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3 years ago

A koala bear eats at a steady rate 15 leaves in 2 hours. In how many minutes does the koala eat 10 leaves?

Mathematics

1 answer:

ZanzabumX [31]3 years ago

8 0

Answer:

80 minutes

Step-by-step explanation:

1 hour=60 minutes

2 hours=120 minutes

-------------------------------

15/120=10/x

cross product

120*10=15*x

1200=15x

x=1200/15

x=80

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Connor has a collection of dimes and quarters with a total value of $6.30. The number of dimes is 14 more than the number of qua

Anarel [89]

Connor has 14 quarters and 28 dimes.

Step-by-step explanation:

Given,

Worth of coins = $6.30 = 6.30*100 = 630 cents

We know that,

1 quarter = 25 cents

1 dime = 10 cents

Let,

Quarter = x

Dime = y

According to given statement;

25x+10y=630 Eqn 1

The number of dimes is 14 more than the number of quarters

y = x+14 Eqn 2

Putting value of y from Eqn 2 in Eqn 1

$25x+10(x+14)=630\\25x+10x+140=630\\35x=630-140\\35x=490$

Dividing both sides by 35

$\frac{35x}{35}=\frac{490}{35}\\x=14$

Putting x=14 in Eqn 2

$y=14+14\\y=28$

Connor has 14 quarters and 28 dimes.

Keywords: linear equation, substitution method

Learn more about substitution method at:

#LearnwithBrainly

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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,

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

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

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

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

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Answer:

A) <

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Step-by-step explanation:

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