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3 years ago

Express tan B as a fraction in simplest form.

Mathematics

1 answer:

liberstina [14]3 years ago

6 0

Answer:

4/3

Step-by-step explanation:

Tan B is defined as the opposite side divided by the adjacent side. 5 is the hypotenuse, 4 is the opposite side of angle B, and 3 is the adjacent side of angle B

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Laurie is trying to stay within 10 feet of her current diving depth of -30 feet (with regard to sea level) so that the light is

frez [133]

Answer:

It should be the first one

Step-by-step explanation:

since positive ten would represent going further up and negative ten going further down, the diver would be within ten feet of her current position

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3 years ago

GIVE ME THE ANSWER* due in 10 minutes

Vlad1618 [11]

Hello from MrBillDoesMath!

Answer:

35x - 600

Discussion:

Area of unshaed rectangle =

area of full rectangle - area of shaded rectangle =

35x - 600

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3 0

3 years ago

Read 2 more answers

A rectangle measures 15 cm by 4 cm what's it's area

schepotkina [342]

The way you find area is multiplying width times height, so you have to multiply 15 by 4 which equals 60. Therefore the answer is 60

5 0

3 years ago

Y=-x^2+2x+10 y=x+2 Substitution Please show your work Need ASAP

erik [133]

Answer:

The solutions of the system of equations are the points

$(\frac{1-\sqrt{33}} {2},\frac{5-\sqrt{33}} {2})$

$(\frac{1+\sqrt{33}} {2},\frac{5+\sqrt{33}} {2})$

Step-by-step explanation:

we have

$y=-x^{2} +2x+10$ ----> equation A

$y=x+2$ ----> equation B

Solve the system by substitution

substitute equation B in equation A

$x+2=-x^{2} +2x+10$

solve for x

$-x^{2} +2x+10-x-2=0$

$-x^{2} +x+8=0$

The formula to solve a quadratic equation of the form

$ax^{2} +bx+c=0$

is equal to

$x=\frac{-b\pm\sqrt{b^{2}-4ac}} {2a}$

in this problem we have

$-x^{2} +x+8=0$

$a=-1\\b=1\\c=8$

substitute in the formula

$x=\frac{-1\pm\sqrt{1^{2}-4(-1)(8)}} {2(-1)}$

$x=\frac{-1\pm\sqrt{33}} {-2}$

$x=\frac{-1+\sqrt{33}} {-2}$ -----> $x=\frac{1-\sqrt{33}} {2}$

$x=\frac{-1-\sqrt{33}} {-2}$ -----> $x=\frac{1+\sqrt{33}} {2}$

Find the values of y

For $x=\frac{1-\sqrt{33}} {2}$

$y=x+2$

$y=\frac{1-\sqrt{33}} {2}+2$ ----> $y=\frac{5-\sqrt{33}} {2}$

For $x=\frac{1+\sqrt{33}} {2}$

$y=x+2$

$y=\frac{1+\sqrt{33}} {2}+2$ ----> $y=\frac{5+\sqrt{33}} {2}$

therefore

The solutions of the system of equations are the points

$(\frac{1-\sqrt{33}} {2},\frac{5-\sqrt{33}} {2})$

$(\frac{1+\sqrt{33}} {2},\frac{5+\sqrt{33}} {2})$

5 0

3 years ago

Mrs.Johnsons new car has an 18 gallon gas tank. If she can travel 400 miles on one tank of gasoline, what is her miles per gallo

serious [3.7K]

Answer:

Step-by-step explanation:

400/18=22.222222222

Round to 22

5 0

3 years ago