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2 years ago

Find the HCF of 15 and 10945

Mathematics

2 answers:

densk [106]2 years ago

8 0

Answer:

Step-by-step explanation:

GCF is the greatest common factor; it's basically the biggest number that can divide the two numbers given. Here, our two numbers are 15 and 10945. In order to figure out their GCF, we need to find each of these two numbers' prime factorization. Prime factorization is when we break up numbers into the product of their prime factors.

15 = 3 * 5 (3 and 5 are both prime numbers)

10945 = 5 * 11 * 199 (5, 11, and 199 are all prime numbers)

Just looking at these prime factorizations, we can tell that the only, and biggest, number they share is 5.

So, 5 is the GCF.

Hope this helps!

baherus [9]2 years ago

3 0

Answer:

Step-by-step explanation:

15 : 3×5

10945: 5 × 11 × 199

HCF: 5

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Ethan bought 4 packages of pencils. After he gave 8 pencils to his friends, he had 40 pencils

LiRa [457]

He had 40 pencils left after he gave away 8, so originally he had 40 + 8 pencils, which is 48.

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To determine the solution arithmetically, first add 8 to 40, then divide 48 by 4.

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2 years ago

Seth is using the figure shown below to prove the Pythagorean Theorem using triangle similarity: In the given triangle PQR, angl

Naily [24]

Answer:

Part A: $\triangle RPQ \sim \triangle RSP$

Part B. All angles are same, so the triangles are similar.

Part C. RP = 8

Step-by-step explanation:

We are given a right angled triangle $\triangle RPQ$ with $\angle P = 90^\circ$ .

PS is perpendicular to the hypotenuse RQ of $\triangle RPQ$ and S lies on RQ.

Part A:

To identify the pair of similar triangles.

$\triangle RPQ \sim \triangle RSP$ .

Part B:

To identify the type of similarity.

Kindly refer to the image attached in the answer area.

Let us consider the triangles $\triangle RPQ \ and\ \triangle RSP$ .

$\angle RSP =\angle RPQ =90^\circ$

Also, $\angle R$ is common to both the triangles under consideration.

Now, we can see that two angles of two triangles are equal.

So, third angle of the two triangles will also be same.

i.e. All three angles of two triangles $\triangle RPQ \ and\ \triangle RSP$ are equal to each other.

So, by A-A-A (Angle - Angle - Angle) similarity, we can say that $\triangle RPQ \sim \triangle RSP$ .

Part C:

RS = 4

RQ = 16, Find RP.

There is one property of similar triangles that:

The ratio of corresponding sides of two similar triangles is equal.

i.e.

$\dfrac{RS}{RP} = \dfrac{RP}{RQ}\\\Rightarrow RP ^2 = RS \times RQ\\\Rightarrow RP ^2 = 4 \times 16\\\Rightarrow RP ^2 = 64\\\Rightarrow \bold{RP = 8\ units}$

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2 years ago

Which of these conditions might be true if polygons ABCD and KLMN are similar? The measures of corresponding angles of ABCD and

AlexFokin [52]

Answer: The correct option is

(A) The measures of corresponding angles of ABCD and KLMN are equal, but the lengths of corresponding sides of ABCD are half those of KLMN.

Step-by-step explanation: We are given to select the correct condition that might be true if polygons ABCD and KLMN are similar.

Two polygons are said to be SIMILAR if corresponding angles are congruent and corresponding sides are proportional.

So, options (B), (C), (D) and (E) are not correct because they contradict conditions of similarity.

In option (A), we have

The measures of corresponding angles of ABCD and KLMN are equal. So, they must be congruent.

And the lengths of corresponding sides of ABCD are half those of KLMN.

So, we can write

$AB=\dfrac{1}{2}KL,~~BC=\dfrac{1}{2}LM,~~CD=\dfrac{1}{2}MN,~~AD=\dfrac{1}{2}KN\\\\\\\Rightarrow \dfrac{AB}{KL}=\dfrac{BC}{LM}=\dfrac{CD}{MN}=\dfrac{AD}{KN}=\dfrac{1}{2}.$