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2 years ago

Parallelogram ABCD is similar to parallelogram QRST

Mathematics

1 answer:

k0ka [10]2 years ago

7 0

Answer:

2. reflect parrellelogram ABCD across the x-axis and dilate the result by a scale factor of two centered at the orgigin

Step-by-step explanation:

to correctly map the parallelograms together you need to dialate the shape, because we already know that its two times bigger

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Find the equations of the tangent lines to x2+y2=1 at x=0, x=1 and x=35.

Ivanshal [37]

Assuming the equation is x^2 + y^2 = 1, then that's a circle, with radius 1, centered on the origin [0,0].

So there are two tangents at x = 0. They are y = 1, and y = -1 (horizontal lines).

There is one tangent at x = 1. It is x = 1 (a vertical line).

There is no tangent at x = 35, because the original equation has no solution at x = 35.

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3 years ago

17. If a + b = c, which of the following statements is true?

Scilla [17]

Answer:

c: c-a = b

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3 years ago

Read 2 more answers

PLEASE HELP ASAP! D:

artcher [175]

The solution to a system of equations is the point where the 2 lines of a graph cross each other.

On the pictured graph, the lines cross each other at x = -6 and Y = -2.

The answer would be (-6,-2)

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3 years ago

If a factory continuously dumps pollutants into a river at the rate of the quotient of the square root of t and 45 tons per day,

julsineya [31]

<h2>Hello!</h2>

The answer is:

The first option, the amount dumped after 5 days is 0.166 tons.

To solve the problem, we need to integrate the given expression and evaluate using the given time.

So, integrating we have:

$\int\limits^5_0 {\frac{\sqrt{t} }{45} } \, dt=\int\limits^5_0 {\frac{1}{45} (t)^{\frac{1}{2} } } \, dt\\\\\int\limits^5_0 {\frac{1}{45} (t)^{\frac{1}{2} } } \ dt=\frac{1}{45}\int\limits^5_0 {t^{\frac{1}{2} } } } \ dt\\\\\frac{1}{45}\int\limits^5_0 {t^{\frac{1}{2} } } } \ dt=(\frac{1}{45}*\frac{t^{\frac{1}{2}+1} }{\frac{1}{2} +1})/t(5)-t(0)\\\\(\frac{1}{45}*\frac{t^{\frac{1}{2}+1} }{\frac{1}{2} +1})/t(5)-t(0)=(\frac{1}{45}*\frac{t^{\frac{3}{2}} }{\frac{3}{2}})/t(5)-t(0)$

$(\frac{1}{45}*\frac{t^{\frac{3}{2}} }{\frac{3}{2}})/t(5)-t(0)=(\frac{1}{45}*\frac{2}{3}*t^{\frac{3}{2} })/t(5)-t(0)\\\\(\frac{1}{45}*\frac{2}{3}*t^{\frac{3}{2} })/t(5)-t(0)=(\frac{2}{135}*t^{\frac{3}{2}})/t(5)-t(0)\\\\(\frac{2}{135}*t^{\frac{3}{2}})/t(5)-t(0)=(\frac{2}{135}*5^{\frac{3}{2}})-(\frac{2}{135}*0^{\frac{3}{2}})\\\\(\frac{2}{135}*5^{\frac{3}{2}})-(\frac{2}{135}*0^{\frac{3}{2}})=\frac{2}{135}*11.18-0=0.1656=0.166$

Hence, we have that the amount dumped after 5 days is 0.166 tons.

Have a nice day!

5 0

3 years ago

100 megatons+234 help please

arlik [135]

Answer:

we know,

1 megaton=1000kg

100 megaton=100000

now,

total weight=100000+234

=100234.

6 0

3 years ago