Please help me with this triangle problem

THIS IS URGENT I NEED HELP ASAP.(9x 10^-3)^2

joja [24]

(9x 10^-3)^2= $\frac{81x^{2}}{1000000}$ , so d 8.1x10^-6

Evaluate 1/2a^-4b^2 for a =-2 and b=4

$\frac{1}{2} (-2)^{-4} *4^{2}\\\\\frac{1}{2} *\frac{1}{(-2)^{4}} *4^{2}\\\\\frac{1}{2} *\frac{1}{2^{4}}*4^{2}\\\\\frac{1}{2} *\frac{1}{16}*4^{2}\\\\\frac{1}{2} *\frac{1}{16}*16=\frac{1}{2}$

A number raised to a negative exponent is sometimes negative.

Hope this helps, now you know the answer and how to do it. HAVE A BLESSED AND WONDERFUL DAY! As well as a great Valentines Day! :-)

- Cutiepatutie ☺❀❤

8 0

3 years ago

A seat on a Ferris wheel is level with the center of the wheel. The diameter of the wheel is 210 feet (ft). If the wheel

frutty [35]

Answer:

The change in the height of the seat of the Ferris wheel is a reduction in height of approximately 74.25 feet

Step-by-step explanation:

The given information are;

The position of the seat on the Ferris wheel = The center level of the wheel

The diameter of the Ferris wheel = 210 feet

The angle of rotation of the wheel = 45° counterclockwise

The height of a point on the Ferris wheel is given by the relation;

h = A·cos(b·x + c) + d

With regards to the question, we have;

h = The height of the seat of the Ferris wheel

A = The amplitude = 1/2 × The diameter of the wheel = 210/2 = 105 feet

360/b = The period

c = The phase shift = 90°

d = The mid line = 0

For a rotation of 45° counterclockwise or-45° clockwise, we have;

b·x = 45°, therefore;

h = 105 × cos(45° + 90°) + 0 = 105 × cos(135°) + 0

h = 105 × cos(135°) ≈ -74.25 feet

Therefore, the height of the seat of the Ferris wheel reduces by 74.246 feet.

6 0

3 years ago

A regular hexagon has sides of 3 feet. What is the area of the hexagon?

frosja888 [35]

Answer:

$13.5\sqrt{3}\:\text{ft}^{2}$

Step-by-step explanation:

Given: The side of a regular hexagon is 3 feet.

To find: Area of the hexagon

Solution:

It is given that the side of a regular hexagon is 3 feet.

We know that the area of a regular hexagon whose side is a units is $\frac{3\sqrt{3} }{2} a^{2}$

Here, the side is 3 feet

So, area of the regular hexagon

$=\frac{3\sqrt{3} }{2} \times3^{2}$

$=\frac{3\sqrt{3} }{2} \times9$

$=\frac{27\sqrt{3} }{2}$

$=13.5\sqrt{3}\:\text{ft}^{2}$

Hence, area of the regular hexagon is $13.5\sqrt{3}\:\text{ft}^{2}$

6 0

3 years ago

What is the y-intercept of the function, represented by the table of values

AleksandrR [38]

Answer:

9

Step-by-step explanation:

The table is given below

$\left|\begin{array}{c|cc}x&y\\--&--\\-2&15\\1&6\\2&3\\4&-3\\7&-12\end{array}\right|$

We can see that the table represents a linear function since by checking its slope.

Using:

Points (-2,15) and (1,6)
Points (1,6) and (2,3)

$Slope =\dfrac{6-15}{1-(-2)} =\dfrac{3-6}{2-1}=-3$

Therefore, the values represent a linear function.

A linear equation is of the form y=mx+b

Therefore:

y=-3x+b

Using the points (7,-12)

$-12=-3(7)+b\\-12=-21+b\\b=-12+21\\b=9$

Therefore, the y-intercept of the function, represented by the table of values is 9.

6 0

3 years ago

What are the steps to solve <br><img src="https://tex.z-dn.net/?f=%20%5Csqrt%7B20%20%7Bk%7D%5E%7B3%7D%20%7D%20" id="TexFormula1"

vlada-n [284]

$\bf \sqrt{20k^3}\qquad 
\begin{cases}
20=2\cdot 2\cdot 5\\
\qquad 2^2\cdot 5\\
k^3=k^{2+1}\\
\qquad k^2\cdot k^1
\end{cases}\implies \sqrt{2^2\cdot 5\cdot k^2\cdot k}\implies 2k\sqrt{5k}$

3 0

4 years ago