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fgiga [73]
2 years ago
11

The discount on a new coat is $175. This is a discount of 20% what is the original selling price of the coat?

Mathematics
1 answer:
DanielleElmas [232]2 years ago
4 0

Answer:

$875

Step-by-step explanation:

175/.20

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Find the 10th term of the following geometric sequence: 2, 8, 32, 128.        A. 164,357   B. 621,325   C. 524,288   D. 248,221
Galina-37 [17]
Common ratio = 8/2 = 32/8 = 4

10th term  = a1*r^(n - 1)   where a1 = 2 , r = 4 and n = 10
                = 2 * 4^9
                =  524,288
4 0
3 years ago
Read 2 more answers
How would I start to solve this? <br> Y=-1/3-1
statuscvo [17]

Answer:

<em>Y = -4/3</em>

Step-by-step explanation:

Subtract 1 from -1/3.

Y = -1/3 - 1

Y = -1/3 - 3/3

Y = -4/3

6 0
2 years ago
Graph y = f(x) = 3(2)x. Adjust the equation to make the y-intercept (0, −3). Which response is not correct?
Sauron [17]

Answer:

Replace x with −x.

f(x) = 3(2)−x is not correct because the y-intercept is (0, 3).

the answer is B

Step-by-step explanation:

5 0
3 years ago
Read 2 more answers
Compute the surface area of the portion of the sphere with center the origin and radius 4 that lies inside the cylinder x^2+y^2=
Tom [10]

Answer:

16π

Step-by-step explanation:

Given that:

The sphere of the radius = x^2 + y^2 +z^2 = 4^2

z^2 = 16 -x^2 -y^2

z = \sqrt{16-x^2-y^2}

The partial derivatives of Z_x = \dfrac{-2x}{2 \sqrt{16-x^2 -y^2}}

Z_x = \dfrac{-x}{\sqrt{16-x^2 -y^2}}

Similarly;

Z_y = \dfrac{-y}{\sqrt{16-x^2 -y^2}}

∴

dS = \sqrt{1 + Z_x^2 +Z_y^2} \ \ . dA

=\sqrt{1 + \dfrac{x^2}{16-x^2-y^2} + \dfrac{y^2}{16-x^2-y^2} }\ \ .dA

=\sqrt{ \dfrac{16}{16-x^2-y^2}  }\ \ .dA

=\dfrac{4}{\sqrt{ 16-x^2-y^2}  }\ \ .dA

Now; the region R = x² + y² = 12

Let;

x = rcosθ = x; x varies from 0 to 2π

y = rsinθ = y; y varies from 0 to \sqrt{12}

dA = rdrdθ

∴

The surface area S = \int \limits _R \int \ dS

=  \int \limits _R\int  \ \dfrac{4}{\sqrt{ 16-x^2 -y^2} } \ dA

= \int \limits^{2 \pi}_{0} } \int^{\sqrt{12}}_{0} \dfrac{4}{\sqrt{16-r^2}} \  \ rdrd \theta

= 2 \pi \int^{\sqrt{12}}_{0} \ \dfrac{4r}{\sqrt{16-r^2}}\ dr

= 2 \pi \times 4 \Bigg [ \dfrac{\sqrt{16-r^2}}{\dfrac{1}{2}(-2)} \Bigg]^{\sqrt{12}}_{0}

= 8\pi ( - \sqrt{4} + \sqrt{16})

= 8π ( -2 + 4)

= 8π(2)

= 16π

4 0
3 years ago
How do i graph this inequality y &lt; -3
iris [78.8K]
Put a point on -3 on y-axis and graph a  straight line that is parallel to x-axis then color under the line because -3 is greater :)))
i hope this is helpful 
have a nice day 
8 0
3 years ago
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