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choli [55]
2 years ago
9

What is the name of the relationship between ∠3 ​ and ∠7?

Mathematics
2 answers:
deff fn [24]2 years ago
7 0
The answer is.. B. Corresponding Angles.

Hope this helps!!
lara31 [8.8K]2 years ago
6 0

answer is B) Corresponding angles

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H(x)= ⎩ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎨ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎧ ​ x−5 1 ​ −(x−4) 2 3x−2 ​ , , , ​ x=0 x=7 x  ​ =0,7 ​ h(6)=h(6)=h, left parenthesis
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H (5) - piw 0x-34 :))))))))))
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A regular triangular pyramid has a base length of 4.6 centimeters and a slant height of 4 centimeters.
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Answer:

hsjsbdjkdowjd nskdkndbdndkckdbnd

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What is 32.6 rounded to the nearest hundredth?
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This the answer in decimal form as well good luck hope this help (my bad if it didn’t )

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3 years ago
Algebra 2B Review of Trig/Statistics<br><br> I really need help
alekssr [168]

\sin B is the ratio of the length of the leg opposite angle B to the length of the hypotenuse:

\sin B=\dfrac{21}{29}

It will be the same as \cos A, which is the ratio of the length of the leg adjacent to angle A to the lenght of the hypotenuse.

3 0
4 years ago
The coordinates of a quadrilateral are (1,3), (7,-3), (1,-9), and (-5,-3). What shape is formed? Select all that apply. -Kite -R
chubhunter [2.5K]

Answer: Sqaure

Step-by-step explanation:

We obtain the figure by either plotting or finding the distances between the points:

\begin{aligned}&\text { Distance } b / w \quad A B =\sqrt{(1-7)^{2}+(3-(-3))^{2}}\\&=\sqrt{6^{2}+6^{2}}\\&A B=\sqrt{72} \text { units }\end{aligned}

\text { Distance } b / w \quad BC =\sqrt{(7-1)^{2}+(-3-(-9))^{2}} \\BC=\sqrt{72} \text { units }

\begin{aligned}\text { Distance } b / w \  C D &=\sqrt{(1-(-5))^{2}+(-9-(-3))^{2}} \\C D=\sqrt{72} \text { units }\end{aligned}

$$ Distance $b/w \ A D=\sqrt{(1-(-5))^{2}+(3(-3))^{2}}$A D=\sqrt{72} \text { units }

Since all the distances of the four sides are equal, we have to check if the shape is either a rhombus or a square by the measuring the length of the diagonals

\begin{aligned}A C &=\sqrt{(1-1)^{2}+(3-(-9))^{2}} \\&=\sqrt{12^{2}} \\A C &=12 \text { units } \\B D &=\sqrt{(7-(-5))^{2}+(-3-(-3))^{2}} \\&=\sqrt{12^{2}} \\A C &=12 \text { units }\end{aligned}

The diagonals are also equal; therefore, the figure is a square

7 0
2 years ago
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