Simplify.<br> 1/2(4x-10)

A state park has two pools. The olympic size pool holds 8.12 x 10^5 gallons of water and the smaller pool holds 5.27 x 10^5 gall

Ainat [17]

B) 1.339 x 10^6 gallons. When you add 812,000 to 527,000 you get 1,339,000, and to write that in scientific notation you need to make the number less than 10 and more than 1, which is 1.339 and to get the 1,339,000 you need to multiply 1.339 by 10^6.

4 0

3 years ago

Please help me ASAP. This is for Trigonometry, Math

Rufina [12.5K]

Answer: C) exactly one triangle

<u>Step-by-step explanation:</u>

Given: ∠A = 45°, ∠B = 65°, side c = 4 cm

By the Triangle Sum Theorem, ∠C = 70°

Now you have a proportion so you can use the Law of Sines to find the exact length of side a and of side b.

$\dfrac{\sin\angle A}{a}=\dfrac{\sin\angle B}{b}=\dfrac{\sin\angle C}{c}\\\\\\\dfrac{\sin 45^o}{a}=\dfrac{\sin65^o}{b}=\dfrac{\sin70^o}{4}$

Thus, there is exactly one triangle.

5 0

4 years ago

?

Snezhnost [94]

Each girl will get 4 tangerines

8 0

3 years ago

Can you please answer the question?

Roman55 [17]

The <em>trigonometric</em> expression $\frac{\tan^{2} \alpha}{\tan \alpha - 1} + \frac{\cot^{2} \alpha}{\cot \alpha - 1}$ is equivalent to the <em>trigonometric</em> expression $\sec \alpha \cdot \csc \alpha + 1$ .

<h3>How to prove a trigonometric equivalence</h3>

In this problem we must prove that <em>one</em> side of the equality is equal to the expression of the <em>other</em> side, requiring the use of <em>algebraic</em> and <em>trigonometric</em> properties. Now we proceed to present the corresponding procedure:

$\frac{\tan^{2} \alpha}{\tan \alpha - 1} + \frac{\cot^{2} \alpha}{\cot \alpha - 1}$

$\frac{\tan^{2}\alpha}{\tan \alpha - 1} + \frac{\frac{1}{\tan^{2}\alpha} }{\frac{1}{\tan \alpha} - 1 }$

$\frac{\tan^{2}\alpha}{\tan \alpha - 1} - \frac{\frac{1 }{\tan \alpha} }{\tan \alpha - 1}$

$\frac{\frac{\tan^{3}\alpha - 1}{\tan \alpha} }{\tan \alpha - 1}$

$\frac{\tan^{3}\alpha - 1}{\tan \alpha \cdot (\tan \alpha - 1)}$

$\frac{(\tan \alpha - 1)\cdot (\tan^{2} \alpha + \tan \alpha + 1)}{\tan \alpha\cdot (\tan \alpha - 1)}$

$\frac{\tan^{2}\alpha + \tan \alpha + 1}{\tan \alpha}$

$\tan \alpha + 1 + \cot \alpha$

$\frac{\sin \alpha}{\cos \alpha} + \frac{\cos \alpha}{\sin \alpha} + 1$

$\frac{\sin^{2}\alpha + \cos^{2}\alpha}{\cos \alpha \cdot \sin \alpha} + 1$

$\frac{1}{\cos \alpha \cdot \sin \alpha} + 1$

$\sec \alpha \cdot \csc \alpha + 1$

The <em>trigonometric</em> expression $\frac{\tan^{2} \alpha}{\tan \alpha - 1} + \frac{\cot^{2} \alpha}{\cot \alpha - 1}$ is equivalent to the <em>trigonometric</em> expression $\sec \alpha \cdot \csc \alpha + 1$ .

To learn more on trigonometric expressions: brainly.com/question/10083069

#SPJ1

6 0

2 years ago

The box-and-whisker plots show the points scored in each Super Bowl by the AFC and NFC. About how many points higher is the uppe

NemiM [27]

The upper quartile for NFC is 3 points higher than that of AFC.

<h3>What is the difference in the upper quartiles?</h3>

A box plot is used to study the distribution and level of a set of scores. The scores are distributed into 4 groups. Each group has a value of 25%

On the box, the third line on the box represents the upper quartile. 75% of the scores represents the upper quartile.

Upper quartile for AFC : 26

Upper quartile for NFC : 23

Difference = 26 - 23 = 3

To learn more about box plots, please check: brainly.com/question/27215146

#SPJ1

4 0

2 years ago

Simplify. 1/2(4x-10)