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ra1l [238]
3 years ago
7

Peter reflected parallelogram ABCD across the y-axis. If angle A is 115° and angle B is 65°, what is the degree measurement of a

ngle A'?
65

50

115

150
Mathematics
2 answers:
Sedaia [141]3 years ago
8 0

Answer:

∠A' = 115°

C is correct

Step-by-step explanation:

Peter reflected parallelogram ABCD across the y-axis.

∠A is 115° and ∠B is 65°

In reflection the property of figure doesn't change.

The given figure is parallelogram ABCD.

Reflection about y-axis.

Change in coordinate of ABCD.

The length of side of parallelogram and angle of parallelogram would be same.

Therefore, the measurement of ∠A' = ∠A

∠A' = 115°

Hence, the degree measurement of ∠A' is 115°

IgorLugansk [536]3 years ago
6 0
All angle measure is preserved, therefore measure of A' is 115
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Add 4 kg 250 g, 2 kg 90 g and 19kg 3 g.
GarryVolchara [31]

Answer:

=25343 grams or 25 kg 343 g.

Step-by-step explanation:

1 kg = 1000 g

∴ 4 kg = 4000 grams

and the 4 kg 250 g would equal = 4250 grams.

∴ 2 kg = 1000 g

2 kg 90 g = 2000 grams + 90 g = 2090 grams

∴ 19 kg = 19000 grams

and 19 kg 3 grams = 19000 grams + 3 grams = 19003 grams

now we will add all of the values together to get the final answer.

4250 grams + 2090 grams + 19003 grams = 25343 grams or 25 kg 343 g.

Hope this helps!!

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3 years ago
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butalik [34]
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Put the following equation below in function notation and evaluate, using x as the independent
Flura [38]

Answer:

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Step-by-step explanation:

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4 0
3 years ago
Read 2 more answers
Need help simplifying this
diamong [38]

The simplified answer is \frac{\left(12 x^{2}+7 x y-4 y z-3 x z+3 y^{2}\right)}{6 x^{2}+3 x z+2 x y+y z}.

<u>Step-by-step explanation:</u>

$\frac{3 y+2 x}{z+2 x}-\frac{2 y-3 x}{3 x+y}-\frac{2 z(y+3 x)}{6 x^{2}+3 x z+2 x y+y z}

To add or subtract denominators of the fraction must be same.

If it is not the same, we must take LCM of the denominators. and so we can add the fractions.

To make the denominator same multiply the 1st term (\frac{3x+y}{3x+y}) and 2nd term by (\frac{z+2x}{z+2x})

= \frac{(3 y+2 x)(3 x+y)}{(z+2 x)(3 x+y)}-\frac{(2 y-3 x)(z+2 x)}{(3 x+y)(z+2 x)}-\frac{2 z(y+3 x)}{6 x^{2}+3 x z+2 x y+y z}

LCM of the denominators is 6x²+ 3xz + 2xy +yz.

Multiply the factors in the numerator.

= \frac{\left(6 x^{2}+3 y^{2}+11 x y\right)}{(z+2 x)(3 x+y)}-\frac{\left(2 y z+4 x y-3 x z-6 x^{2}\right)}{(3 x+y)(z+2 x)}-\frac{2 z y+6 x z}{6 x^{2}+3 x z+2 x y+y z}

Now, the denominators are same, you can subtract it.

= \frac{\left(6 x^{2}+6 x^{2}+11 x y-4 x y-2 y z-2 y z+3 x z-6 x z+3 y^{2}\right)}{6 x^{2}+3 x z+2 x y+y z}

= \frac{\left(12 x^{2}+7 x y-4 y z-3 x z+3 y^{2}\right)}{6 x^{2}+3 x z+2 x y+y z}

Thus the simplified solution is  \frac{\left(12 x^{2}+7 x y-4 y z-3 x z+3 y^{2}\right)}{6 x^{2}+3 x z+2 x y+y z}

4 0
3 years ago
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