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3 years ago

What is 258/300 as a percent

2 answers:

4 0

To transform it into percent we want to have 100 into denominator. The easiest way is divide numerator and denominator by 3 (thats because than it doesnt change the value of fraction), so:
$\frac{258}{300}= \frac{258:3}{300:3}= \frac{86}{100} = 86%$ - its the result

katrin2010 [14]3 years ago

3 0

$258\ \ |\ \ :2\\129\ \ |\ \ :3\\43\ \ \ \ |\ \ :43\\1\\\\258=6\cdot43\\\\300\ \ |\ \ :2\\150\ \ |\ \ :2\\75\ \ \ \ |\ \ :3\\25\ \ \ \ | \ \ : 5\\5\ \ \ \ \ |\ \ :5\\1\\\\300=6\cdot50\\\\ \frac{\big{258}}{\big{300}} =\frac{\big{6\cdot43}}{\big{6\cdot50}} =\frac{\big{43}}{\big{50}}$

$\\\\\frac{\big{43}}{\big{50}} =\frac{\big{43}}{\big{50}} \cdot 100\%=\frac{\big{43}}{\big{50}} \cdot 50\cdot2\%=43\cdot2\%=86\%$

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Passes through (1,12) and has a vertex (10,-4)

agasfer [191]

Answer: 16/81 (x-10)^2 -4

Step-by-step explanation:

To write a vertex equation with just a point and the vertex, you have to figure out the variables.

In vertex form, the equation is y = a (x-h)^2 + k

Your y is 12, x = 1, h = 10, and k = -4

Plug everything into equation

12 = a (1 - 10)^2 -4

12 = a (-9)^2 - 4

12 = 81a - 4

16 = 81a

16/81 = a

Now you know what the 'a' value is.

If you graph 16/81 (x-10)^2 -4 , you will get a point at (1,12) and a vertex of (10,-4)!

I hope this helps!

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3 years ago

45% as fraction in the simplest form

Kobotan [32]

Answer:9/20

Step-by-step explanation: 45/100=9/20, 45÷5=9, 100÷5=20 both the denominater and the numerator are divisible by 5.

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When two variables have an inverse relationship, the slope is?

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When two variables have an inverse relationship, the slope is negatively correlated.

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3 years ago

i f N(-7,-1) is a point on the terminal side of ∅ in standard form, find the exact values of the trigonometric functions of ∅.

Natali [406]

Answer:

$sin\ \varnothing = \frac{-1}{10}\sqrt 2$

$cos\ \varnothing = \frac{-7}{10}\sqrt 2$

$tan\ \varnothing = \frac{1}{7}$

$cot\ \varnothing = 7$

$sec\ \varnothing = \frac{-5}{7}\sqrt 2$

$csc\ \varnothing = -5\sqrt 2$

Step-by-step explanation:

Given

$N = (-7,-1)$ --- terminal side of $\varnothing$

Required

Determine the values of trigonometric functions of $\varnothing$ .

For $\varnothing$ , the trigonometry ratios are:

$sin\ \varnothing = \frac{y}{r}$ $cos\ \varnothing = \frac{x}{r}$ $tan\ \varnothing = \frac{y}{x}$

$cot\ \varnothing = \frac{x}{y}$ $sec\ \varnothing = \frac{r}{x}$ $csc\ \varnothing = \frac{r}{y}$

Where:

$r^2 = x^2 + y^2$

$r = \sqrt{x^2 + y^2$

In $N = (-7,-1)$

$x = -7$ and $y = -1$

So:

$r = \sqrt{(-7)^2 + (-1)^2$

$r = \sqrt{50$

$r = \sqrt{25 * 2$

$r = \sqrt{25} * \sqrt 2$

$r = 5 * \sqrt 2$

$r = 5 \sqrt 2$

<u>Solving the trigonometry functions</u>

$sin\ \varnothing = \frac{y}{r}$

$sin\ \varnothing = \frac{-1}{5\sqrt 2}$

Rationalize:

$sin\ \varnothing = \frac{-1}{5\sqrt 2} * \frac{\sqrt 2}{\sqrt 2}$

$sin\ \varnothing = \frac{-\sqrt 2}{5*2}$

$sin\ \varnothing = \frac{-\sqrt 2}{10}$

$sin\ \varnothing = \frac{-1}{10}\sqrt 2$

$cos\ \varnothing = \frac{x}{r}$

$cos\ \varnothing = \frac{-7}{5\sqrt 2}$

Rationalize

$cos\ \varnothing = \frac{-7}{5\sqrt 2} * \frac{\sqrt 2}{\sqrt 2}$

$cos\ \varnothing = \frac{-7*\sqrt 2}{5*2}$

$cos\ \varnothing = \frac{-7\sqrt 2}{10}$

$cos\ \varnothing = \frac{-7}{10}\sqrt 2$

$tan\ \varnothing = \frac{y}{x}$

$tan\ \varnothing = \frac{-1}{-7}$

$tan\ \varnothing = \frac{1}{7}$

$cot\ \varnothing = \frac{x}{y}$

$cot\ \varnothing = \frac{-7}{-1}$

$cot\ \varnothing = 7$

$sec\ \varnothing = \frac{r}{x}$

$sec\ \varnothing = \frac{5\sqrt 2}{-7}$

$sec\ \varnothing = \frac{-5}{7}\sqrt 2$

$csc\ \varnothing = \frac{r}{y}$

$csc\ \varnothing = \frac{5\sqrt 2}{-1}$

$csc\ \varnothing = -5\sqrt 2$

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