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3 years ago

Which triangles must be congruent?

Mathematics

1 answer:

blsea [12.9K]3 years ago

6 0

Answer:

Δ ABC ≅ Δ FDE ≅ Δ GIH

Step-by-step explanation:

Between Δ ABC and Δ FDE, given that

(i) AB = DF

(ii) BC = HI and DE = HI, hence, BC = DE

(iii) ∠ B = ∠ D

Hence, by Side-Angle -Side i.e. SAS criteria Δ ABC ≅ Δ FDE

Again, between Δ ABC and Δ GIH, given that

(i) AB = GI

(ii) BC = HI and

(iii) ∠ B = ∠ I.

Hence, by Side-Angle-Side i.e. SAS criteria Δ ABC ≅ Δ GIH.

Therefore, Δ ABC ≅ Δ FDE ≅ Δ GIH (Answer)

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The answer is A) measures things using quantities or numbers. Hope this helps!

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2 years ago

What is 0.0592 in scientific notation

DiKsa [7]

Answer:

5.92 times 10 to the 2nd power.

Step-by-step explanation:

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3 years ago

Read 2 more answers

The temperature at 5pm is 20 degrees. The temperature at 10pm is -5 degrees. What is the equation

marusya05 [52]

The equation is y=-5x + 20, x being the hours that go by after 5pm.

8 0

2 years ago

you are buying wood to build a gate for your backyard. if your car trunk has the following dimensions (47 inches by 33 inches by

riadik2000 [5.3K]

The longest piece of wood that could fit in your trunk to build a gate for the backyard is 60.65 inches.

<h3>How to measure diagonal of the cuboid?</h3>

The length of the diagonal of the cuboid is found out using the following formula.

$D=\sqrt{l^2+b^2+h^2}$

Here, (l) is the length of the cuboid, (w) is the width, and (h) is the height of the cuboid.

The wood has to buy to build a gate for backyard. The car trunk has the following dimensions (47 inches by 33 inches by 19.5 inches). Here, we have,

Length (l)=47 in
Width (w)=33 in
Height (h)=19.5 in

The longest piece of wood which fits is equal to the length of diagonal of cuboid. Put the values in the formula,

$D=\sqrt{47^2+33^2+19.5^2}\\D=\sqrt{3678.25}\\D=60.65\rm\; in$

Thus, the longest piece of wood that could fit in your trunk to build a gate for the backyard is 60.65 inches.

Learn more about the cuboid here;

brainly.com/question/22694657

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1 year ago

A box at a yard sale contains 3 different china dinner sets, each consisting of 5 plates. A customer will randomly select 2 plat

marshall27 [118]

Answer:

$\dfrac{2}{7}$

Step-by-step explanation:

3 different china dinner sets, each consisting of 5 plates consist of 15 plates.

A customer can select 2 plates in

$C^{15}_2=\dfrac{15!}{2!(15-2)!}=\dfrac{15!}{13!\cdot 2!}=\dfrac{13!\cdot 14\cdot 15}{2\cdot 13!}=7\cdot 15=105$

different ways.

2 plates can be selected from the same dinner set in

$3\cdot C^5_2=3\cdot \dfrac{5!}{2!(5-2)!}=3\cdot \dfrac{3!\cdot 4\cdot 5}{2\cdot 3!}=3\cdot 2\cdot 5=30$

different ways.

Thus, the probability that the 2 plates selected will be from the same dinner set is

$Pr=\dfrac{30}{105}=\dfrac{6}{21}=\dfrac{2}{7}$

7 0

3 years ago