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Hatshy [7]
3 years ago
14

Help !! Which transformation can not be used to prove that ABC is congruent to DEF ?

Mathematics
2 answers:
Vesnalui [34]3 years ago
8 0

The answer is D since dialation only beating to increase in size

Over [174]3 years ago
4 0

D

the answer is D. dilation cannot be used to prove this

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hich of the following expressions is equivalent to |x + 4| < 5? A. –5 > x + 4 < 5 B. –5 < x + 4 < 5 C. x + 4 <
Airida [17]

Answer:

option B

Given : |x + 4| < 5

A. –5 > x + 4 < 5

B. –5 < x + 4 < 5

C. x + 4 < 5 and x + 4 < –5

D. x + 4 < 5 or x + 4 < –5

In general , |x|< n where n is positive

Then we translate to -n < x < n

|x + 4| < 5

5 is positive, so we translate the given absolute inequality to

-5 < x+4 < 5

So option B is correct

5 0
3 years ago
What is the value of x in the equation 4x+8y=40, when y=0.8?
lubasha [3.4K]
4x+8y=40
0.8*8=6.4
4x+6.4=40
40-6.4=33.6
4x=33.6
33.6\4=8.4
X=8.4
6 0
3 years ago
In 2010, the world's largest pumpkin weighed 1,810 kilograms. An average-sized pumpkin weighs 5,000 grams.
jekas [21]

Answer:

1,805 kg.

Step-by-step explanation:

We have been given that in 2010, the world's largest pumpkin weighed 1,810 kilograms. An average-sized pumpkin weighs 5,000 grams. We are asked to find the how much the world's largest pumpkin weighs than an average pumpkin.

First of all, we will convert the weight of average pumpkin in kilograms by dividing 5,000 by 1000 as 1 kg equals 1,000 gm.

\text{The weight of average pumpkin}=\frac{5000\text{ gm}}{\frac{\text{1000 gm}}{\text{ 1 kg}}}

\text{The weight of average pumpkin}=5000\text{ gm}\times \frac{\text{ 1 kg}}{\text{1000 gm}}

\text{The weight of average pumpkin}=5\text{ kg}

Now, we will subtract the weight of average pumpkin from world's largest pumpkin's weight.

1,810\text{ kilograms}-5\text{ kilograms}=1,805\text{ kilograms}

Therefore, the 2010 world-record pumpkin weighs 1,805 kilograms more than an average-sized pumpkin.

3 0
3 years ago
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Evaluate: 6q + m − m; use m = 8, and q = 3
adelina 88 [10]

Answer:

your answer is 18

Step-by-step explanation:

4 0
3 years ago
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Find two consecutive even integers such that the sum of the larger and twice the smaller is 62
kolbaska11 [484]
20 and 22

You would create an equation that would simulate this situation and solve for x to find the first even integer. Then add 2 to x to find the second even integer.

4 0
3 years ago
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