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3 years ago

Please help me! I’m desperate! Help asap!

Mathematics

2 answers:

JulijaS [17]3 years ago

7 0

Answer:

scale factor - stretch by 6

Step-by-step explanation:

lilavasa [31]3 years ago

7 0

Answer:

1/3

Step-by-step explanation:

Look at two corresponding lengths that are easy to measure.

BC and B'C' are both along horizontal lines, so you can easily count squares to find their lengths.

BC = 3

B'C' = 1

B'C' is the image, and BC is the original.

The scale factor is obtained by dividing a length of the image by the corresponding length of the original.

scale factor = B'C'/BC = 1/3

Answer: 1/3

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Aye can yall help me right quick

sladkih [1.3K]

Answer:

10 for the empty box

Step-by-step explanation:

When your placing them on the graph the one on the left side is x axis (the horizontal line) and the one on the right is the y axis (the vertical line)

Hope this helps!

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3 years ago

FIND AC! PLS HELP!!

Anna007 [38]

Answer:

AC = 7.3 in

Step-by-step explanation:

tan(51°) = 9/AC

1.24 = 9/AC

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3 years ago

a construction worker needs to put a rectangular window in the side of a building. he knows from measuring that the top and bott

Cloud [144]

The other diagonal would be 13 feet as well because it is a rectangle.

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4 years ago

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Hi guys, can anyone help me with this triple integral? Many thanks:)

Crank

Another triple integral. We're integrating over the interior of the sphere

$x^2+y^2+z^2=2^2$

Let's do the outer integral over z. z stays within the sphere so it goes from -2 to 2.

For the middle integral we have

$y^2=4-x^2-z^2$

x is the inner integral so at this point we conservatively say its zero. That means y goes from $-\sqrt{4-z^2}$ and $+\sqrt{4-z^2}$

Similarly the inner integral x goes between $\pm-\sqrt{4-y^2-z^2}$

So we rewrite the integral

$\displaystyle \int_{-2}^{2} \int_{-\sqrt{4-z^2}}^{\sqrt{4-z^2}} \int_{-\sqrt{4-y^2-z^2}}^{\sqrt{4-y^2-z^2}} (x^2+xy+y^2)dx \; dy \; dz$

Let's work on the inner one,

$\displaystyle\int_{-\sqrt{4-y^2-z^2}}^{\sqrt{4-y^2-z^2}} (x^2+xy+y^2)dz$

There's no z in the integrand, so we treat it as a constant.

$=(x^2+xy+y^2)z \bigg|_{z=-\sqrt{4-y^2-z^2}}^{z=\sqrt{4-y^2-z^2}}$

So the middle integral is

$\displaystyle\int_{-\sqrt{4-z^2}}^{\sqrt{4-z^2}}2(x^2+xy+y^2)\sqrt{4-y^2-z^2} \ dy$

I gotta go so I'll stop here, sorry.

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3 years ago

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How many sides,angle and verticles are there in the following polygons

tino4ka555 [31]

Answer:

Here are the answers:

1.5

2.0

7 0

3 years ago

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