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Genrish500 [490]
3 years ago
14

What is the answer to this

Mathematics
2 answers:
ivann1987 [24]3 years ago
6 0

Answer:

AC=BD

BY using the rules of parallelograms we see that since AB is l l to DC and AD is l l to BC, then the length of the diagonals BD and AC must be equal

Step-by-step explanation:

svetoff [14.1K]3 years ago
5 0
The answer is d :):)
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1.Solve the equation 2x^2+10x=0 . Check your solution(s) and state the final solution set.
Setler [38]
2x^2  + 10x = 0

we can take out x from both terms and we get:
x(2x + 10) = 0

now either x has to be equal to 0 or 2x + 10 has to be equal to zero in order for equation to be equal to 0. 

one solution is x=0 as stated above

other is:
2x + 10 = 0
2x = -10
x = -10/2 = -5

Answers are x=0 and x = -5.
3 0
2 years ago
Find the missing number. 1, 2, 6, 24, 120,?
vfiekz [6]
The answer is A. Each time it is multiplied by a higher number so 1 x 2 = 2 . 2 x 3 = 6 . 6 x 4 = 24 . 24 x 5 = 120 . 120 x 6 = 720
8 0
3 years ago
PLEASE HELP !! BRAINLIEST AND 10 POINTS
Bumek [7]
POINT A I had the same question in my math class !!!
8 0
3 years ago
What is 2xy-3yz+5+2yz-xy
melomori [17]

To simplify this expression, we are going to combine like terms.


First, we can see that we have two terms with the variables xy being multiplied together, 2xy and -xy. Together, these add to xy, so our expression is now:

xy - 3yz + 5 + 2yz


We also have two terms with yz, -3yz and 2yz. Together, these add to -yz, so our expression is now:

xy - yz + 5


Our final expression is xy - yz + 5.

6 0
3 years ago
SOMEONE HELP ME IM FREAKING OUT I LITERALLY CANT WITH THIS QUESTION IM PRAYING PLEASE HELP ME IM SO SERIOUS IM GONNA END IT PLS
antiseptic1488 [7]

Answer:

\sf -11+7\sqrt{2}

Step-by-step explanation:

Given expression:

\sf \dfrac{3-\sqrt{32}}{1+\sqrt{2} }

Rewrite 32 as 16 · 2:

\sf \implies \dfrac{3-\sqrt{16 \cdot 2}}{1+\sqrt{2} }

Apply radical rule \sf \sqrt{a \cdot b}=\sqrt{a}\sqrt{b}

\sf \implies \dfrac{3-\sqrt{16}\sqrt{2}}{1+\sqrt{2} }

As \sf \sqrt{16}=4:

\sf \implies \dfrac{3-4\sqrt{2}}{1+\sqrt{2} }

Multiply by the conjugate:

\sf \implies \dfrac{3-4\sqrt{2}}{1+\sqrt{2} } \times \dfrac{1-\sqrt{2} }{1-\sqrt{2} }

\sf \implies \dfrac{(3-4\sqrt{2})(1-\sqrt{2})}{(1+\sqrt{2})(1-\sqrt{2})}

\sf \implies \dfrac{3-3\sqrt{2}-4\sqrt{2}+4\sqrt{2}\sqrt{2}}{1-\sqrt{2}+\sqrt{2}-\sqrt{2}\sqrt{2}}

As \sf \sqrt{2}\sqrt{2}=\sqrt{4}=2:

\sf \implies \dfrac{3-3\sqrt{2}-4\sqrt{2}+4 \cdot 2}{1-\sqrt{2}+\sqrt{2}-2}

\sf \implies \dfrac{3-7\sqrt{2}+8}{1-2}

\sf \implies \dfrac{11-7\sqrt{2}}{-1}

\sf \implies -11+7\sqrt{2}

7 0
2 years ago
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