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astra-53 [7]
3 years ago
5

I need help on a and b

Mathematics
1 answer:
stepladder [879]3 years ago
8 0

Problem A

A rectangle has equal diagonals

So 9x + 20 = 10x + 8 Subtract 8 from both sides.

9x + 20 - 8 = 10x

9x + 12 = 10x Subtract 9x from both sides.

12 = 10x - 9x

12 = x Answer <<<<<

Problem B

The diagonals of a rhombus bisect the angle vertex that the diagonal is connect between. All that language says is that

2x + 15 = 5x + 3 Subtract 2x from both sides.

15 = 5x - 2x + 3

15 = 3x + 3 Subtract 3 from both sides.

15 - 3 = 3x

12 = 3x Divide by 3

12/3 = x

x = 4 <<<< Answer B

When x = 4 the two angles are equal.

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Step-by-step explanation:

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\tt{}1.440.000 \\

Step-by-step explanation:

\tt{}1.200 \times 1.200  \:  \:  \:  \:  \:  \:  \:  \:  \:  \: \\  \\ \tt{}((12 \times 12) + (0000)) \\  \\ 1.440.000 \:  \:  \:  \:  \:  \:  \:  \:  \:  \:  \:  \:  \:  \:  \:  \:  \:  \:  \:  \:  \:  \\  \\  \\  \\

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3 0
2 years ago
7. In A.ABC. JB and KA are medians, JK = 10x - 12, AB = 9x + 18, JM = 21, KM = 23. AJ = 60. and
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A) JK and AB =

[tex] \boxed{( \theta) = \f{12}{( \theta)} } So, =》 \dfrac{10}{5} = \dfrac{1}{ \( \theta) } =》 \( \theta) = \dfrac{5}{2}[/tex]

B) AABC

[tex] \boxed{( \theta) = {1}{( \theta)} } So, 23}{-18=\\Ans } \boxed

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[tex] e \fbox{: } The value of k for which \sin(jm) = \cos(x) is \dfrac{\p}{2} ans[/tex]

----------

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