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2 years ago

What is the solution of 2.5z = 27.5

Mathematics

2 answers:

belka [17]2 years ago

5 0

Z=11 is the answer for this problem

s344n2d4d5 [400]2 years ago

3 0

Let's solve your equation step-by-step.

2.5z=27.5

Step 1: Divide both sides by 2.5.

2.5z/ 2.5 = 27.5/ 2.5

z=11

Answer:

z=11

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x+1/4x² + x+1/x²

The above can be simply as follow:

Find the least common multiple (LCM) of 4x² and x². The result is 4x²

Now Divide the LCM by the denominator of each term and multiply the result with the numerator as show below:

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Therefore,

x+1/4x² + x+1/x² = (x + 1)/4x² + 4(x + 1)/4x²

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3 years ago

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2 years ago

Prove that the diagonals of a parallelogram bisect each other

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

♕ And we're done! Hurrayyy! ;)

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☄ Hope I helped! ♡

☃ Let me know if you have any questions! ♪

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3 years ago