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2 years ago

30 POINTS SEND ASAP OR I WONT SEE IT! Do both

Mathematics

1 answer:

Kobotan [32]2 years ago

5 0

Answer:

ITS "B"

Step-by-step explanation:

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What are the factors of x2 – 100?

Nadya [2.5K]

Answer:

${x }^{2} - 100$

${x}^{2} - {10}^{2}$

$(x - 10)(x + 10)$

this is the answer

hope this is helpful for you.

3 0

2 years ago

What is the length of the diameter of a circle if the endpoints of the diameter are at the points (4, 2) and (–6, –1)?

Oksana_A [137]

Answer:

d= 10.44030651

Step-by-step explanation:

The diameter is the length between the endpoints. We can find it using the distance formula.

d= sqrt((x2-x1)^2+(y2-y1)^2 )

d = sqrt((-6-4)^2+ (-1-2)^2)

d = sqrt((-10)^2+(-3)^2)

d= sqrt(100+9)

d = sqrt(109)

d= 10.44030651

3 0

2 years ago

Somebody help me please ASAP!!!!!!!!!!!!!!

mixer [17]

Answer:

V=-15

Step-by-step explanation:

What you do is you take the negative 3 and multiply it by the 5 at the top which makes -15/v

7 0

3 years ago

Read 2 more answers

If I'm 16 years old, how many months, days, hours, minutes, and seconds old am I?

irakobra [83]

16*12=months
16*365=days
Then take that answer and multiply it by 24 for hours then take that number and multiply it by 60 for minutes and then take that number and multiply it by 60 for seconds.

8 0

2 years ago

Solve the equation on the interval [0,2π]

WARRIOR [948]

$\bf 16sin^5(x)+2sin(x)=12sin^3(x)
\\\\\\
16sin^5(x)+2sin(x)-12sin^3(x)=0
\\\\\\
\stackrel{common~factor}{2sin(x)}[8sin^4(x)+1-6sin^2(x)]=0\\\\
-------------------------------\\\\
2sin(x)=0\implies sin(x)=0\implies \measuredangle x=sin^{-1}(0)\implies \measuredangle x=
\begin{cases}
0\\
\pi \\
2\pi 
\end{cases}\\\\
-------------------------------$

$\bf 8sin^4(x)+1-6sin^2(x)=0\implies 8sin^4(x)-6sin^2(x)+1=0$

now, this is a quadratic equation, but the roots do not come out as integers, however it does have them, the discriminant, b² - 4ac, is positive, so it has 2 roots, so we'll plug it in the quadratic formula,

$\bf 8sin^4(x)-6sin^2(x)+1=0\implies 8[~[sin(x)]^2~]^2-6[sin(x)]^2+1=0
\\\\\\
~~~~~~~~~~~~\textit{quadratic formula}
\\\\
\begin{array}{lcccl}
& 8 sin^4& -6 sin^2(x)& +1\\
&\uparrow &\uparrow &\uparrow \\
&a&b&c
\end{array} 
\qquad \qquad 
sin(x)= \cfrac{ - b \pm \sqrt { b^2 -4 a c}}{2 a}
\\\\\\
sin(x)=\cfrac{-(-6)\pm\sqrt{(-6)^2-4(8)(1)}}{2(8)}\implies sin(x)=\cfrac{6\pm\sqrt{4}}{16}
\\\\\\
sin(x)=\cfrac{6\pm 2}{16}\implies sin(x)=
\begin{cases}
\frac{1}{2}\\\\
\frac{1}{4}
\end{cases}$

$\bf \measuredangle x=
\begin{cases}
sin^{-1}\left( \frac{1}{2} \right)
sin^{-1}\left( \frac{1}{4} \right)
\end{cases}\implies \measuredangle x=
\begin{cases}
\frac{\pi }{6}~,~\frac{5\pi }{6}\\
----------\\
\approx~0.252680~radians\\
\qquad or\\
\approx~14.47751~de grees\\
----------\\
\pi -0.252680\\
\approx 2.88891~radians\\
\qquad or\\
180-14.47751\\
\approx 165.52249~de grees
\end{cases}$

3 0

3 years ago