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Darina [25.2K]
3 years ago
11

A desktop is shaped like a semicircle with a diameter of 28 what is the area of the desktop?

Mathematics
1 answer:
OlgaM077 [116]3 years ago
8 0

Answer:

307.72

Step-by-step explanation:

if the diameter is 28 then the radius is 14 (half of the diameter)

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The dimensions of a box that is a rectangular prism are 3 inches, 4 inches, and 6 inches. What is the length of the diagonal fro
Neporo4naja [7]

Answer:

7.8 in

Step-by-step explanation:

Given:-

- The dimension of rectangular prism:

                                   (3 x 4 x 6) inches

Find:-

What is the length of the diagonal from the point R to point S, to the nearest tenth of an inch?

Solution:-

- We will set up an origin with coordinates ( 0 , 0 , 0 ) at point R. Then the coordinates of point S would be ( 3 , 4 , 6 ).

- Then we will use the distance between two points formula in cartesian coordinate system:

                    Distance = \sqrt{(x_1 - x_2)^2 + (y_1 - y_2)^2 + (z_1 - z_2)^2}

Where the coordinates of two points ( x1 , y1 , z1 ) & ( x2 , y2 , z2 ).

- Using R( 0 , 0 , 0 ) & S( 3 , 4 , 6 ), the distance |RS| would be:

                   |RS| = \sqrt{(0 - 3)^2 + (0 - 4)^2 + (0 - 6)^2}\\\\|RS| = \sqrt{ 9 + 16 + 36}\\\\|RS| = \sqrt{61} = 7.81024

- The distance |RS| to nearest 10th of an inch is = 7.8 in  

4 0
3 years ago
The measure of arc QS is (4x – 18)°.
Zigmanuir [339]

Answer: 49.5

Step-by-step explanation:

7 0
3 years ago
Read 2 more answers
HI can you help me with this problem?The quantity demanded x for a certain brand of MP3 players is 300 units when the unit price
Vlad [161]

The demand equation is P = -0.05x + 105

STEP - BY - STEP EXPLANATION

What to find?

<em>The demand equation.</em>

The following information is used to solve the problem.

If (x₁ , y₁) and (x₂, y₂) are any dinstinct points on a line a non-vertical line L, then the equation of the line that has a slope(m) and passes through the point (x₁ , y₁) is given by the formula below:

y-y_1=_{}\frac{y_2-y_1}{x_2-x_1}(x-x_1)

Assume p, be the unit price and x be the demand when the price is $p.

From the question given, when the unit price x = 300 units, the demand p = $90 and when the unit price x = 1300 units, the demand p = $40

This means the demand curve passes through the points (300, 90) and (1300, 40).

Assume the demand equation is linear, then the line of the equation passes through the points (300, 90) and (1300, 40).

Now;

y=p x₁ =300 y₁ = 90 x₂ = 1300 y₂=40

Substitute the values into the function and simplify.

p-90=\frac{40-90}{1300-300}(x-300)

Simplify the above.

p-90=\frac{-50}{1000}(x-300)p-90=-0.05(x-300)

Open the parenthesis.

P- 90 = -0.05x + 15

Add 90 to both-side of the equation.

P = -0.05x + 15 + 90

P = -0.05x + 105

Hence, the demand function is P = -0.05x + 105

6 0
1 year ago
Let T(x)=−6x4+x2−1 ....
jolli1 [7]

Answer:

D is the correct answer.

Step-by-step explanation:

Distribute (−6x^4 + x^2 − 1)(2x^3 - 4x + 3)

-12x^7 + 24x^5 - 18x^4 + 2x^5 - 4x^3 + 3x^2 - 2x^3 + 4x - 3

Combine Like Terms

-12x^7 + 26x^5 - 18x^4  - 6x^3 + 3x^2 + 4x - 3

5 0
4 years ago
Exemplos de problemas com porcentagem (resolvido)
Marizza181 [45]
12 de 24 seria

12÷24 = 0.5
0.5=
50% entonces 12 de 24 es 50%

igual con 18 de 20

18÷20= 0.9
0.9=
90% entonces 18 de 20 es 90%
5 0
3 years ago
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