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4 years ago

Plz help

1 answer:

8 0

$\bf \begin{array}{lllll}
&x_1&y_1&x_2&y_2\\
% (a,b)
&({{ 3}}\quad ,&{{7}})\quad 
% (c,d)
&({{ 4}}\quad ,&{{ 9}})
\end{array}
\\\\\\
% slope = m
slope = {{ m}}= \cfrac{rise}{run} \implies 
\cfrac{{{ y_2}}-{{ y_1}}}{{{ x_2}}-{{ x_1}}}\implies \cfrac{9-7}{4-3}\implies \cfrac{2}{1}\implies 2$

$\bf \stackrel{\textit{point-slope form}}{y-{{ y_1}}={{ m}}(x-{{ x_1}})}\implies y-7=2(x-3)\implies y-7=2x-6
\\\\\\
y=2x+1\qquad \qquad \qquad \qquad \qquad \qquad \stackrel{standard~form}{-2x+y=1}$

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Find x and Y pls help

lisabon 2012 [21]

Answer:

Step-by-step explanation:

Comment

The angle supplementary to x and 3x are equal.

Supplementary angles = 180

Solution

x + 3x = 180

4x = 180 Divide by 4

4x/4 = 180/4

x = 45

The angle supplementary to x is 3x

3x = 3*45

3x = 135

This angle is vertically opposite y. Vertically opposite angles are equal. So y = 135

Answer

x = 45

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2 years ago

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3 years ago

A given field mouse population satisfies the differential equation dp dt = 0.5p − 410 where p is the number of mice and t is the

ohaa [14]

Answer:

a) $t = 2 *ln(\frac{82}{5}) =5.595$

b) $t = 2 *ln(-\frac{820}{p_0 -820})$

c) $p_0 = 820-\frac{820}{e^6}$

Step-by-step explanation:

For this case we have the following differential equation:

$\frac{dp}{dt}=\frac{1}{2} (p-820)$

And if we rewrite the expression we got:

$\frac{dp}{p-820}= \frac{1}{2} dt$

If we integrate both sides we have:

$ln|P-820|= \frac{1}{2}t +c$

Using exponential on both sides we got:

$P= 820 + P_o e^{1/2t}$

Part a

For this case we know that p(0) = 770 so we have this:

$770 = 820 + P_o e^0$

$P_o = -50$

So then our model would be given by:

$P(t) = -50e^{1/2t} +820$

And if we want to find at which time the population would be extinct we have:

$0=-50 e^{1/2 t} +820$

$\frac{820}{50} = e^{1/2 t}$

Using natural log on both sides we got:

$ln(\frac{82}{5}) = \frac{1}{2}t$

And solving for t we got:

$t = 2 *ln(\frac{82}{5}) =5.595$

Part b

For this case we know that p(0) = p0 so we have this:

$p_0 = 820 + P_o e^0$

$P_o = p_0 -820$

So then our model would be given by:

$P(t) = (p_o -820)e^{1/2t} +820$

And if we want to find at which time the population would be extinct we have:

$0=(p_o -820)e^{1/2 t} +820$

$-\frac{820}{p_0 -820} = e^{1/2 t}$

Using natural log on both sides we got:

$ln(-\frac{820}{p_0 -820}) = \frac{1}{2}t$

And solving for t we got:

$t = 2 *ln(-\frac{820}{p_0 -820})$

Part c

For this case we want to find the initial population if we know that the population become extinct in 1 year = 12 months. Using the equation founded on part b we got:

$12 = 2 *ln(\frac{820}{820-p_0})$

$6 = ln (\frac{820}{820-p_0})$

Using exponentials we got:

$e^6 = \frac{820}{820-p_0}$

$(820-p_0) e^6 = 820$

$820-p_0 = \frac{820}{e^6}$

$p_0 = 820-\frac{820}{e^6}$

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4 years ago

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Arisa [49]

You know the price of a good,

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3 years ago

Read 2 more answers

Square root of 550 simplified. Please show work. 10 pts and brainliest answer, for whoever is right and shows good enough work.

Svetlanka [38]

Factor 550: 550=5×5×2×11
to square root a number, you look for pairs of factors. Here you have a pair of 5s, so when you square root them, you get one 5. whatever factor that does not make a pair stays inside the square root sign.
√550=√(5^2×22)=5√22

4 0

3 years ago