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3 years ago

What is the prime factor of 8

Mathematics

1 answer:

ExtremeBDS [4]3 years ago

5 0

Prime factor is a factor that is a prime number.
Prime factors of :
8 : 2 × 2 × 2. Also can be written as
${2}^{3}$
12 : 2 × 2 × 3. Also can be written as
${2}^{2} \times 3$

20 : 2 × 2 × 5. Also can be written as
${2}^{2} \times 5$

30 : 2 × 3 × 5

56 : 2 × 2 × 2 ×7. Also can be written as
${2}^{3} \times 7$

70 : 2 × 5 × 7

You can find all of them by using Factor Tree [ as shown in the picture] or Dividing by prime numbers.

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Help! Give brainliest!

dsp73

Initial population=145,201

Total years=2037-2001=36

$\\ \sf\longmapsto S=P\left(1+\dfrac{r}{100}\right)^t$

$\\ \sf\longmapsto S=145201\left(1+\dfrac{5}{100}\right)^{36}$

$\\ \sf\longmapsto S=145201\left(\dfrac{105}{100}\right)^{36}$

$\\ \sf\longmapsto S=145201(1.05)^{36}$

$\\ \sf\longmapsto S=145201(5.79)$

$\\ \sf\longmapsto S\approx 840713$

4 0

3 years ago

Read 2 more answers

Using the given zero, find one other zero of f(x). Explain the process you used to find your solution.

IRISSAK [1]

Answer:

One other zero is 2+3i

Step-by-step explanation:

If 2-3i is a zero and all the coefficients of the polynomial function is real, then the conjugate of 2-3i is also a zero.

The conjugate of (a+b) is (a-b).

The conjugate of (a-b) is (a+b).

The conjugate of (2-3i) is (2+3i) so 2+3i is also a zero.

Ok so we have two zeros 2-3i and 2+3i.

This means that (x-(2-3i)) and (x-(2+3i)) are factors of the given polynomial.

I'm going to find the product of these factors (x-(2-3i)) and (x-(2+3i)).

(x-(2-3i))(x-(2+3i))

Foil!

First: x(x)=x^2

Outer: x*-(2+3i)=-x(2+3i)

Inner: -(2-3i)(x)=-x(2-3i)

Last: (2-3i)(2+3i)=4-9i^2 (You can just do first and last when multiplying conjugates)

---------------------------------Add together:

x^2 + -x(2+3i) + -x(2-3i) + (4-9i^2)

Simplifying:

x^2-2x-3ix-2x+3ix+4+9 (since i^2=-1)

x^2-4x+13 (since -3ix+3ix=0)

So x^2-4x+13 is a factor of the given polynomial.

I'm going to do long division to find another factor.

Hopefully we get a remainder of 0 because we are saying it is a factor of the given polynomial.

x^2+1

---------------------------------------

x^2-4x+13| x^4-4x^3+14x^2-4x+13

-( x^4-4x^3+ 13x^2)

------------------------------------------

x^2-4x+13

-(x^2-4x+13)

-----------------

So the other factor is x^2+1.

To find the zeros of x^2+1, you set x^2+1 to 0 and solve for x.

$x^2+1=0$

$x^2=-1$

$x=\pm \sqrt{-1}$

$x=\pm i$

So the zeros are i, -i , 2-3i , 2+3i

7 0

3 years ago

The 250 gallons of maple sap will be

Darya [45]

Answer:

7.5 gallons

Step-by-step explanation:

3 x 250/100

= 7.5

8 0

3 years ago

Pls help with this question

kirill115 [55]

Answer: first answer choice

Step-by-step explanation:

8 0

3 years ago

The given matrix is the augmented matrix for a linear system. Use technology to perform the row operations needed to transform t

shtirl [24]

Answer:

$x_{1} = \frac{176}{127} + \frac{71}{127}x_{4}\\\\ x_{2} = \frac{284}{127} + \frac{131}{254}x_{4}\\\\x_{3} = \frac{845}{127} + \frac{663}{254}x_{4}\\$

Step-by-step explanation:

As the given Augmented matrix is

$\left[\begin{array}{ccccc}9&-2&0&-4&:8\\0&7&-1&-1&:9\\8&12&-6&5&:-2\end{array}\right]$

Step 1 :

$r_{1}$ ↔ $r_{1} - r_{2}$

$\left[\begin{array}{ccccc}1&-14&6&-9&:10\\0&7&-1&-1&:9\\8&12&-6&5&:-2\end{array}\right]$

Step 2 :

$r_{3}$ ↔ $r_{3} - 8r_{1}$

$\left[\begin{array}{ccccc}1&-14&6&-9&:10\\0&7&-1&-1&:9\\0&124&-54&77&:-82\end{array}\right]$

Step 3 :

$r_{2}$ ↔ $\frac{r_{2}}{7}$

$\left[\begin{array}{ccccc}1&-14&6&-9&:10\\0&1&-\frac{1}{7} &-\frac{1}{7} &:\frac{9}{7} \\0&124&-54&77&:-82\end{array}\right]$

Step 4 :

$r_{1}$ ↔ $r_{1} + 14r_{2}$ , $r_{3}$ ↔ $r_{3} - 124r_{2}$

$\left[\begin{array}{ccccc}1&0&4&-11&:-8\\0&1&-\frac{1}{7} &-\frac{1}{7} &:\frac{9}{7} \\0&0&- \frac{254}{7} &\frac{663}{7} &:-\frac{1690}{7} \end{array}\right]$

Step 5 :

$r_{3}$ ↔ $\frac{r_{3}. 7}{254}$

$\left[\begin{array}{ccccc}1&0&4&-11&:-8\\0&1&-\frac{1}{7} &-\frac{1}{7} &:\frac{9}{7} \\0&0&1&-\frac{663}{254} &:-\frac{1690}{254} \end{array}\right]$

Step 6 :

$r_{1}$ ↔ $r_{1} - 4r_{3}$ , $r_{2}$ ↔ $r_{2} + \frac{1}{7} r_{3}$

$\left[\begin{array}{ccccc}1&0&0&-\frac{71}{127} &:\frac{176}{127} \\0&1&0&-\frac{131}{254} &:\frac{284}{127} \\0&0&1&-\frac{663}{254} &:\frac{845}{127} \end{array}\right]$

∴ we get

$x_{1} = \frac{176}{127} + \frac{71}{127}x_{4}\\\\ x_{2} = \frac{284}{127} + \frac{131}{254}x_{4}\\\\x_{3} = \frac{845}{127} + \frac{663}{254}x_{4}\\$

6 0

3 years ago