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Rudiy27
3 years ago
9

The vertex of the parabola is at (4,-3) which of the following could be its equation?

Mathematics
1 answer:
kolbaska11 [484]3 years ago
6 0
\bf \qquad \textit{parabola vertex form}\\\\
\begin{array}{llll}
y=a(x-{{ h}})^2+{{ k}}\\\\
\boxed{x=a(y-{{ k}})^2+{{ h}}}
\end{array} \qquad\qquad  vertex\ ({{ h}},{{ k}})\\\\
-------------------------------\\\\
\begin{cases}
h=4\\
k=-3
\end{cases}\implies x=a(y-(-3))^2+4\implies x=a(y+3)^2+4
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Does anyone knows this please!<br><br> -2x x 5x<br> -(8x)^2
satela [25.4K]

Answer:

-74x^2

Step-by-step explanation:

-2x x 5x -(8x)^2

-10x^2-(8x)^2

-10x^2-64x^2

-74x^2

7 0
3 years ago
Given a quadrilateral with vertices ????(0, 5), ????(6, 5), ????(4, 0), and H(−2, 0):
tankabanditka [31]

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3 years ago
If a/b &lt; c/d with b &gt; 0, d &gt; 0, prove that a+c / b+d lies between the two fractions a/b and c/d .​
Tems11 [23]

Divide through everything by <em>b</em> :

\dfrac{a+c}{b+d} = \dfrac{\dfrac ab + \dfrac cb}{1 + \dfrac db}

Since <em>a/b</em> < <em>c/d</em>, it follows that

\dfrac{a+c}{b+d} < \dfrac{\dfrac cd+\dfrac cb}{1 + \dfrac db}

Multiply through everything on the right side by <em>b/d</em> to get

\dfrac{a+c}{b+d} < \dfrac{\dfrac{bc}{d^2}+\dfrac cd}{\dfrac bd+1} = \dfrac{\dfrac cd\left(\dfrac bd+1\right)}{\dfrac bd+1} = \dfrac cd

and so (<em>a</em> + <em>c</em>)/(<em>b</em> + <em>d</em>) < <em>c/d</em>.

For the other side, you can do something similar and divide through everything by <em>d</em> :

\dfrac{a+c}{b+d} = \dfrac{\dfrac ad + \dfrac cd}{\dfrac bd + 1}

and <em>a/b</em> < <em>c/d</em> tells us that

\dfrac{a+c}{b+d} > \dfrac{\dfrac ad + \dfrac ab}{\dfrac bd + 1}

Then

\dfrac{a+c}{b+d} > \dfrac{\dfrac ab + \dfrac{ad}{b^2}}{1 + \dfrac db} = \dfrac{\dfrac ab\left(1+\dfrac db\right)}{1 + \dfrac db} = \dfrac ab

and so (<em>a</em> + <em>c</em>)/(<em>b</em> + <em>d</em>) > <em>a/b</em>.

Then together we get the desired inequality.

5 0
3 years ago
<img src="https://tex.z-dn.net/?f=%7C2x-3%7C%5Cleq%209" id="TexFormula1" title="|2x-3|\leq 9" alt="|2x-3|\leq 9" align="absmiddl
MA_775_DIABLO [31]

Answer:

Step-by-step explanation:

|2x-3|≤9

so -9≤2x-3≤9

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3 years ago
What is the general form of the equation of the given circle with center A
Nonamiya [84]
When the centre A(xa,ya) and radius are given, the general equation is

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5 0
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