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JulsSmile [24]
3 years ago
15

Triangles FAD and DCE are translations of triangle ABC,select all the statements that must be true

Mathematics
2 answers:
Tatiana [17]3 years ago
7 0

Answer: A and C

Step-by-step explanation:

coldgirl [10]3 years ago
3 0

Answer:

    a c

Step-by-step explanation:

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Math. please help me!!!
d1i1m1o1n [39]

Part A.

In this part we have to find the increasing interval .

Increasing interval is that interval where graph goes up. And from the graph we can say that the graph goes up in the interval

(-1,0)U (1, \infty)

And that's the increasing interval .

Part B.

Correct option is B

x and y intercepts

x intercept is the point where graph touches the x axis. And the graph touches the x axis at x=-1 and 1 .

So the x intercepts are

(-1,0) ,(1,0)

y intercepts are the point, where the graph touches or crosses the y axis.

Therefore the y intercept is (0,1)

So the correct option is D .

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3 years ago
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A cell phone company recently noticed a 26% sales decrease in the month of July compared to June. Let n represent June’s sales.
Ivanshal [37]

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Step-by-step explanation:

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3 years ago
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Which equation represents an ellipse? Help please!
Tems11 [23]

Answer:

2

Step-by-step explanation:

2

4 0
3 years ago
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What is 2.45 written as a percent
soldier1979 [14.2K]

Answer:

0.0245

Step-by-step explanation:

1. Divide 2.45 by 100 since you are finding it as a percent, which is 100.

3 0
4 years ago
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In triangle JKL, tan(b°) = 3/4 and cos(b°) =4/5. If triangle JKL is dilated by a scale factor of 1/2, what is sin(b°)?
zysi [14]

Answer:

\sin (b^\circ)=\dfrac{3}{5}.

Step-by-step explanation:

It is given that,

\tan (b^\circ)=\dfrac{3}{4}

\cos (b^\circ)=\dfrac{4}{5}

If a figure is dilated, then the image is similar to the figure. It means the corresponding angles of figure and image are congruent.  

So, the value of sin(b°) after dilation is equal to the value of sin(b°) before dilation.

We know that,

\dfrac{\sin \theta}{\cos \theta}=\tan \theta

\dfrac{\sin (b^\circ)}{\cos (b^\circ)}=\tan (b^\circ)

\sin (b^\circ)=\tan (b^\circ)\times \cos (b^\circ)

\sin (b^\circ)=\dfrac{3}{4}\times \dfrac{4}{5}

\sin (b^\circ)=\dfrac{3}{5}

Therefore, \sin (b^\circ)=\dfrac{3}{5}.

5 0
4 years ago
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