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2 years ago

On the graph, trapezoid A has rotated 180 degrees about the origin. Compare the areas of the trapezoids. Justify your answer.

Mathematics

2 answers:

Nikitich [7]2 years ago

6 0

Answer:

D) The areas are equal. A rotation is a rigid transformation, which does not change the size of the original trapezoid. Therefore, the area is preserved.

Step-by-step explanation:

Rigid transformations are transformations that maintain congruence. They do not change the size or shape of a figure.

Rigid transformations include:

Translations (slide the figure);

Reflections (create a mirror image of the figure);

Rotations (turn the figure).

Since rotations are rigid transformations, this means the measures of all of the sides will be the same; this also means the area will be the same.

telo118 [61]2 years ago

5 0

Answer:

Step-by-step explanation:

they are equal

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I need help please with this question

uranmaximum [27]

Answer:

around like 33.50$

Step-by-step explanation:

All you have to do is take the number of hours times the amount of money per hour ( so like the equation would look like y=5x) and since the 45 minutes isn't a full hour it shouldn't matter. Just in case though we'll consider the 45 minutes and make it a .7 and multiply that by 5 as well and we get 3.5. Now we just take 5*6+3.5 and bang.

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2 years ago

Which of the following statements is true of a transversal?

kolbaska11 [484]

Answer:

The answer here is the third box(C)

I hope this helps you!

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2 years ago

Please help me, worth LOTS of points

Aliun [14]

Answer:

a) $91\,+\,0.14 \,x \leq 140$ where " $x$ " stands for the number of miles driven.

b) He can drive as far as 250 miles to keep the rental cost limited to $140.

Step-by-step explanation:

a) Robert wants to make sure that the addition of the costs coming from the car rental per week ($91) plus the amount paid for the coverage of "x" number of miles (which goes as $0.14 times x) does not exceed $140 (which is the same as saying that this total cost must be smaller than or equal to $140.

In math terms, such is written as:

$91\,+\,0.14 \,x \leq 140$

where "x" stands for the number of miles driven.

b) the total number of miles (x) he is allowed to cover given the $140 restriction is obtained by solving for "x" (the number of driven miles) in the inequality of part a):

$91\,+\,0.14 \,x \leq 140\\0.14\,x\leq 140-91\\0.14\,x\leq 49\\x\leq \frac{49}{0.14} \\x\leq 350$

which tells us that the number of driven miles (x) has to be smaller or equal to 350 miles. Then he can drive as far as 250 miles to keep the rental cost limited to $140.

4 0

2 years ago

The coefficient of 8.2 N is 8 2 16

Aliun [14]

Answer:

Step-by-step explanation:

A coefficient is the number placed before the variable. The variable gets multiplied by the coefficient.

The format is not exactly clear with the expression in the question.

Assuming the expression is '8 * 2n', the coefficient would be 2.

'n' is getting multiplied by '2'.

'8' does not have a variable next to it, so it is just a term in the expression.

The product of 8 and 2 is 16. 16 is not in the expression.

If the expression was '8.2n', 8.2 would be the coefficient.

Brainilest Appreciated.

4 0

2 years ago

Read 2 more answers

PLEAZZZZZZZZZZ HELP ASAP

stich3 [128]

QUESTION 1

GIVEN:

$long \: base = 3 \times (short \: base)$

$height = (short \: base) - 2$

$area = 30 {in}^{2}$

IMPORTANT EQUATIONS

$trapezoid \: area = \frac{sum \: of \: bases}{2} \times height$

SOLVE:

$30 = \frac{short + long}{2} (short - 2)$
$30 = \frac{(short + 3short)}{2} \times (short - 2)$

$30 = \frac{4short}{2} (short - 2)$

$30 = 2{(short)}^{2} - 4(short)$

${short}^{2} - 2(short) - 15 = 0$

factor:

$(short - 5)(short + 3) = 0$

$short = 3 \: and \: - 5$
Negative five is not a reasonable answer, so we focus on the positive three and say that is the length of the short base.

so, the answers:

$short \: base = 3 \: in$

$long \: base = 3 \times 3 = 9 \: in$

$height = 3 - 2 = 1 \: in$

QUESTION 2

GIVEN
convert 3ft to inches.

$l = 32 + w$

$area = lw = 130 {in}^{2}$
or

$w = \frac{130}{l}$

substitute

$l = 32 + \frac{130}{l}$

solve for l:

${l}^{2} - 32 l- 130 = 0$
QUESTION 3

$x(x + 2) = 80$

solve for X

${x}^{2} + 2x - 80 = 0$
$(x - 10)(x + 8) = 0$

$x = 10 \: and \: - 8$

take the positive value since it's the only one that makes sense in this context.

so the two values are:

$10 \: and \: 8$

8 0

2 years ago