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3 years ago

One hundred fifty is what percent of 200?

1 answer:

8 0

$Method\ 1:\\\\\dfrac{150}{200}\cdot100\%=\dfrac{15}{20}\cdot100\%=15\cdot5\%=75\%\\\\Method\ 2:\\\\200\ -\ 100\%\\150\ -\ p\%\\\\cross\ multiply\\\\200\cdot p=150\cdot100\\200p=15,000\ \ \ |divide\ both\ sides\ by\ 200\\\boxed{p=75\ (\%)}$

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Sketch the graph of y =sine3x

IgorLugansk [536]

The graph of $\mathbf{y=\sin{3x}}$ is given by,

Here given the function is $y=\sin{3x}$

when $x=0\Rightarrow y=\sin{3\times 0}=\sin0=0$ , then it passes through the origin (0,0).

When $x=\frac{\pi}{6}\Rightarrow y=\sin{3\times\frac{\pi}{6}}=\sin\frac{\pi}{2}=1$

Since we know that, $-1\leq\sin{x}\leq1,\forall x$ , then the function has maximum height at $x=\frac{\pi}{6}$ .

Again when $x=\frac{\pi}{3}\Rightarrow y=\sin{(3\times\frac{\pi}{3})}=\sin\pi=0$ , again the function is 0.

So clearly the function is the Oscillating function.

Using graphing calculator we get the graph of function $y=\sin3x$ ,

Learn more about Trigonometric Function here -

brainly.com/question/1143565

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8 0

2 years ago

Which product has 4 zeros after the digit 3?

sveticcg [70]

Answer:

0.03×104=3.12

0.3×104=31.2

0.3×10=3

0.03×109=3.27

these are the products

hope it helps!

4 0

2 years ago

A certain radioactive isotope has leaked into a small stream. Three hundred days after the leak, 3% of the original amount of t

sesenic [268]

If the half-life is t, then every t days, the amount of the radioactive isotope will be cut in half.
(1/2)^(number of half-lives) = 3%
number of half-lives = ln(0.03) / ln(0.5)
This gives the number of half-lives as 5.06.
Then 300 days = (5.06)(length of 1 half-life)
length of 1 half-life = 300 / 5.06 = 59.29 days

6 0

3 years ago

0.5a − 0.3 = 5 (solve for a)

Alla [95]

0.5a - 0.3 = 5

Add 0.3 to both sides:

0.5a = 5.3

Divide both sides by 0.5:

a = 10.6

3 0

3 years ago

Read 2 more answers

With reference to the figure, sin X equals?<br> 1.) 0.250<br> 2.)0.447<br> 3.)0.894<br> 4.)1

Flura [38]

Answer:

3.) 0.894

Step-by-step explanation:

✔️First, find BD using Pythagorean Theorem:

BD² = BC² - DC²

BC = 17.89

DC = 16

Plug in the values

BD² = 17.89² - 16²

BD² = 64.0521

BD = √64.0521

BD = 8.0 (nearest tenth)

✔️Next, find AD using the right triangle altitude theorem:

BD = √(AD*DC)

Plug in the values into the equation

8 = √(AD*16)

Square both sides

8² = AD*16

64 = AD*16

Divide both sides by 16

4 = AD

AD = 4

✔️Find AB using Pythagorean Theorem:

AB = √(BD² + AD²)

AB = √(8² + 4²)

AB = √(64 + 16)

AB = √(80)

AB = 8.9 (nearest tenth)

✔️Find sin x using trigonometric ratio formula:

Reference angle = x

Opposite side = BD = 8

Hypotenuse = AB = 8.944

Thus:

$sin(x) = \frac{opp}{hyp} = \frac{8}{8.944} = 0.894$ (nearest thousandth)

6 0

3 years ago