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2 years ago

Image Which of the following statements is true for ∠a and ∠b in the diagram?

Mathematics

1 answer:

saveliy_v [14]2 years ago

3 0

Answer:

You have not added a picture soi can not help answer this question.

Step-by-step explanation:

Add the picture by either taking a photo or screenshot of the question then upload it.

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Solve 2/3 plus 5/6 and put answer in simplest form. A 9/6 B 2/3 C 7/6 D 1 1/2

klio [65]

Answer:

A. is the answer

Step-by-step explanation:

5 0

3 years ago

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Lanny’s credit card has an apr of 33%, calculated on the previous monthly balance. His credit card record for the last 7 months

Softa [21]

The answer is A ( $83.39 )

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3 years ago

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Use p = 2 and p = 8 to determine if the two expressions are equivalent. 4(2 + p) 14 + p Complete the statements.

Maru [420]

Answer:

The expressions are not equivalent.

Step-by-step explanation:

Here we have two expressions:

4*(2 + p)

14 + p

We want to evaluate these two expressions in two different values of p to check if the expressions are equivalent or not.

First, we evaluate both of them in p = 2, this is just replace p by 2 in each expression and then solve it.

4*(2 + 2) = 4*4 = 16

14 + 2 = 16

In this case, we can see that both expressions yield the same number, so we could think that the expressions are equivalent, now let's try with other value of p.

Now let's do it with p = 8

4*(2 + 8) = 4*10 = 40

14 + 8 = 22

Now we can see that the results are different, then we can conclude that the expressions are not equivalent.

4 0

3 years ago

What does DE equal to ?

guajiro [1.7K]

The answer to your question should be EF

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3 years ago

The probability of an event - picking teams. About (a) 10 kids are randomly grouped into an A team with five kids and a B team w

LuckyWell [14K]

Answer: The required probability is $\dfrac{4}{9}$

Step-by-step explanation:

Since we have given that

Number of kids = 10

Number of kids in Team A = 5

Number of kids in Team B = 5

There are three kids in the group, Alex and his two best friends Jose and Carl.

So, number of favourable outcome is given by

$2(\dfrac{8!}{3!\times 5!})$

Total number of outcomes is given by

$\dfrac{10!}{5!\times 5!}$

So, the probability that Alex ends up on the same team with at least one of his two best friends is given by

$\dfrac{2(\dfrac{8!}{5!\times 3!)}}{\dfrac{10!}{5!\times 5!}}\\\\=2\times \dfrac{8!}{3!}\times \dfrac{5!}{10!}\\\\=\dfrac{4}{9}$

Hence, the required probability is $\dfrac{4}{9}$

3 0

3 years ago