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Kipish [7]
3 years ago
5

Point Alies on Quadrant IV. If it is rotated 180° about the origin which quadrant will the image A' lie on.

Mathematics
1 answer:
horsena [70]3 years ago
6 0

Answer:

Quadrant 1, I think.

Step-by-step explanation:

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Albert bought some fish at a grocery store at $2 per pound for salmon and $4 per pound for catfish. He spent a total of $12 on s
Gemiola [76]

Albert bought 2 pounds of catfish and 2 pounds of salmon

Let c represent the amount of catfish in pounds and s represent the amount of salmon in pounds.

He spent a total of $12 on salmon and catfish and bought a total of 4 pounds.  Hence:

c + s = 4    (1)

4c + 2s = 12    (2)

Solving equations 1 and 2 simultaneously gives:

c = 2, s = 2

Albert bought 2 pounds of catfish and 2 pounds of salmon

Find out more on equation at: brainly.com/question/2972832

7 0
2 years ago
A right circular cylinder is inscribed in a sphere with diameter 4cm as shown. If the cylinder is open at both ends, find the la
SOVA2 [1]

Answer:

8\pi\text{ square cm}

Step-by-step explanation:

Since, we know that,

The surface area of a cylinder having both ends in both sides,

S=2\pi rh

Where,

r = radius,

h = height,

Given,

Diameter of the sphere = 4 cm,

So, by using Pythagoras theorem,

4^2 = (2r)^2 + h^2   ( see in the below diagram ),

16 = 4r^2 + h^2

16 - 4r^2 = h^2

\implies h=\sqrt{16-4r^2}

Thus, the surface area of the cylinder,

S=2\pi r(\sqrt{16-4r^2})

Differentiating with respect to r,

\frac{dS}{dr}=2\pi(r\times \frac{1}{2\sqrt{16-4r^2}}\times -8r + \sqrt{16-4r^2})

=2\pi(\frac{-4r^2+16-4r^2}{\sqrt{16-4r^2}})

=2\pi(\frac{-8r^2+16}{\sqrt{16-4r^2}})

Again differentiating with respect to r,

\frac{d^2S}{dt^2}=2\pi(\frac{\sqrt{16-4r^2}\times -16r + (-8r^2+16)\times \frac{1}{2\sqrt{16-4r^2}}\times -8r}{16-4r^2})

For maximum or minimum,

\frac{dS}{dt}=0

2\pi(\frac{-8r^2+16}{\sqrt{16-4r^2}})=0

-8r^2 + 16 = 0

8r^2 = 16

r^2 = 2

\implies r = \sqrt{2}

Since, for r = √2,

\frac{d^2S}{dt^2}=negative

Hence, the surface area is maximum if r = √2,

And, maximum surface area,

S = 2\pi (\sqrt{2})(\sqrt{16-8})

=2\pi (\sqrt{2})(\sqrt{8})

=2\pi \sqrt{16}

=8\pi\text{ square cm}

4 0
3 years ago
At a local coffee shop, three coffees and two muffins cost a total of $7.00. Two coffees and four
rewona [7]

Answer:

$10.25

Step-by-step explanation:

Set up a system of equations, so 3c + 2m = 7 and 2c + 4m = 8, where c = the cost of coffees, and m = cost of muffins

To solve multiply the first equation by -2(3c + 2m=7) so you get -6c -4m = -14 now add the two equations

-6c - 4m = -14

2c + 4m = 8   notice that -6c + 2c = -4c and -4m + 4m cancels, then -14+8=-6

so we have -4c = -6 so c = $1.50

To solve for m, substitute c = 1.50 so 3(1.50) + 2m = 7, so solve for m so

4.5 + 2m = 7, 2m = 2.5, so m = $1.25

So now solve the final problem 6(1.50) + 1(1.25) = $10.25

6 0
3 years ago
Outline the steps for taking cross sections of a rectangular prism.
fgiga [73]

Answer:

  • find a suitable cutting tool
  • cut the prism on the plane of interest

Step-by-step explanation:

A cross section is the intersection of a cut plane with the object of interest. In a classroom setting, we often try to do the cross sectioning mentally or with diagrams, rather than physically cutting anything. Sometimes, there is no substitute for actually performing the cut to see what the cross section looks like.

For certain samples that don't take kindly to cutting, we sometimes encase them in a block of material that helps them hold their shape during the process. Sometimes the "cutting" is performed by grinding away the portion of the material on one side of the cut plane.

7 0
3 years ago
what �is the equation of a line that is perpendicular to the line  y = − 2 5 x − 5 y = - 2 5 x - 5 and passes thr
gladu [14]

Answer:

y=\frac{1}{25}x-\frac{31}{25}

Step-by-step explanation:

y=-25x-5

You can get the gradient of the line perpendicular to this as you know the products of their gradients will be -1.

-25*m_{perpendicular}=-1\\m_{perpendicular}=\frac{1}{25}

We have a gradient, and a point (6, 1), so we can now substitute our known values into the line equation.

(y-y_0)=m(x-x_0)

y-1=\frac{1}{25} (x-6)

Rearrange for y and collect terms.

y=\frac{1}{25}x-\frac{6}{25}-1\\\\y=\frac{1}{25}x-\frac{31}{25}

8 0
4 years ago
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