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Alla [95]
3 years ago
12

Help please and I only need help on 3 and also explain your reasoning!!!!!! If you don’t you already know what’s gonna happen! T

hanks!

Mathematics
2 answers:
vova2212 [387]3 years ago
7 0

Answer:

pick any 2 properties listed from 1 - 3

Step-by-step explanation:

properties of a parallelogram:

1) Opposite sides are congruent.

2) Opposite angels are congruent.

3) Consecutive angles are supplementary.

4) If one angle is right, then all angles are right.

5) The diagonals of a parallelogram bisect each other.

now we see which don't apply:

1 - doesn't apply

2 - doesn't apply

3 - doesn't apply

4 - irrelevant to this problem (don't use)

5 - applies

kherson [118]3 years ago
5 0
Hopefully that helps
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Please help,, will mark brainliest!!
avanturin [10]

Answer:

87 degrees

Step-by-step explanation:

because 87 and 2 are diagonal and they are going to be the same. :)

7 0
3 years ago
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Is y=4x-1 proprtional
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3 years ago
Simplify....................
ELEN [110]

( 25-10 / 17 - 3.4 ) - 1
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7 0
3 years ago
9(6u+5)+2(u-4) need solved
Mazyrski [523]

Answer:

Step-by-step explanation:

Subtract  

6 u  from  2 u .

− 4 u + 9 = 19

Move all terms not containing  u  to the right side of the equation.

− 4 u = 10

Divide each term by  − 4  and simplify.

u = −  \frac{5}{2}

The result can be shown in multiple forms.

Exact Form:

u = −  \frac{5}{2}

Decimal Form:

u =− 2.5

Mixed Number Form:

u = −  2 \frac{1}{2}

8 0
3 years ago
Find the sum of the first 100<br> terms of the arithmetic<br> sequence.<br> a₁ 15 and a100=307
lara31 [8.8K]

The sequence is arithmetic, so there is a constant difference d between consecutive terms such that

a_{100} = a_{99} + d

a_{100} = (a_{98} + d) + d = a_{98} + 2d

a_{100} = (a_{97} + d) + 2d = a_{97} + 3d

and so on, down to

a_{100} = a_1 + 99d

(notice the pattern of 100 = 99 + 1 = 98 + 2 = 97 + 3 = … = 1 + 99)

Solve for d :

307 = 15 + 99d \implies 99d = 292 \implies d = \dfrac{292}{99}

Now, the sum of the first 100 terms of this sequence is

\displaystyle \sum_{n=1}^{100} a_n = \sum_{n=1}^{100} \left(15 + \frac{292}{99}(n-1)\right) = \boxed{16100}

which follows from the well-known sums

\displaystyle \sum_{n=1}^N 1 = N

\displaystyle \sum_{n=1}^N n = \frac{N(N+1)}2

3 0
2 years ago
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