year

Step-by-step explanation:

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A cereal box says that it now contains 25% more cereal. If it originally had 12 ounces of cereal, how much does it have now?
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15

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Find the value(s) of "k" that will cause the equation 4x^2 +kx + 4 to have one real solution.
Andreas93 [3]

Answer:  k = {8, -8}

<u>Step-by-step explanation:</u>

In order for a quadratic equation to have exactly one solution, the discriminant must equal zero.       →        b² - 4ac = 0

  4x² + kx + 4 = 0

  ↓       ↓      ↓

a=4    b=k    c=4

b² - 4ac = 0

k² - 4(4)(4) = 0

k²             = 64

           k = √64

           k = ± 8

6 0
3 years ago
Johnny can read 10 pages of a book in 15 minutes.
Svetach [21]

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Its C

Step-by-step explanation:

Im really good with answers

5 0
3 years ago
Save me the headache
maxonik [38]

(9\sin2x+9\cos2x)^2=81

Taking the square root of both sides gives two possible cases,

9\sin2x+9\cos2x=9\implies\sin2x+\cos2x=1

or

9\sin2x+9\cos2x=-9\implies\sin2x+\cos2x=-1

Recall that

\sin(\alpha\pm\beta)=\sin\alpha\cos\beta\pm\cos\alpha\sin\beta

If \alpha=2x and \beta=\dfrac\pi4, we have

\sin\left(2x+\dfrac\pi4\right)=\dfrac{\sin2x+\cos2x}{\sqrt2}

so in the equations above, we can write

\sin2x+\cos2x=\sqrt2\sin\left(2x+\dfrac\pi4\right)=\pm1

Then in the first case,

\sqrt2\sin\left(2x+\dfrac\pi4\right)=1\implies\sin\left(2x+\dfrac\pi4\right)=\dfrac1{\sqrt2}

\implies2x+\dfrac\pi4=\dfrac\pi4+2n\pi\text{ or }\dfrac{3\pi}4+2n\pi

(where n is any integer)

\implies2x=2n\pi\text{ or }\dfrac\pi2+2n\pi

\implies x=n\pi\text{ or }\dfrac\pi4+n\pi

and in the second,

\sqrt2\sin\left(2x+\dfrac\pi4\right)=-1\implies\sin\left(2x+\dfrac\pi4\right)=-\dfrac1{\sqrt2}

\implies2x+\dfrac\pi4=-\dfrac\pi4+2n\pi\text{ or }-\dfrac{3\pi}4+2n\pi

\implies2x=-\dfrac\pi2+2n\pi\text{ or }-\pi+2n\pi

\implies x=-\dfrac\pi4+n\pi\text{ or }-\dfrac\pi2+n\pi

Then the solutions that fall in the interval [0,2\pi) are

x=0,\dfrac\pi4,\dfrac\pi2,\dfrac{3\pi}4,\pi,\dfrac{5\pi}4,\dfrac{3\pi}2,\dfrac{7\pi}4

5 0
3 years ago
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