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4 years ago

Prime factorization of 39

1 answer:

4 0

Lets quickly run through the prime numbers and see what it is not divisible by

39 is not divisible by 2
39 IS divisible by 3

So lets stop right here.

What is 39 divided by 3? 13 !

And 13 is a prime number too!

So that is all we can do.

So prime factorization of 39 is <u>3 x 13</u><u><em /></u>

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What is p=6-2q, p=1+q, how do I use on graph

Novosadov [1.4K]

1] ⬇
p - (6 - 2(q))➡0
p + 2q - 6➡0
q = 3 because 6÷2 = 0
p = 6 because 6 ÷ 1 = 6
Slope ➡ 0.5
-----------------------------------------------------------------
2] ⬇
p - (1 + q)➡0
p - q- 1➡0
Q = -1 because 1 ÷ -1 ='-1
P = 1 because 1 ÷ 1 = 1
Slope ➡ 1

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4 years ago

Greta is trying to determine the portion of green candies in various bags of green and yellow candies. Using the information bel

IrinaK [193]

Answer: a) $\dfrac{1}{3}$ b) $\dfrac{71}{100}$ c) $\dfrac{5}{9}$

Step-by-step explanation:

Since we have given that

There are green and yellow candies in each bag.

Bag A: Two thirds of the candies are yellow. What portion of the candies is green?

Part of yellow candies in bag A = $\dfrac{2}{3}$

Part of green candies in bag A would be

$1-\dfrac{2}{3}\\\\=\dfrac{3-2}{3}\\\\=\dfrac{1}{3}$

Bag B: 29 % of the candies are yellow. What portion of the candies is green?

Percentage of candies are yellow = 29%

Portion of candies are green is given by

$1-\dfrac{29}{100}\\\\=1-0.29\\\\=0.71\\\\=\dfrac{71}{100}$

Bag C: 4 out of every 9 candies are yellow. What portion of the candies is green?

Portion of yellow candies = $\dfrac{4}{9}$

Portion of green candies would be

$1-\dfrac{4}{9}\\\\=\dfrac{9-4}{9}\\\\=\dfrac{5}{9}$

Hence, a) $\dfrac{1}{3}$ b) $\dfrac{71}{100}$ c) $\dfrac{5}{9}$

6 0

3 years ago

To find the extreme values of a function f(x,y) on a curve xequalsx(t), yequalsy(t), treat f as a function of the single vari

pychu [463]

Answer:

Absolute maximum is 2

Absolute minimum at -2

Step-by-step explanation:

The given parametric functions are:

$x=2\cos t,y=2\sin t$

By the chain rule:

$f'(t)=\frac{\frac{dy}{dt} }{\frac{dx}{dt} }$

$\frac{df}{dt} =\frac{2 \cos t}{-2\sin t} =-\cot(t)$

At fixed points, $f'(t)=0$

$\implies -\cot (t)=0$

This gives $t=\frac{\pi}{2} ,\frac{3\pi}{2}$ on $0\le t\le 2\pi$

This implies that the extreme points are $(2\cos \frac{\pi}{2}, 2\sin \frac{\pi}{2})=(0,2)$ and $(2\cos \frac{3\pi}{2}, 2\sin \frac{3\pi}{2})=(0,-2)$

By eliminating the parameter, we have $x^2+y^2=4$

This is a circle with radius 2, centered at the origin.

Hence (0,2) is an absolute maximum ,at $t=\frac{\pi}{2}$ and (0,-2) is an absolute minimum at $t=\frac{3\pi}{2}$

7 0

3 years ago

How much greater is 4/9 than 2/9. Please answer.

kodGreya [7K]

Step-by-step explanation:

A better way to word this is how much more is 4/9 than 2/9 which is 2/9 more because 2/9 plus 2/9 is 4/9 and the denominator stays the same

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2 years ago

Which sequence of transformations will verify that EFGH and quadrilateral EFGH and quadrilateral E'F'G'H are congruent?

levacccp [35]

D) EFGH moved onto E'F'G'H after rotating 180 counterclockwise around the origin and the reflecting across the y-axis.

<h3>How to carry out transformations?</h3>

From online resources gotten about this question, for quadrilateral EFGH and quadrilateral E'F'G'H to be congruent, what we must do first is to rotate 180° counterclockwise around the origin and then move EFGH onto E'F'G'H'.

The last step to get this proof of congruency is to reflect across the y-axis.

Read more about transformations at; brainly.com/question/4289712

#SPJ1

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2 years ago