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2 years ago

Three prime numbers that can be multiplied to get 825

Mathematics

2 answers:

Leona [35]2 years ago

5 0

Answer:

Prime factorization: 825 = 3 × 5 × 5 × 11, which can be written 825 = 3 × 5² × 11.

Step-by-step explanation:

Hope this helps!

Ksivusya [100]2 years ago

4 0

Answer: 3, 5, and 11

3 * 5 * 5 * 11

Step-by-step explanation:

Make a factor tree for 825

825

/ \

5 165

/ \

5 33

/ \

3 11

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irina [24]

Answer:

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Step-by-step explanation:

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3 years ago

Point J is located at (-4,4).<br> Select all of the following that are 4 units from point J.

lara [203]

Answer:

<h2><em>All of the above!</em></h2>

Step-by-step explanation:

They all are 4 points away from (-4, 4)..

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4 0

3 years ago

The perimeter of a rectangle is 28 cm and<br>its length is 8 cm. Find its breadth.

Yakvenalex [24]

Answer:
The width of the rectangle is 6cm.

Explanation:
Perimeter=2(length+width)
28=2(8+w)
Distribute 2 into the parentheses
28=16+2w
Combine like terms
28-16=2w
12=2w
Divide both sides by 2
6=w

I hope this helps! Please comment if you have any questions.

7 0

3 years ago

2(3/5x+3). How would I get that answer

Airida [17]

Answer: $\frac{6}{5}$ x + 6

Step-by-step explanation:

We will use the order of operations, also known as PEMDAS. See attached. Please note that we cannot do what is in the parentheses first because we have a variable, so we will distribute the 2 into it. Since this is an expression, I do not believe we can fully solve for x, but can only simplify.

Given:

2(3/5x+3)

Distribute:

2 * (3/5x) + (2 * 3)

6/5x + 6

Answer:

$\frac{6}{5}$ x + 6

5 0

2 years ago

3x^2 - 4x = -2<br> Solve

amid [387]

Answer:

$x = \bigg \{\frac{ 2 - \sqrt{2} \: i }{3}, \: \: \frac{ 2 + \sqrt{2} \: i }{3} \bigg \}$

Step-by-step explanation:

$3 {x}^{2} - 4x = - 2 \\ 3 {x}^{2} - 4x + 2 = 0 \\ equating \: it \: with \\ a {x}^{2} + bx + c = 0 \\ a = 3 \: \: b = - 4 \: \: c = 2 \\ {b}^{2} - 4ac \\ = {( - 4)}^{2} - 4 \times 3 \times 2 \\ = 16 - 24 \\ = - 8 \\ \ {b}^{2} - 4ac < 0 \\ \therefore \: given \: quadratic \: equation \: have \: \\ imaginary \: solutios. \\ \\ x = \frac{ - b \pm \sqrt{{b}^{2} - 4ac } }{2a} \\ = \frac{ - ( - 4) \pm \sqrt{ - 8} }{2 \times 3} \\ = \frac{ 4 \pm 2\sqrt{2} \: i }{2 \times 3} \\ = \frac{ 2 \pm \sqrt{2} \: i }{3} \\ \therefore \: x = \frac{ 2 - \sqrt{2} \: i }{3} \: or \: x = \frac{ 2 + \sqrt{2} \: i }{3} \: \\ \\ x = \bigg \{\frac{ 2 - \sqrt{2} \: i }{3}, \: \: \frac{ 2 + \sqrt{2} \: i }{3} \bigg \}$

7 0

3 years ago