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sergiy2304 [10]

3 years ago

11

What is the relationship between two coplanar lines that are perpendicular to the same line?

2 answers:

Charra [1.4K]3 years ago

6 0

Answer:

I'm sorry for this but I'll help you if you help me?

grigory [225]3 years ago

3 0

Answer:

If two lines are perpendicular to the same line, then the two lines are parallel, and if two lines are parallel, then their slopes are equal

Step-by-step explanation:

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A sandwich store charges $20 to have three turkey subs delivered and $26 to have four delivered. How much does the store charge

sweet [91]

$6 per turkey and $2 delivery fee. Do da math.

7 0

3 years ago

A given field mouse population satisfies the differential equation dp/dt=.4p-450 where p is the number of mice and t is the time

Flura [38]

Answer:

a) 7.78 months

b) 1116

Step-by-step explanation:

Given -

$\frac{dP}{dt} = 0.4p-450$

Integrating the above equation with respect to time, we get -

$\int\ \frac{dP}{0.4p-450} = \int\ dt\\$

let us define new variable x

$x = 0.4p - 450 \\dx = 0.4 dy$

substituting these values in above integral equation, we get -

$\frac{1}{0.4} \int\ \frac{dx}{x} = \int\ dt\\ln x = 0.4 t + C\\x = ce^{0.4t}$

$P (t) = \frac{C}{0.4} e^{0.4t} +1125\\$

at t = 0, P (0) = 1075, using this condition, we get -

$\frac{c}{0.4} = -50\\P(t) = 50 e^{0.4t} +1125 \\t = 7.78\\P(0) = 1125 - \frac{1125}{e^{4.8}} = 1116$

7 0

3 years ago

What is 9 divided by 2 over 7?

andreev551 [17]

2 over 7 =0,28
9/0,28=32,14

Hope it helped

6 0

3 years ago

5. What is the solution set for the equation x? + 5x3 = 0?

blsea [12.9K]

Answer:

x = 0

Step-by-step explanation:

3 0

3 years ago

Solve the simultaneous equations<br> y = 9 - X<br> y = 2x2 + 4x + 6

kenny6666 [7]

Answer:

$\mathrm{Therefore,\:the\:final\:solutions\:for\:}y=9-x,\:y=2x^2+4x+6\mathrm{\:are\:}$

$\begin{pmatrix}x=\frac{1}{2},\:&y=\frac{17}{2}\\ x=-3,\:&y=12\end{pmatrix}$

Step-by-step explanation:

Given the simultaneous equations

$y=9-x$

$y\:=\:2x^2\:+\:4x\:+\:6$

Subtract the equations

$y=9-x$

$-$

$\underline{y=2x^2+4x+6}$

$y-y=9-x-\left(2x^2+4x+6\right)$

$\mathrm{Refine}$

$x\left(2x+5\right)=3$

$\mathrm{Solve\:}\:x\left(2x+5\right)=3$

$2x^2+5x=3$ ∵ $\mathrm{Expand\:}x\left(2x+5\right):\quad 2x^2+5x$

$\mathrm{Subtract\:}3\mathrm{\:from\:both\:sides}$

$2x^2+5x-3=3-3$

$\mathrm{Solve\:with\:the\:quadratic\:formula}$

$\mathrm{Quadratic\:Equation\:Formula:}$

$\mathrm{For\:a\:quadratic\:equation\:of\:the\:form\:}ax^2+bx+c=0\mathrm{\:the\:solutions\:are\:}$

$x_{1,\:2}=\frac{-b\pm \sqrt{b^2-4ac}}{2a}$

$\mathrm{For\:}\quad a=2,\:b=5,\:c=-3:\quad x_{1,\:2}=\frac{-5\pm \sqrt{5^2-4\cdot \:2\left(-3\right)}}{2\cdot \:2}v\\$

$x=\frac{-5+\sqrt{5^2-4\cdot \:2\left(-3\right)}}{2\cdot \:2}$

$=\frac{-5+\sqrt{5^2+4\cdot \:2\cdot \:3}}{2\cdot \:2}$

$=\frac{-5+\sqrt{49}}{2\cdot \:2}$

$=\frac{-5+\sqrt{49}}{4}$

$=\frac{-5+7}{4}$

$=\frac{2}{4}$

$=\frac{1}{2}$

Similarly,

$x=\frac{-5-\sqrt{5^2-4\cdot \:2\left(-3\right)}}{2\cdot \:2}:\quad -3$

$\mathrm{The\:solutions\:to\:the\:quadratic\:equation\:are:}$

$x=\frac{1}{2},\:x=-3$

$\mathrm{Plug\:the\:solutions\:}x=\frac{1}{2},\:x=-3\mathrm{\:into\:}y=9-x$

$\mathrm{For\:}y=9-x\mathrm{,\:subsitute\:}x\mathrm{\:with\:}\frac{1}{2}:\quad y=\frac{17}{2}$

$\mathrm{For\:}y=9-x\mathrm{,\:subsitute\:}x\mathrm{\:with\:}-3:\quad y=12$

$\mathrm{Therefore,\:the\:final\:solutions\:for\:}y=9-x,\:y=2x^2+4x+6\mathrm{\:are\:}$

$\begin{pmatrix}x=\frac{1}{2},\:&y=\frac{17}{2}\\ x=-3,\:&y=12\end{pmatrix}$

3 0

4 years ago