Ask question

Ask question

All categories

2 years ago

14

What is 15% from $850

2 answers:

Sergio039 [100]2 years ago

5 0

Multiply 850x.15 which equals 127.5 and then subtract that from 850 and you get 722.5

ivolga24 [154]2 years ago

4 0

You could easily multiple .15 by 850 to get

.15(850) =127.5

You might be interested in

Plz help people who are good at math

seropon [69]

Answer:

<ABC

Step-by-step explanation:

When you use three points, two points must be on the rays of the anlge. The middle point must be the vertex.

Answer: <ABC

6 0

2 years ago

Read 2 more answers

Find the perimeter of a triangle 20cm and 29cm

bagirrra123 [75]

Answer:

Unclear.

To find the perimeter you need all three sides of the triangle. You have only given two.

All you need to do is add up all three sides and you have your perimeter.

Formula: <u>P=a+b+c</u>

5 0

1 year ago

Read 2 more answers

To simplify expressions, we often ___ like terms.

tekilochka [14]

Answer: combine

Step-by-step explanation:

When you combine like terms, it shortens or simplified the equation

Example: 2x-5+15

Combine actual numbers -5 and 15.

2x+10

6 0

2 years ago

Mr. Mole left his burrow and started digging his way down at a constant rate.

Art [367]

Mr. Mole's burrow was at an altitude of 6 meters below the ground.

Step-by-step explanation:

Step 1:

We need to determine the distance that Mr. Mole covers in a single minute.

To do that we divide the difference in values of altitude by the difference in the time periods.

For the first case, Mr. Mole had traveled -18 meters in 5 minutes.

We also have, he traveled -25.2 meters in 8 minutes.

Step 2:

The distance he covered in 1 minute $= \frac{-25.2-(-18)}{8-5} = \frac{-25.2+18}{3},$

$\frac{-25.2+18}{3} = \frac{-7.2}{3} = -2.4.$

So with every minute, Mr. Mole digs down an additional 2.4 meters below the surface.

To determine where Mr. Mole's burrow is we subtract the distance traveled in 5 minutes from -18.

The altitude of Mr. Mole's burrow $= -18 - 5(-2.4) = -18+12 = -6.$

So Mr. Mole's burrow was at an altitude of 6 meters below the ground i.e. -6 meters.

5 0

2 years ago

Read 2 more answers

Given tan theta =9, use trigonometric identities to find the exact value of each of the following:_______

Ludmilka [50]

Answer:

$(a)\ \sec^2(\theta) = 82$

$(b)\ \cot(\theta) = \frac{1}{9}$

$(c)\ \cot(\frac{\pi}{2} - \theta) = 9$

$(d)\ \csc^2(\theta) = \frac{82}{81}$

Step-by-step explanation:

Given

$\tan(\theta) = 9$

Required

Solve (a) to (d)

Using tan formula, we have:

$\tan(\theta) = \frac{Opposite}{Adjacent}$

This gives:

$\frac{Opposite}{Adjacent} = 9$

Rewrite as:

$\frac{Opposite}{Adjacent} = \frac{9}{1}$

Using a unit ratio;

$Opposite = 9; Adjacent = 1$

Using Pythagoras theorem, we have:

$Hypotenuse^2 = Opposite^2 + Adjacent^2$

$Hypotenuse^2 = 9^2 + 1^2$

$Hypotenuse^2 = 81 + 1$

$Hypotenuse^2 = 82$

Take square roots of both sides

$Hypotenuse =\sqrt{82}$

So, we have:

$Opposite = 9; Adjacent = 1$

$Hypotenuse =\sqrt{82}$

Solving (a):

$\sec^2(\theta)$

This is calculated as:

$\sec^2(\theta) = (\sec(\theta))^2$

$\sec^2(\theta) = (\frac{1}{\cos(\theta)})^2$

Where:

$\cos(\theta) = \frac{Adjacent}{Hypotenuse}$

$\cos(\theta) = \frac{1}{\sqrt{82}}$

So:

$\sec^2(\theta) = (\frac{1}{\cos(\theta)})^2$

$\sec^2(\theta) = (\frac{1}{\frac{1}{\sqrt{82}}})^2$

$\sec^2(\theta) = (\sqrt{82})^2$

$\sec^2(\theta) = 82$

Solving (b):

$\cot(\theta)$

This is calculated as:

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

Where:

$\tan(\theta) = 9$ ---- given

So:

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

$\cot(\theta) = \frac{1}{9}$

Solving (c):

$\cot(\frac{\pi}{2} - \theta)$

In trigonometry:

$\cot(\frac{\pi}{2} - \theta) = \tan(\theta)$

Hence:

$\cot(\frac{\pi}{2} - \theta) = 9$

Solving (d):

$\csc^2(\theta)$

This is calculated as:

$\csc^2(\theta) = (\csc(\theta))^2$

$\csc^2(\theta) = (\frac{1}{\sin(\theta)})^2$

Where:

$\sin(\theta) = \frac{Opposite}{Hypotenuse}$

$\sin(\theta) = \frac{9}{\sqrt{82}}$

So:

$\csc^2(\theta) = (\frac{1}{\frac{9}{\sqrt{82}}})^2$

$\csc^2(\theta) = (\frac{\sqrt{82}}{9})^2$

$\csc^2(\theta) = \frac{82}{81}$

4 0

2 years ago