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2 years ago

Can someone help me with this problem?

Mathematics

1 answer:

Nadusha1986 [10]2 years ago

8 0

Increasing is= [0,90], [270,360]

Decreasing is=[90,270]

Concave up=[180,360]

Concave down=[0,180]

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Is this equation an identity? z = -10 − z

Umnica [9.8K]

Answer:

Step-by-step explanation:

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2 years ago

Use distributive property to simplify. 10(-8s-k) just need to check if -80s-10k is correct.

Nitella [24]

10(-8s-k)
(10) (-8s + -k)
(10) (-8s) + (10) (-k)
= -80s - 10k
= -10k-80s

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3 years ago

Help help help me with this it is an emergency for you guys to help me plz ASAP

Andru [333]

☻ ~~~~~~~~~~~~ Hello! ~~~~~~~~~~~~~~~ ☻

➴ The correct answer is B. the temperature in F on a summer day

B is correct because temperatures in the summer (in Fahrenheit) are high and positive. All the other answer choices are negative numbers.

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2 years ago

Read 2 more answers

What is the area of this polygon? Pleasee help mee I will mark brainliest!

Ira Lisetskai [31]

Answer:

60 squared inches.

Step-by-step explanation:

You can break this polygon into two parts. The rectangle part of the polygon is 55 squared inches. The triangle part is 5 squared inches.

The rectangle:

$5(11)=55$

The triangle:

$2(5)=10\\10/2=5$

Therefore, the whole polygon is 60 squared inches.

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3 years ago

2 tan 30° II 1 + tan- 300

shusha [124]

Question:

$\frac{2tan30^{\circ}}{1 + tan^2(30^{\circ})}$

Answer:

$\frac{2tan30^{\circ}}{1 + tan^2(30^{\circ})}$ $= sin(60^{\circ})$

Step-by-step explanation:

Given

$\frac{2tan30^{\circ}}{1 + tan^2(30^{\circ})}$

Required

Simplify

In trigonometry:

$tan(30^{\circ}) = \frac{1}{\sqrt{3}}$

So, the expression becomes:

$\frac{2tan30^{\circ}}{1 + tan^2(30^{\circ})}$ $= \frac{2 * \frac{1}{\sqrt{3}}}{1 + (\frac{1}{\sqrt{3}})^2}$

Simplify the denominator

$\frac{2tan30^{\circ}}{1 + tan^2(30^{\circ})}$ $= \frac{2 * \frac{1}{\sqrt{3}}}{1 + \frac{1}{3}}$

$\frac{2tan30^{\circ}}{1 + tan^2(30^{\circ})}$ $= \frac{\frac{2}{\sqrt{3}}}{1 + \frac{1}{3}}$

$\frac{2tan30^{\circ}}{1 + tan^2(30^{\circ})}$ $= \frac{\frac{2}{\sqrt{3}}}{ \frac{3+1}{3}}$

$\frac{2tan30^{\circ}}{1 + tan^2(30^{\circ})}$ $= \frac{\frac{2}{\sqrt{3}}}{ \frac{4}{3}}$

Express the fraction as:

$\frac{2tan30^{\circ}}{1 + tan^2(30^{\circ})}$ $= \frac{2}{\sqrt 3} / \frac{4}{3}$

$\frac{2tan30^{\circ}}{1 + tan^2(30^{\circ})}$ $= \frac{2}{\sqrt 3} * \frac{3}{4}$

$\frac{2tan30^{\circ}}{1 + tan^2(30^{\circ})}$ $= \frac{1}{\sqrt 3} * \frac{3}{2}$

$\frac{2tan30^{\circ}}{1 + tan^2(30^{\circ})}$ $= \frac{3}{2\sqrt 3}$

Rationalize

$\frac{2tan30^{\circ}}{1 + tan^2(30^{\circ})}$ $= \frac{3}{2\sqrt 3} * \frac{\sqrt{3}}{\sqrt{3}}$

$\frac{2tan30^{\circ}}{1 + tan^2(30^{\circ})}$ $= \frac{3\sqrt{3}}{2* 3}$

$\frac{2tan30^{\circ}}{1 + tan^2(30^{\circ})}$ $= \frac{\sqrt{3}}{2}$

In trigonometry:

$sin(60^{\circ}) = \frac{\sqrt{3}}{2}$

Hence:

$\frac{2tan30^{\circ}}{1 + tan^2(30^{\circ})}$ $= sin(60^{\circ})$

3 0

3 years ago