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2 years ago

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If the data below were represented with a stem-and -leaf plot, a stem of 3 would have how many leaves? 53, 34, 16, 93, 72, 30, 2

3, 48, 31, 83 A. 4, B. 5, C. 2, D. 3

1 answer:

Nikitich [7]2 years ago

3 0

Answer:

D) 3

Step-by-step explanation

The stem of 3 would have 3 leaves of (0, 1, and 4) from the numbers -> (34, 30, and 31)

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Answer:

$SA=54\pi\ units^2$

Step-by-step explanation:

we know that

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$SA=\pi r^{2}+\pi rl$

where

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substitute

$SA=\pi (3)^{2}+\pi (3)(15)$

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Answer:

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2 years ago

How do i find each product by factoring the tens? 3x2, 3x20, and 3x200

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2 years ago

Read 2 more answers

Liono4ka [1.6K]

Answer:

Part 1: The polygon ABCDE reflected across y-axis to get the polygon MNOPQ. So, the polygon ABCDE congruent to polygon MNOPQ.

Part 2: The vertices of polygon VWXYZ are V(1, 3), W(-1, 7), X(-7, 7), Y(-5, 3), and Z(-3, 1).

Step-by-step explanation:

Part 1:

The vertices of the polygon ABCDE are A(2, 8), B(4, 12), C(10, 12), D(8, 8), and E(6, 6).

The vertices of the polygon MNOPQ are M(-2, 8), N(-4, 12), O(-10, 12), P(-8, 8), and Q(-6, 6).

We need to find the transformation or sequence of transformations that can be performed on polygon ABCDE to show that it is congruent to polygon MNOPQ.

The relation between the vertices of ABCDE and MNOPQ are defined as

$(x,y)\rightarrow (-x,y)$

It means the polygon ABCDE reflected across y-axis to get the polygon MNOPQ. So, the polygon ABCDE congruent to polygon MNOPQ.

Part 2:

If polygon MNOPQ is translated 3 units right and 5 units down, then

$(x,y)\rightarrow (x+3,y-5)$

$M(-2,8)\rightarrow V(-2+3,8-5)=V(1,3)$

$N(-4,12)\rightarrow W(-4+3,12-5)=W(-1,7)$

$O(-10,12)\rightarrow X(-10+3,12-5)=X(-7,7)$

$P(-8,8)\rightarrow Y(-8+3,8-5)=Y(-5,3)$

$Q(-6,6)\rightarrow Z(-6+3,6-5)=Z(-3,1)$

Therefore the vertices of polygon VWXYZ are V(1, 3), W(-1, 7), X(-7, 7), Y(-5, 3), and Z(-3, 1).

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Answer:

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Step-by-step explanation:

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2 years ago

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