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2 years ago

Please can someone help??????

Mathematics

1 answer:

bearhunter [10]2 years ago

6 0

Answer: I’m not sure if I’m right but I think it’s C because there’s only 2 triangles and only one has the same measurement

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There are 45 students in a speech contest. Yesterday, 1/3 of them gave their speeches. Today, 2/5 of the remaining students gave

viva [34]

1/3 of 45 is 15, meaning 15 students gave their speech on day 1

Today 2/5 of the remaining students gave their speech. That’s 12 more

So a total of 27 students gave their speeches. That means 18 students remain. The answer is 18

7 0

2 years ago

Winston typically spends about $200 per year playing the lottery. If he took that same

Dvinal [7]

Answer:

point!!!!!!!

Step-by-step explanation:

8 0

3 years ago

An animal shelter has 9 puppies. If the puppies are 36% of the total dog and cat population, how many dogs and cats are in the a

Luda [366]

Answer:

Let total number of dog and cat population in animal shelter be x.

As per the statement:

An animal shelter has 9 puppies.

⇒ Total number of puppies = 9

It is also given that if the puppies are 36% of the total dog and cat population.

⇒ $9 = 36\% \text{of}{\text{total number of dogs and Cat}$

$9 = \frac{36}{100} \times x$

By cross multiply we get;

$900 = 36 x$

Divide both sides by 36 we get;

$25 = x$

Therefore, total number of dogs and cat population in animal shelter are 25

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3 years ago

Write an inequality to represent the graph.

Westkost [7]

Answer:

y ≥ −4x − 2

Step-by-step explanation:

8 0

3 years ago

In a parallelogram ABCD point K belongs to diagonal BD so that BK:DK=1:4. If the extension of AK meets BC at point E, what is th

olya-2409 [2.1K]

Answer:

$\frac{BE}{EC} =\frac{1}{3}$

Step-by-step explanation:

In the diagram below we have

ABCD is a parallelogram. K is the point on diagonal BD, such that

$\frac{BK}{CK} =\frac{1}{4}$

And AK meets BC at E

now in Δ AKD and Δ BKE

∠AKD =∠BKE ( vertically opposite angles are equal)

since BC ║ AD and BD is transversal

∠ADK = ∠KBE ( alternate interior angles are equal )

By angle angle (AA) similarity theorem

Δ ADK and Δ EBK are similar

so we have

$\frac{AD}{BE} =\frac{DK}{BK}$

$\frac{AD}{BE} =\frac{4}{1}$

$\frac{BC}{BE}=\frac{4}{1}$ ( ABCD is parallelogram so AD=BC)

$\frac{BE+EC}{BE}=\frac{4}{1}$ ( BC= BE+EC)

$\frac{BE}{BE} +\frac{EC}{BE}=\frac{4}{1}$

$1+\frac{EC}{BE}=4$

$\frac{EC}{BE}=3$ ( subtracting 1 from both side )

$\frac{EC}{BE}=\frac{3}{1}$

taking reciprocal both side

$\frac{BE}{EC} =\frac{1}{3}$

8 0

3 years ago