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nikitadnepr [17]

2 years ago

13

Whats 326482745445-453637+938464846-3845+784657

2 answers:

Burka [1]2 years ago

8 0

327421537466 and yes i did all of this that fast

lubasha [3.4K]2 years ago

7 0

When you do the first step which would be subtracting the 453637 from 326482745445 you get 3.264822918(x10) to the 11 power. Then you add 938464846 you get 3.274207565(x10) to the 11 power. Then you subtract 3845 from that and get 3.274207528(x10) to the 11 power. Finally you add 784657 and get 3.274215375(x10) to the 11 power.

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Masteriza [31]

Answer:

It's letter C

Step-by-step explanation:

6 0

3 years ago

If the endpoints of the diameter of a circle are (−8, −6) and (−4, −14), what is the standard form equation of the circle?

kondaur [170]

Equation of the circle is $(x+6)^{2}+(y+10)^{2}=20$ .

Solution:

The endpoints of the diameter of a circle are (–8, –6) and (–4, –14).

Center of the circle = Mid point of the diameter

Mid point formula:

$$P(x, y)=\left(\frac{x_{1}+x_{2}}{2}, \frac{y_{1}+y_{2}}{2}\right)$

Here, $x_1=-8, y_1=-6, x_2=-4, y_2=-14$

$$P(x, y) =\left(\frac{-8-4}{2}, \frac{-6-14}{2}\right)$

$$P(x, y) =\left(\frac{-12}{2}, \frac{-20}{2}\right)$

$$P(x, y) =(-6, -10)$

Center of the circle = (–6, –10)

Radius is the distance between center and any endpoint of the diameter.

To calculate the radius using distance formula.

$r=\sqrt{\left(x_{2}-x_{1}\right)^{2}+\left(y_{2}-y_{1}\right)^{2}}$

Here, $x_1=-6, y_1=-10, x_2=-8, y_2=-6$

$r=\sqrt{\left(-8-(-6)\right)^{2}+\left(-6-(-10)}\right)^{2}}$

$r=\sqrt{(-8+6)^{2}+(-6+10)^{2}}$

$r=\sqrt{(-2)^{2}+(4)^{2}}$

$r=\sqrt{20}$ units

The standard form of the equation of a circle is

$(x-a)^{2}+(y-b)^{2}=r^{2}$ , where (a, b) are center and r is the radius.

Here, center = (–6, –10) and $r=\sqrt{20}$

$(x-(-6))^{2}+(y-(-10))^{2}={(\sqrt{20})} ^{2}$

$(x+6)^{2}+(y+10)^{2}=20$

Equation of the circle is $(x+6)^{2}+(y+10)^{2}=20$ .

4 0

2 years ago

HelloI need some assistance with this question. I provided an answer, but it was incorrect. The images are enclosed.

ycow [4]

Given:

There are given that the zeroes and degrees of the polynomial:

$\begin{gathered} \text{zeros:}-2,2,4 \\ \text{Degres:}3 \end{gathered}$

Explanation:

From the concept of a polynomial:

A polynomial has a as zero if and only if (x -a) is a factor of the polynomial.

Then,

From the given polynomial:

$f(x)=(x+2)(x-2)(x-4)$

Then,

$\begin{gathered} f(x)=(x+2)(x-2)(x-4) \\ f(x)=(x^2-2x+2x-4)(x-4) \\ f(x)=(x^2-4)(x-4) \\ f(x)=x^3-4x^2-4x+16 \end{gathered}$

Final answer:

Hence, the polynomial is shown below:

$f(x)=x^3-4x^2-4x+16$

6 0

1 year ago

Expand.<br><br>Your answer should be a polynomial in standard form.<br><br>(x-1)(x+4)=

Montano1993 [528]

Answer:

X^2+3x-4

Step-by-step explanation:

This is the answer

8 0

2 years ago

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Which mixed number is equivalent to the improper fraction 55/4

Lunna [17]

The mixed number will be 13 3/4.

8 0

2 years ago

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