Ask question

Ask question

All categories

3 years ago

8

What is the center of a circle whose equation

1 answer:

miss Akunina [59]3 years ago

5 0

Come believe the answer is c

You might be interested in

What is 60/8 in simplest form?

Montano1993 [528]

Answer:

7 1/2

Step-by-step explanation:

Sven and a half once you simplify

6 0

4 years ago

Read 2 more answers

Solve the equation on the interval [0,2π). cos^4x=cos^4xcscx

omeli [17]

$\cos^4x=\cos^4x\csc x$
$\cos^4x-\cos^4x\csc x=0$
$\cos^4x(1-\csc x)=0$
$\implies \begin{cases}\cos^4x=0\\1-\csc x=0\end{cases}\implies x=\dfrac\pi2,\dfrac{3\pi}2$

7 0

4 years ago

Find the general solution of the differential equation y′=e6x−7x.

andre [41]

Consider this option:
1) dy/dx=e⁶ˣ-7x; => dy=(e⁶ˣ-7x)dx;
2) $y= \int\ {(e^{6x}-7x)} \, dx; \ =\ \textgreater \ \ y= \frac{1}{6}e^{6x}- \frac{7}{2}x^2+C.$

3 0

3 years ago

Is 83,391,946 divisible by 4?<br><br> YE or NO<br><br> HELP ME PLZ<br><br> T^T

Vanyuwa [196]

Answer:

No

Step-by-step explanation:

Reason for this is because when you divide 4 into 83,391,946 it ends up as 20847986.5 and since the number has to be whole its not divisible

4 0

3 years ago

The quadratic equation 8x²+12x-14 has two real roots. What is the sum of the squares of these roots?

mars1129 [50]

Answer:

The real roots are

$x=\frac{(-3+\sqrt{37})}{4}$ and $x=\frac{(-3-\sqrt{37})}{4}$

The sum of the squares of these roots is $\frac{-3}{2}$

Step-by-step explanation:

The given quadratic equation is $8x^2+12x-14$ has two real roots.

To find the roots .

$8x^2+12x-14=0$

Dividing the above equation by 2

$\frac{1}{2}(8x^2+12x-14)=\frac{0}{2}$

$4x^2+6x-7=0$

For quadratic equation $ax^2+bx+c=0$ the solution is $x=\frac{-b\pm\sqrt{b^2-4ac}}{2a}$

Where a and b are coefficents of $x^2$ and x respectively, c is a constant.

For given quadratic equation

a=4, b=6, c=-7

$x=\frac{-6\pm\sqrt{6^2-4(4)(-7)}}{2(4)}$

$=\frac{-6\pm\sqrt{36+112}}{8}$

$=\frac{-6\pm\sqrt{148}}{8}$

$=\frac{-6\pm\sqrt{37\times 4}}{8}$

$=\frac{-6\pm\sqrt{37}\times\sqrt{4}}{8}$

$=\frac{-6\pm\sqrt{37}\times 2}{8}$

$=2\frac{(-3\pm\sqrt{37})}{8}$

$=\frac{-3\pm\sqrt{37}}{4}$

$x=\frac{(-3\pm\sqrt{37})}{4}$

The real roots are

$x=\frac{(-3+\sqrt{37})}{4}$ and $x=\frac{(-3-\sqrt{37})}{4}$

Now to find the sum of the squares of these roots

$\left[\frac{-3+\sqrt{37}}{4}+\frac{(-3-\sqrt{37})}{4}\right]^2=\frac{-3+\sqrt{37}-3-\sqrt{37}}{4}$

$=\frac{-6}{4}$

$=\frac{-3}{2}$

$\left[\frac{-3+\sqrt{37}}{4}+\frac{(-3-\sqrt{37})}{4}\right]^2=\frac{-3}{2}$

Therefore the sum of the squares of these roots is $\frac{-3}{2}$

3 0

3 years ago