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3 years ago

Please give answer ofHCF of 15 and 18

1 answer:

6 0

Answer: The prime factorization of 15 is: 3 x 5. The prime factorization of 18 is: 2 x 3 x 3. The prime factors and multiplicities 15 and 18 have in common are: 3. 3 is the gcf of 15 and 18.

Step-by-step explanation:

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The equation, 3x2 + 6x + 3y2 + 7y + 4 = 0, represents a conic.

Leno4ka [110]

Answer: Circle

Step-by-step explanation:

Check the variables' degrees and their coefficients' signs to determine the type of the conic.

7 0

3 years ago

Read 2 more answers

How do I solve for x

Alchen [17]

Greetings!

To find the length of any side of a right triangle, you can use the Pythagorean Thereom. It states that the squares of two sides are equal to the square of the hypotenuse:
$a^2+b^2=c^2$

Input the information from the diagram into the formula:
$(x)^2+(x+7)^2=(13)^2$

Expand each term:
$(x)^2+(x+7)^2=(13)^2$

$x^2+((x+7)(x+7))=169$

$x^2+(x(x+7)+7(x+7))=169$

$x^2+x^2+7x+7x+49=169$

Combine like terms:
$2x^2+14x+49=169$

Add -169 to both sides:
$(2x^2+14x+49)+(-169)=(169)+(-169)$

$2x^2+14x-120=0$

Factor out the Common Term (2):
$2(x^2+7x-60)=0$

Factor the Complex Trinomial:
$2(x^2-5x+12x-60)=0$

$2(x(x-5)+12(x-5))=0$

$2(x-5)(x+12)=0$

Set Factors to equal 0:
$x-5=0$

$x=5$

or

$x+12=0$

$x=-12$

However, since we are solving for the side length, the only possible answer is 5 (a shape can't have a side with a negative length.)

The Solution Is:
$\boxed{x=5}$

I hope this helped!
-Benjamin

6 0

3 years ago

Perform the indicated operation. 4/11 * 10/8

Sliva [168]

So basically what you wanna do is multiply across giving you 40/88 which you can simplify down to 5/11
Hope this helps :D

4 0

3 years ago

Read 2 more answers

Expand in logarithmic expression

olchik [2.2K]

$ln( \frac{m^{5}}{n^{2}})^{3} =3ln( \frac{m^{5}}{n^{2}})= 3(ln({m^{5})-ln({n^{2}))=3(5ln(m)-2ln(n) = 15*ln(m)-6*ln(n)

$

The answer is 15*ln(m)-6*ln(n)

3 0

3 years ago

Our school's girls volleyball team has 14 players, including a set of 3 triplets: Missy, Lauren, and Liz. In how many ways can w

Sonja [21]

A - not all 3 triplets can be in the starting lineup

A' - all 3 triplets can be in the starting lineup

$\displaystyle |\Omega|=\binom{14}{6}=\dfrac{14!}{6!8!}=\dfrac{9\cdot10\cdot11\cdot12\cdot13 \cdot14}{2\cdot3\cdot4\cdot5\cdot6}=3003\\ |A'|=\binom{11}{3}=\dfrac{11!}{3!8!}=\dfrac{9\cdot10\cdot11}{2\cdot3}=165\\ |A|=3003-165=2838\\\\ P(A)=\dfrac{2838}{3003}=\dfrac{86}{91}\approx95\%$

3 0

3 years ago

Please give answer ofHCF of 15 and 18​

Please give answer ofHCF of 15 and 18