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Marina CMI [18]

3 years ago

6

Need help on 7 simplest form

2 answers:

Olegator [25]3 years ago

6 0

Hopefully the picture is clear enough but I got 1 and 1/20

Ksju [112]3 years ago

4 0

I think it's 1 and 1/20

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Need to know these two problems

Arturiano [62]

Multiple 16 by 1 and then divide by 8 which makes 2.

Multiply 36 by 1 and then divide by 9 which makes 4.

5 0

3 years ago

Read 2 more answers

Find an equivalent expression for each expression by simplifying. (3x - 5y) + 7y

sertanlavr [38]

This question is quite simple really. They are basically asking you to combine like terms. You have two different figures with the variable y, so you just combine the them. It would then be 3x + 2y. You cannot simplify this equation any further since there is nothing to solve for or anymore like terms to combine.

4 0

2 years ago

I need help to these questions. Convert these fractions to a decimal 7/8 , 4 3/8. Thanks

ehidna [41]

Simple...

you have: $\frac{7}{8}$ and $4 \frac{3}{8}$

Now, the first one is easy just divide it...

$\frac{7}{8} =0.875$

Now....

$4 \frac{3}{8}$ needs to be converted...

Multiply the whole number*denominator...

4*8=32

Now add the numerator....

32+3=35

Now use the original denominator...

$\frac{35}{8}$

Now divide...

$\frac{35}{8}=4.375
$

Thus, your answer.

7 0

2 years ago

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Hey guys plz help with this one i beg

kvv77 [185]

Answer:

102

Step-by-step explanation:

i think it is the ans.

6 0

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

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

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

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

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

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

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

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

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

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

$cot\ \varnothing = 7$

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

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

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

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

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

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

3 0

2 years ago