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3 years ago

If ∠BAC ≅ ∠DCA, is ABE ≅ CDE? Explain.

Mathematics

1 answer:

Tanzania [10]3 years ago

7 0

Answer:

The answer is A.

Step-by-step explanation:

It is A because both sides match up except for the last pair, where they are the same. Hope this helped!

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What is 15 percent of 60

maria [59]

Answer:

9 %

Step-by-step explanation:

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3 years ago

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Help I'll give extra cred if u do

Schach [20]

I think it’d be 2 and 3 if you’re rounding 2.984

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2 years ago

1. John’s age is three times that of his sister. When you add John’s age to his sister's age, you get 36. How old are John and h

iren [92.7K]

Answer:

sister = 6 years

John=18 years

Step-by-step explanation:

A(S) + 12 = 3 * A(S), subtracting A(S) from both sides one gets 12 = 2 * A(S). Now dividing both sides by 2 we get A(S) = 12/2 = 6 . Therefore, the age of the sister is 6 years when her brother John is three times as old, that is 18 years.

6 0

3 years ago

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Crawling Crayfish!

boyakko [2]

Answer: Crayfish, 25 s

Step-by-step explanation:

Given

The length of race is 60 cm

Flicker gets a 10 cm head-start

Cray fish covers 6 cm in 5 s i.e. its speed is

$\Rightarrow \dfrac{6}{5}=1.2\ cm/s$

Flicker covers 4 cm in 5 s, its speed is

$\Rightarrow \dfrac{4}{5}=0.8\ cm/s$

Time taken by Cray fish to cover the race is

$\Rightarrow t_1=\dfrac{60}{1.2}\\\\\Rightarrow t_1=50\ s$

Time taken by flicker to cover race with 10 cm head start

$\Rightarrow t_2=\dfrac{60-10}{0.8}\\\\\Rightarrow t_2=62.5\ s$

Time taken by crayfish is less. Hence, crayfish wins the race

When they both covers the same distance, they tied momentarily i.e.

$\Rightarrow 1.2t=10+0.8t\\\Rightarrow 0.4t=10\\\\\Rightarrow t=\dfrac{10}{0.4}\\\\\Rightarrow t=25\ s$

After 25 s, they tied the race.

5 0

3 years ago

Find the product of z1 and z2, where z1 = 7(cos 40° + i sin 40°) and z2 = 6(cos 145° + i sin 145°).

Oduvanchick [21]

For two complex numbers $z_1=re^{i\theta}=r(\cos\theta+i\sin\theta)$ and $z_2=se^{i\varphi}=s(\cos\varphi+i\sin\varphi)$ , the product is

$z_1z_2=rse^{i(\theta+\varphi)}=rs(\cos(\theta+\varphi)+i\sin(\theta+\varphi))$

That is, you multiply the moduli and add the arguments. You have $z_1=7e^{i40^\circ}$ and $z_2=6e^{i145^\circ}$ , so the product is

$z_1z_2=7\times6(\cos(40^\circ+145^\circ)+i\sin(40^\circ+145^\circ)=42(\cos185^\circ+i\sin185^\circ)=42e^{i185^\circ}$

3 0

3 years ago

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