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3 years ago

Please help me with this question

Mathematics

1 answer:

jeyben [28]3 years ago

4 0

Answer:

add all of them together then divide it by two

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Pls pls pls pls pls pls pls pls pls help pls

lapo4ka [179]

So
C=10
3c+c/2c means
3(10)+10/2(10)
30+10/20
40/20=2

5 0

2 years ago

Kelly works at her own nail

Sergeeva-Olga [200]

Answer:

(x×11)÷52.25

Step-by-step explanation:

if 11=52.25 and y=x then x11÷52.25 i think that is the answer

6 0

3 years ago

Question 31 only please

Brut [27]

Answer:

this is answer

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3 0

3 years ago

I need not just the answer but the work written out

Mekhanik [1.2K]

Answer:

A(3) = -6 + (3 - 1) (5)

-6 + (2)(5)

-6 + (10)

A(4) = -6 + (4 - 1) (5)

-6 + (3)(5)

-6 + 15

A(10) = -6 + (10 - 1) (5)

-6 + (9)(5)

-6 + 45

6 0

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

♕ And we're done! Hurrayyy! ;)

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5 0

3 years ago