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3 years ago

What is the value of X in the figure below? In this diagram.

Mathematics

1 answer:

kipiarov [429]3 years ago

3 0

<h3>Answer: E. 45/4</h3>

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Work Shown:

Because we have similar triangles, we can form the proportion below to isolate x.

(AC)/(BC) = (CD)/(AC)

(15)/(20) = (x)/(15)

15*15 = 20x .... cross multiply

225 = 20x

20x = 225

x = 225/20

x = (5*45)/(5*4)

x = 45/4

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Nastasia [14]

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3 years ago

Prove that the diagonals of a parallelogram bisect each other

Nady [450]

Answer:

[ See the attached picture ]

The diagonals of a parallelogram bisect each other.

✧ Given : ABCD is a parallelogram. Diagonals AC and BD intersect at O.

✺ To prove : AC and BD bisect each other at O , i.e AO = OC and BO = OD.

Proof : $\begin{array}{ |c| c | c | } \hline \tt{SN}& \tt{STATEMENTS} & \tt{REASONS}\\ \hline 1& \sf{In \: \triangle ^{s} \:AOB \: and \: COD } \\ \sf{(i)}& \sf{ \angle \: OAB = \angle \: OCD\: (A)}& \sf{AB \parallel \: DC \: and \: alternate \: angles} \\ \sf{(ii)} &\sf{AB = DC(S)}& \sf{Opposite \: sides \: of \: a \: parallelogram} \\ \sf{(iii)} &\sf{ \angle \: OBA= \angle \: ODC(A)} &\sf{AB \parallel \:DC \: and \: alternate \: angles} \\ \sf{(iv)}& \sf{ \triangle \:AOB\cong \triangle \: COD}& \sf{A.S.A \: axiom}\\ \hline 2.& \sf{AO = OC \: and \: BO = OD}& \sf{Corresponding \: sides \: of \: congruent \: triangle}\\ \hline 3.& \sf{AC \: and \: BD \: bisect \: each \: other \: at \: O}& \sf{From \: statement \: (2)}\\ \\ \hline\end{array}. Proved ✔$

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3 years ago

What is the square root of 2x plus the square root of 2x

WITCHER [35]

Answer:

i just took the test, so its b. 84

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3 years ago

What is 56 divided by 8

aleksklad [387]

56 ÷ 8 = 7

8 goes into 56 7 times

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-AaronWiseIsBae

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2 years ago

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How many solutions does the equation y=mx have

kherson [118]

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