What is the vertex of the graph of the function below? y=x^2-10x+24

2 answers:

4 0

$\bf \textit{vertex of a vertical parabola, using coefficients}
\\\\
\begin{array}{llcccl}
y = & 1x^2& -10x& +24\\
&\uparrow &\uparrow &\uparrow \\
&a&b&c
\end{array}
\qquad 
\left(-\cfrac{ b}{2 a}\quad ,\quad c-\cfrac{ b^2}{4 a}\right)
\\\\\\
\left( -\cfrac{-10}{2(1)}~~,~~24-\cfrac{(-10)^2}{4(1)} \right)\implies \left(5~~,~~24-25 \right)\implies (5~,~-1)$

cupoosta [38]3 years ago

3 0

The vertex of the equation is (5,-1). Rewrite in standard form. (h,k)

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Need help thank you:)

inna [77]

Answer:

$F = 1.8C + 32$

Step-by-step explanation:

Given

The attached table

Required

The relationship between both temperatures

<em>This implies that we calculate the table equation</em>

First, we calculate the slope (m)

$m = \frac{F_2 - F_1}{C_2 - C_1}$

Where:

$(C_1,F_1) = (0,32)$

$(C_2,F_2) = (20,68)$

So, we have:

$m = \frac{68 - 32}{20 - 0}$

$m = \frac{36}{20}$

$m = 1.8$

The equation is then calculated using:

$F = m(C - C_1) + F_1$

This gives:

$F = 1.8(C - 0) + 32$

$F = 1.8(C ) + 32$

$F = 1.8C + 32$

8 0

3 years ago

Solve using square roots 3x^2+25=73

seropon [69]

Heya !

Given expression -

$3 {x}^{2} + 25 = 73$

Subtracting 25 both sides ,

$3 {x}^{2} + 25 - 25 = 73 - 25 \\ \\ 3 {x}^{2} = 48$

Dividing by 3 on both sides ,

$\frac{3 {x}^{2} }{3} = \frac{48}{3} \\ \\ {x}^{2} = 16$

Therefore ,
$x = + \: 4 \: \: \: \: or \: \: - 4 \: \: \: \: \: \: \: Ans.$