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3 years ago

Explain this strategy: 1.4 to the power of 4 times .75 squared equals 2.1609

Mathematics

1 answer:

yaroslaw [1]3 years ago

3 0

What do u think the answer is

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Explain and Prove that the sides of the triangles below are a right triangle.

sukhopar [10]

3.) 3, 4, 5
&
4.) 6, 8, 10
Those two are right triangles because when you add the squares of the first two it should give you the square root of the third number

a^2 + b^2 = c^2

So
3^2 + 4^2 = 5^2
9 + 16 = 25
25 = 25

6^2 + 8^2 = 10^2
36 + 64 = 100
100 = 100

Therefore they are right triangles

But 1 & 2 aren’t right triangles because when you add the squares of the first two number, it doesn’t equal to the square root of the third number.

Hope this helps!!

3 0

3 years ago

2 tan 30°<br>II<br>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

Use the graph below to determine a1 and d for the sequence.

Rama09 [41]

A1 would be the first y value, which is 3 and d is the ratio ( distance between each y value) which is 3 ( add 3 to the y value to get the next one)

The answer is a1= 2, d = 3

4 0

3 years ago

you want to put 520 quarters in coin wrappers. you need one wrapper for every $10 in quarters. write an equation you can use to

Lisa [10]

40c=520
This is the proper equation because there are 40 quarters in $10. Thus, if we could only fit $10 worth of quarters in each wrapper, we can only fit 40quarters in each wrapper. C represents the amount of wrappers we will need.

Hope this helps!

6 0

3 years ago

Find the distance between the points (-5, 0) and (-4, 1).

choli [55]

The formula of a distance between two points A and B is :

$AB=\sqrt{(x_{B}- x_{A})^2 +( y_{B}- y_{A})^2 }$

You should this formula by heart.

Then, you just have to apply it ! :D

Here :

$x_{A}=-5$
$y_{A} =0$

$x_{B} =-4$
$y_{B}= 1$

Therefore :

$\sqrt{(-4-(-5))^2+(1-0)^2}$
$= \sqrt{1^2+1^2}$
$= \sqrt{1+1}$
$= \sqrt{2}$

In short, the answer would be the first option : $\sqrt{2}$ .

Hope this helps !

Photon

6 0

3 years ago

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