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stiks02 [169]
3 years ago
11

Need help: write the expression in a complete factored form: x²-9x+xy-9y

Mathematics
1 answer:
Galina-37 [17]3 years ago
4 0
Factor this by grouping.  Group the first two terms together and then the second 2 terms together.  The idea here is that, after grouping, what you can pull out in both groups is the same.  Like this:
( x^{2} -9x)+(xy-9y)
Now work on finding a common thing between the 2 groups:
x(x-9)+y(x-9)
See, what you have in common there now is the (x-9) that you can factor out:
(x-9)(x+y)
And that is factored completely.
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A toy train with a resistance of 10 ohms is powered by four 1.5-volt batteries. What is the power of the toy train?
Tamiku [17]

Answer:

12

Step-by-step explanation:

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Determine the unknown side lengths’ to the nearest tenth. for the first question and also please help with question 6
Levart [38]

Answer:

z =15cm and

y=9.0cm

Step-by-step explanation:

Use sine rule

12/ sin53 = z / sin 90

Z = 12 x sin 90 / sin 53

Z = 15cm

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So

15/sin 90= y/sin37

y= 15 x sin37 /sin 90

y = 9.0cm

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3 years ago
Jennifer has 3 dogs and 6 cats.<br><br> What is the ratio of dogs to cats?
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Answer:

1:2 - one dog to two cats.

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2 years ago
A ball is thrown into the air by a baby alien on a planet in the system of Alpha Centauri with a velocity of 30 ft/s. Its height
Crank

Answer:

a) h = 0.1: \bar v = -11\,\frac{ft}{s}, h = 0.01: \bar v = -10.1\,\frac{ft}{s}, h = 0.001: \bar v = -10\,\frac{ft}{s}, b) The instantaneous velocity of the ball when t = 2\,s is -10 feet per second.

Step-by-step explanation:

a) We know that y = 30\cdot t -10\cdot t^{2} describes the position of the ball, measured in feet, in time, measured in seconds, and the average velocity (\bar v), measured in feet per second, can be done by means of the following definition:

\bar v = \frac{y(2+h)-y(2)}{h}

Where:

y(2) - Position of the ball evaluated at t = 2\,s, measured in feet.

y(2+h) - Position of the ball evaluated at t =(2+h)\,s, measured in feet.

h - Change interval, measured in seconds.

Now, we obtained different average velocities by means of different change intervals:

h = 0.1\,s

y(2) = 30\cdot (2) - 10\cdot (2)^{2}

y (2) = 20\,ft

y(2.1) = 30\cdot (2.1)-10\cdot (2.1)^{2}

y(2.1) = 18.9\,ft

\bar v = \frac{18.9\,ft-20\,ft}{0.1\,s}

\bar v = -11\,\frac{ft}{s}

h = 0.01\,s

y(2) = 30\cdot (2) - 10\cdot (2)^{2}

y (2) = 20\,ft

y(2.01) = 30\cdot (2.01)-10\cdot (2.01)^{2}

y(2.01) = 19.899\,ft

\bar v = \frac{19.899\,ft-20\,ft}{0.01\,s}

\bar v = -10.1\,\frac{ft}{s}

h = 0.001\,s

y(2) = 30\cdot (2) - 10\cdot (2)^{2}

y (2) = 20\,ft

y(2.001) = 30\cdot (2.001)-10\cdot (2.001)^{2}

y(2.001) = 19.99\,ft

\bar v = \frac{19.99\,ft-20\,ft}{0.001\,s}

\bar v = -10\,\frac{ft}{s}

b) The instantaneous velocity when t = 2\,s can be obtained by using the following limit:

v(t) = \lim_{h \to 0} \frac{x(t+h)-x(t)}{h}

v(t) =  \lim_{h \to 0} \frac{30\cdot (t+h)-10\cdot (t+h)^{2}-30\cdot t +10\cdot t^{2}}{h}

v(t) =  \lim_{h \to 0} \frac{30\cdot t +30\cdot h -10\cdot (t^{2}+2\cdot t\cdot h +h^{2})-30\cdot t +10\cdot t^{2}}{h}

v(t) =  \lim_{h \to 0} \frac{30\cdot t +30\cdot h-10\cdot t^{2}-20\cdot t \cdot h-10\cdot h^{2}-30\cdot t +10\cdot t^{2}}{h}

v(t) =  \lim_{h \to 0} \frac{30\cdot h-20\cdot t\cdot h-10\cdot h^{2}}{h}

v(t) =  \lim_{h \to 0} 30-20\cdot t-10\cdot h

v(t) = 30\cdot  \lim_{h \to 0} 1 - 20\cdot t \cdot  \lim_{h \to 0} 1 - 10\cdot  \lim_{h \to 0} h

v(t) = 30-20\cdot t

And we finally evaluate the instantaneous velocity at t = 2\,s:

v(2) = 30-20\cdot (2)

v(2) = -10\,\frac{ft}{s}

The instantaneous velocity of the ball when t = 2\,s is -10 feet per second.

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3 years ago
Help xd <br> -(x+2) =<br> 3(7x - 8)=<br> 7(x+3)-5x<br><br> Thank you!
HACTEHA [7]

Answer:

x=−5

x= 31/21 =1 10/21= 1.476190476

2x+21

hope this helps

3 0
3 years ago
Read 2 more answers
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