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Alex_Xolod [135]
3 years ago
14

HELP ME QUICK: Suppose AB has one endpoint at A(0, 0). If (5, 3) is the midpoint of AB, what are the coordinates of point B?

Mathematics
1 answer:
DiKsa [7]3 years ago
5 0

Answer:

<h3>           B(10, 6)</h3>

Step-by-step explanation:

If P is midpoint of AB and:  A(0,\,0)\,,\quad P(5,\,3)\,,\quad B(x_B,\,y_B)

then:

x_P-x_A=x_B - x_P\qquad\quad\ \wedge\qquad y_P-y_A=y_B - y_P\\\\ 5-0=x_B-5\qquad\quad\wedge\qquad 3-0=y_B -3\\\\ x_B=5+5\qquad\qquad\wedge\qquad\ \ y_B=3+3 \\\\ {}\quad x_B=10\qquad\qquad\wedge\qquad\qquad \ y_B=6

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Answer:

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Step-by-step explanation:

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4 0
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Nataly_w [17]

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8 0
3 years ago
“I think of a number, multiply it by 2, subtract 3 and multiply the result by 4."
Scorpion4ik [409]

Answer:

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5 0
3 years ago
Find a solution x = x(t) of the equation x′ + 2x = t2 + 4t + 7 in the form of a quadratic function of t, that is, of the form x(
Temka [501]
The particular quadratic solution to the ODE is found as follows:

x=at^2+bt+c
x'=2at+b

(2at+b)+2(at^2+bt+c)=t^2+4t+7
2at^2+(2a+2b)t+(b+2c)=t^2+4t+7

\begin{cases}2a=1\\2(a+b)=4\\b+2c=7\end{cases}\implies a=\dfrac12,b=\dfrac32,c=\dfrac{11}4

Note that there's also the fundamental solution to account for, which is obtained from the characteristic equation for the ODE:

x'+2x=0\implies r+2=0\implies r=-2

so that x_c=Ce^{-2t} is a characteristic solution to the ODE, and the general solution would be

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4 0
3 years ago
What is the product of (r + 4) and (r - 3)
dlinn [17]

r + 4(r - 3)

<em><u>We'll use the distributive property to simplify this.</u></em>

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<u><em>Combine like terms</em></u>

r^2 + r - 12 is the simplified form.

6 0
3 years ago
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