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Korvikt [17]
3 years ago
8

What is part of a prime factorization of 630

Mathematics
2 answers:
Mrrafil [7]3 years ago
8 0

Answer:

<u>the prime factors of 630 are 2, 3, 5, 7. Therefore, the product of prime factors = 2 × 3 × 5 × 7 = </u><u>2</u><u>1</u><u>0</u>

Furkat [3]3 years ago
3 0

Step-by-step explanation:

Since, the prime factors of 630 are 2, 3, 5, 7. Therefore, the product of prime factors = 2 × 3 × 5 × 7 = 210.

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emmasim [6.3K]

Answer:

2^8x-3

Step-by-step explanation:

5 0
3 years ago
A) Work out the value of (square root of 5)^2 x (square root of 3)^2?
gulaghasi [49]

Answer:

15

Step-by-step explanation:

Square root of 5=√5

Square of the square root=(√5)^2

=5

Square root of 3=√3

Square of the square root=(√3)^2

=3

So (square root of 5)^2×(square root of 3)^2

5×3

15

Therefore the final answer is 15

6 0
3 years ago
Avani wrote the linear equation y = 5 x + 4. Then, she wrote the equation of the line that is perpendicular to y = 5 x + 4 and t
Margarita [4]

Avani wrote the linear equation y = 5x + 4. Then, she wrote the equation of the line that is perpendicular to y = 5x+ 4 and that passes through (15,–2). If her new equation is in the form  y-\frac{1}{5}x+b, what is the value of b?

\bf y=mx+b

  • m is the slope
  • b is the y-intercept

The equation of a line that is perpendicular to another line always has the opposite reciprocal slope. In the original equation, the slope is 5. The opposite of 5, which refers to the sign, is -5. The reciprocal of -5 is -1/5. That means the slope of the new line is -1/5.

The perpendicular line:

  • Has a slope of -\frac{1}{5}
  • Passes through the point (15, -2)

To find the equation of a line that passes through a specific point, you need to use slope-intercept form.

\bf y-y_{1} =m(x-x_{1})

  • y₁ is the y-value of the point
  • m is the slope
  • x₁ is the x-value of the point

y₁ is -2, m is -1/5,  and x₁ is 15.

Substitute the known values into the equation:

y-(-2)=-\frac{1}{5}(x-15) \rightarrow y+2= -\frac{1}{5}(x-15)

Open the parentheses and distribute:

-\frac{1}{5}(x-15) \rightarrow -\frac{1}{5} \cdot x\ +  -\frac{1}{5} \cdot -15 \rightarrow -\frac{1}{5}x +3

The new equation will be:

y+2=-\frac{1}{5}x+3

Lastly, you need to move the positive 2 to the other side to keep it in y = mx + b form.

y+2-2=-\frac{1}{5}x+3-2 \rightarrow y=-\frac{1}{5}x+1

The equation of the perpendicular line is:

\bf y=-\frac{1}{5}x+1

The value of b is 1.

5 0
3 years ago
Log (4-x)=log (x+8) + log (2x+13)\\[solve for x
dimaraw [331]

Answer:

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Step-by-step explanation:

it requires a calculator

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Two consecutive integers have a sum of 513. Find the smaller of these two numbers
lubasha [3.4K]
A+b=513
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