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3 years ago

Trapezoid ABCD was dilated to create trapezoid A'B'C'D'.

Mathematics

2 answers:

Firdavs [7]3 years ago

3 0

Answer:

True, True, False, False, True

Step-by-step explanation:

Trapezoid ABCD was dilated to create trapezoid A'B'C'D'.

Scale factor is 1/2 according to the coordinates.

<h3>The answer options</h3>

The length of side AD is 8 units.

AD = 4 - (-4) = 8
True

The length of side A'D' is 4 units.

A'D' = 2 - (-2) = 4
True

The image is larger than the pre-image.

False. It is 2 times smaller.

Sides CD and C'D' both have the same slope, 2.

Slopes of CD and C'D' are same: m = (0 - 4)/(4 - 2) = -4/2 = -2

False

The scale factor is One-half.

True

bekas [8.4K]3 years ago

3 0

Answer:

A,B,E

Step-by-step explanation:

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Answer:

Hello the answer is

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Round 82290 times 240

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19,749,600 is the correct answer. according to calculator

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3 years ago

A 55m and 35m broad park is surrounded by a 2.5m wide path.

daser333 [38]

Step-by-step explanation:

☄ $\underline {\underline{ \text{Given}}} :$

Length of a park ( l ) = 55 m
Breadth of a park ( b ) = 35 m
Width running outside the park ( d ) = 2.5 m
Rate of paving the path with stones = Rs 120 per sq.metre

☄ $\underline{ \underline{ \text{To \: find}}}:$

Area of the park
Cosy of paving the path with stones at Rs 130 per sq.metre

☄ $\underline{ \underline{ \text{Solution}}} :$

Part 1 : $\boxed{\text{Area \:of \:a \:path \: running \:outside = 2d(l + b + 2d)}}$

plug the known values and simplify :

⟹ $\sf{2 \times 2.5(55 + 35 + 2 \times 2.5)}$

⟹ $\sf{5 \times 95}$

⟹ $\sf{475 \: {m}^{2} }$

Part 2 : $\boxed{ \sf{Total \: cost = Area \: of \: paths \: \times Rate}}$

⟹ $\sf{475 \times 120}$

⟹ $\sf{Rs \: 57000}$

$\purple{ \boxed{ \boxed{ \sf{ \tt{⟿ \: Our \: final \: answer : (i) = 475 \: {m}^{2} \: and \: (ii ) = Rs \: 57000}}}}}$

Hope I helped ! ♡

Have a wonderful day /night ツ

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If you spend $60 for dinner at the restaurant, what should be your 15% tip?

LUCKY_DIMON [66]

Answer:

.15 * 60 = 9

Step-by-step explanation:

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a car can travel 476 miles on 14 gallons of gas. Write an equation relating the distance d to the number of gallon g. How many g

ch4aika [34]

miles per gallon (x) = d/g

x = 476/14 = x = 34 miles per gallon

so d/x = g

578/34 = 17 gallons are needed

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