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Llana [10]
3 years ago
6

I really dont get this

Mathematics
2 answers:
VikaD [51]3 years ago
7 0
D regular polygon means everything is the same, so the square would be the correct answer.
OverLord2011 [107]3 years ago
5 0
The answer is D. because regular quadrilaterals are perfect squares with equal feet, width, and height. <span />
You might be interested in
Look at all the measurements in model 4. when a number in scientific notation is changed to expanded notation, are any of the ad
zheka24 [161]
Yes, the added zeros are significant. Lets say you are converting 1.23 x 10 to the power of 4. The answer would be 12,300 in which case, if you took away the zeros, the number would be 123, which is an incorrect answer.

Hope this makes sense and helps!
4 0
3 years ago
Read 2 more answers
Mr. Smith received a phone call from the bank manager. His bank balance had dropped to - $15 (he had overdrawn his account by $1
Pani-rosa [81]

Answer:

$10

Step-by-step explanation:

(-15) + 25 = 10

6 0
3 years ago
Read 2 more answers
HURRY PLEASE When 5 and 6 are multiplied by the same factor, how do the ratios compare to the ratio 5:6? The ratios are equivale
erma4kov [3.2K]

Answer:

The ratios are equivalent.

Step-by-step explanation:

If you multiply both 5 and 6, the ratio stays the same.

5:6 = 10:12 = 15:18 = 20:24 etc.

You can try this by dividing the larger number by the smaller number. You'll always get 1.2 as the "relationship", or ratio, is preserved.

24/20 = 1.2

15/18 = 1.2

10/12 = 1.2

6/5 = 1.2

6 0
3 years ago
Can someone help me in this plssss I really need it
kompoz [17]

Answer:

\boxed{-3xy^{2}\sqrt [3] {2x^{2}}}

Step-by-step explanation:

Your expression is

\sqrt [3] {-54x^{5}y^{6}}

Here's how I would simplify it.

\begin{array}{rcll}\sqrt [3] {-54x^{5}y^{6}} & = & \sqrt [3] {(-1)^{3}\times 2 \times 27 \times x^{2} \times x^{3} \times y^{6}} & \text{Factored the cubes}\\& = & \sqrt [3] {(-1)^{3} \times 3^{3}\times x^{3} \times y^{6}\times 2 \times x^{2}} & \text{Grouped the cubes}\\\end{array}

\begin{array}{rcll}& = & \sqrt [3] {(-1)^{3} \times {3^{3}\times x^{3} \times y^{6}}} \times\sqrt [3] { 2 \times x^{2}} & \text{Separated the cubes}\\&=& \mathbf{-3xy^{2}\sqrt [3] {2x^{2}}} & \text{Took cube roots}\\\end{array}

\text{The simplified expression is $\boxed{\mathbf{-3xy^{2}\sqrt [3] {2x^{2}}}}$}

6 0
3 years ago
PLEASE HELP ME FAST!!14 MINUTES LEFT
Irina18 [472]

Answer:

The answer to your question is:  B and D are correct

Step-by-step explanation:

Remember that the sum of the internal angles in a triangle equals 180°.

The sum of supplementary angles equals 180°

Then

                (6x + 1) + (4x + 5) + [180 - (8x + 30)] = 180

                6x + 1 + 4x + 5 + [180 - 8x - 30] = 180

                10x + 6 + 150 - 8x = 180

                2 x = 180 - 156

                2x = 24

                x = 24/2

                x = 12

∠JKL = 4(12) + 5 = 48 + 5 = 53

∠KJL = 6(12) + 1 = 72 + 1 = 73

∠KLJ = 180 - 73 - 53 = 54

∠KLM = 126

7 0
3 years ago
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