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RoseWind [281]
2 years ago
10

Answer to this question?

Mathematics
2 answers:
Liula [17]2 years ago
5 0
The answer would be -4
STatiana [176]2 years ago
3 0

Answer:

-4

3*(-1)-(-1*-1)

-3-(+1)

-3-1

<em><u>-</u></em><em><u>4</u></em><em><u>.</u></em><em><u>.</u></em><em><u>.</u></em><em><u>.</u></em><em><u>.</u></em><em><u> Answer</u></em>

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What is latus rectum?
anyanavicka [17]
Latus rectum is the line segment that passes through the focus, is perpendicular to the axis, and has both endpoints on the curve.
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3 years ago
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Find the center of the equation 4x2 + y2 + 8x-5= 0.
evablogger [386]

Answer:

(-1,0)

Step-by-step explanation:

Plug it into desmos graphing calculator exactly how it is.

3 0
3 years ago
What is the precent of 15 is 12
kati45 [8]
100%/x%=12/15
<span>(100/x)*x=(12/15)*x       - </span>we multiply both sides of the equation by x
<span>100=0.8*x       - </span>we divide both sides of the equation by (0.8) to get x
<span>100/0.8=x </span>
<span>125=x </span>
<span>x=125

so 125%</span>
8 0
3 years ago
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Find the sum of the geometric series 512+256+ . . .+4
mario62 [17]

\bf 512~~,~~\stackrel{512\cdot \frac{1}{2}}{256}~~,~~...4

so, as you can see above, the common ratio r = 1/2, now, what term is +4 anyway?

\bf n^{th}\textit{ term of a geometric sequence}\\\\a_n=a_1\cdot r^{n-1}\qquad \begin{cases}n=n^{th}\ term\\a_1=\textit{first term's value}\\r=\textit{common ratio}\\----------\\r=\frac{1}{2}\\a_1=512\\a_n=+4\end{cases}

\bf 4=512\left( \cfrac{1}{2} \right)^{n-1}\implies \cfrac{4}{512}=\left( \cfrac{1}{2} \right)^{n-1}\\\\\\\cfrac{1}{128}=\left( \cfrac{1}{2} \right)^{n-1}\implies \cfrac{1}{2^7}=\left( \cfrac{1}{2} \right)^{n-1}\implies 2^{-7}=\left( 2^{-1}\right)^{n-1}\\\\\\(2^{-1})^7=(2^{-1})^{n-1}\implies 7=n-1\implies \boxed{8=n}

so is the 8th term, then, let's find the Sum of the first 8 terms.

\bf \qquad \qquad \textit{sum of a finite geometric sequence}\\\\S_n=\sum\limits_{i=1}^{n}\ a_1\cdot r^{i-1}\implies S_n=a_1\left( \cfrac{1-r^n}{1-r} \right)\quad \begin{cases}n=n^{th}\ term\\a_1=\textit{first term's value}\\r=\textit{common ratio}\\----------\\r=\frac{1}{2}\\a_1=512\\n=8\end{cases}

\bf S_8=512\left[ \cfrac{1-\left( \frac{1}{2} \right)^8}{1-\frac{1}{2}} \right]\implies S_8=512\left(\cfrac{1-\frac{1}{256}}{\frac{1}{2}}  \right)\implies S_8=512\left(\cfrac{\frac{255}{256}}{\frac{1}{2}}  \right)\\\\\\S_8=512\cdot \cfrac{255}{128}\implies S_8=1020

7 0
3 years ago
Which answer includes the intrevals that contain the solution to the inequality x^2-1/3x+9&lt;0
victus00 [196]

Answer:

x<±1 and x< -3

Step-by-step explanation:

Given the inequality x^2-1/3x+9<0, we are to find the values of x that satisfies the inequality

x^2-1/3x+9<0

x^2-1/3x+9 (3x+9)²<0*  (3x+9)²

(x^2-1)(3x+9) < 0

x²-1 < 0 and 3x + 9 <0

x²-1 < 0

x²<1

x<±√1

x<±1

Also 3x + 9 <0

3x < -9

x < -9/3

x < -3

Hence the required values of x are x<±1 and x< -3

4 0
3 years ago
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