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2 years ago

Order of operations

Mathematics

2 answers:

lorasvet [3.4K]2 years ago

7 0

PEMDAS

12*(3/4) comes first

8+9=17

zheka24 [161]2 years ago

3 0

You have to tines 12 and 3/4 then add 8

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I need help please. I need the the discount and selling price. Thank you

salantis [7]

Answer:

<h2><u>15.12 is the discount</u></h2><h2><u>68.88 is the sell price</u></h2>

Step-by-step explanation:

When you are taking off percentages from numbers, you first want to make the percent a number. This step is super simple. Just divide the 18 by 100

This gets you 0.18

Now multiply

84 * 0.18 = 15.12

So 15.12 is the discount

Then take 84 and subtract 15.12

84 - 15.12 = 68.88

So 68.88 is the sell price

5 0

2 years ago

I neeeddd helppppppppp

Marina CMI [18]

Answer:

Step-by-step explanation:

A and D are opposite.

A is in the negative side and d is in the positive side.

6 0

2 years ago

What values of c and d make the equation true?

saul85 [17]

Answer:

Third option.

Step-by-step explanation:

You need to cube both sides of the equation. Remember the Power of a power property:

$(a^m)^n=a^{mn$

$\sqrt[3]{162x^cy^5}=3x^2y(\sqrt[3]{6y^d})\\\\(\sqrt[3]{162x^cy^5})^3=(3x^2y(\sqrt[3]{6y^d}))^3\\\\162x^cy^5=27x^6y^36y^d$

According to the Product of powers property:

$(a^m)(a^n)=a^{(m+n)$

Then. simplifying you get:

$162x^cy^5=162x^6y^{(3+d)}$

Now you need to compare the exponents. You can observe that the exponent of "x" on the right side is 6, then the exponent of "x" on the left side must be 6. Therefore:

$c=6$

You can notice that the exponent of "y" on the left side is 5, then the exponent of "x" on the left side must be 5 too. Therefore "d" is:

$3+d=5\\d=5-3\\d=2$

4 0

3 years ago

If cos() = − 2 3 and is in Quadrant III, find tan() cot() + csc(). Incorrect: Your answer is incorrect.

nydimaria [60]

Answer:

$\tan(\theta) \cdot \cot(\theta) + \csc(\theta) = \frac{5 - 3\sqrt 5}{5}$

Step-by-step explanation:

Given

$\cos(\theta) = -\frac{2}{3}$

$\theta \to$ Quadrant III

Required

Determine $\tan(\theta) \cdot \cot(\theta) + \csc(\theta)$

We have:

$\cos(\theta) = -\frac{2}{3}$

We know that:

$\sin^2(\theta) + \cos^2(\theta) = 1$

This gives:

$\sin^2(\theta) + (-\frac{2}{3})^2 = 1$

$\sin^2(\theta) + (\frac{4}{9}) = 1$

Collect like terms

$\sin^2(\theta) = 1 - \frac{4}{9}$

Take LCM and solve

$\sin^2(\theta) = \frac{9 -4}{9}$

$\sin^2(\theta) = \frac{5}{9}$

Take the square roots of both sides

$\sin(\theta) = \±\frac{\sqrt 5}{3}$

Sin is negative in quadrant III. So:

$\sin(\theta) = -\frac{\sqrt 5}{3}$

Calculate $\csc(\theta)$

$\csc(\theta) = \frac{1}{\sin(\theta)}$

We have: $\sin(\theta) = -\frac{\sqrt 5}{3}$

So:

$\csc(\theta) = \frac{1}{-\frac{\sqrt 5}{3}}$

$\csc(\theta) = \frac{-3}{\sqrt 5}$

Rationalize

$\csc(\theta) = \frac{-3}{\sqrt 5}*\frac{\sqrt 5}{\sqrt 5}$

$\csc(\theta) = \frac{-3\sqrt 5}{5}$

So, we have:

$\tan(\theta) \cdot \cot(\theta) + \csc(\theta)$

$\tan(\theta) \cdot \cot(\theta) + \csc(\theta) = \tan(\theta) \cdot \frac{1}{\tan(\theta)} + \csc(\theta)$

$\tan(\theta) \cdot \cot(\theta) + \csc(\theta) = 1 + \csc(\theta)$

Substitute: $\csc(\theta) = \frac{-3\sqrt 5}{5}$

$\tan(\theta) \cdot \cot(\theta) + \csc(\theta) = 1 -\frac{3\sqrt 5}{5}$

Take LCM

$\tan(\theta) \cdot \cot(\theta) + \csc(\theta) = \frac{5 - 3\sqrt 5}{5}$

6 0

3 years ago

What is bigger 0.125 or 0.0125

tangare [24]

Answer:

0.125 is the answer...........

7 0

2 years ago

Read 2 more answers