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3 years ago

What else must you know to prove the triangles here by SAS

Mathematics

1 answer:

Gwar [14]3 years ago

3 0

We need to know that the diagonal AC is congruent to itself and we also need to know that AD and BC are congruent

Hope this helps :)

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Please help me thank you.

7nadin3 [17]

For this question, we want to match the width of the smaller rectangle with the width of the larger rectangle, and we want to match the length of the smaller rectangle with the length of the larger rectangle.

Thus, we have two possible answers.

$\frac{4}{8} = \frac{12}{24}$
OR
$\frac{8}{4} = \frac{24}{12}$

Thus, B is our answer.

5 0

4 years ago

How can a table represent an algebraic relationship between two variables?

frez [133]

Here I have a video that can answer your question

<span>https://youtu.be/iVs4dQsKBGE</span>

5 0

4 years ago

(1.7 x 103) x (2 x 103)

Stells [14]

Answer:

Step-by-step explanation:

First off:

multiply 1.7x103 and you get 175.1 then you multiply 2x103 and you get 206.

multiply 175.1x206 and you get 36070.6

Final answer is 36070.6

5 0

3 years ago

Read 2 more answers

Geometry help needed!

salantis [7]

Answer:

AC Midpoint (2,0)-

CB Midpoint (5,-4)

midlength is 5

Step-by-step explanation:

midpoint formula $(\frac{x_{1}+x_{2} }{2} ,\frac{y_{1} +y_{2} }{2})$

A(2,4) C(2,-4)

$\frac{2+2}{2} =\frac{4}{2}=2\\\frac{4-4}{2} =\frac{0}{2} =0$

(2,0)

C(2,-4) B(8,-4)

$\frac{8+2}{2} =\frac{10}{2} =5$

$\frac{-4-4}{2} =\frac{-8}{2} =-4$ midpoint is (5,-4)

Distance between the 2 midpoints is the midsegment

(2,0) and (5,-4)

$d=\sqrt{(x_{2}-x_{1})^{2} +(y_{2}-y_{1}) ^{2} }$

$d=\sqrt{(5-2)^{2} +(-4-0)^{2} } \\d=\sqrt{3^{2} +(-4^{2})\\ } \\d=\sqrt{x} 9+16\\d=\sqrt{25} \\d=5$

3 0

3 years ago

Which expression can be used to find measure angle R

Novay_Z [31]

Answer:

cos-1(-4/14)

Step-by-step explanation:

4.5=r

3.5=s

2=t

By law of cosines:

$r^{2}=s^{2}+t^{2} -2st cosR\\\\CosR= \frac{r^{2} -s^{2} -t^{2} }{-2st} \\\\R=cos^{-1} (\frac{(4.5)^{2}-3.5^{2} -2^{2} }{-2(3.5)(2)} )\\R= cos^{-1} (-\frac{4}{14} )$

4 0

3 years ago