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3 years ago

Which expression is equivalent to 5(x + 2) - 7(x + 2)?

Mathematics

1 answer:

Harman [31]3 years ago

6 0

The answer is
A: -2x - 4

I did the math and that’s what I got so I hope it helped!

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What does |x| mean. I'm talking about the lines specifically.

Ymorist [56]

Those lines stand for the absolute value, or how far away it is from zero.

If you draw a number line from -3 to 3, and plot the points x = 2 and x = -2, what similarities do you see?
Although one is two units to the right of 0 and one is two units to the left of 0, both x = 2 and x = -2 are two units away from zero. Therefore, the absolute value of both 2 and -2 is 2.

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Minchanka [31]

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Evgesh-ka [11]

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Step-by-step explanation:

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3 years ago

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8 yd 15 yd 4 yd r ay Find the volume of the prism. Round your answer to the nearest tenth. 18.5 ft 8.6 ft h = 8.4 ft

seraphim [82]

L=8yd
B=15yd
H=4yd

Volume:-.

$\\ \tt\longmapsto V=LBH$

$\\ \tt\longmapsto V=8(15)(4)$

$\\ \tt\longmapsto V=480yd^3$

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2 years ago

ASAP! I NEED THIS DONE AS QUICKLY AS POSSIBLE WITH WORKING OUT

Elza [17]

Answer:

See explanation.

Step-by-step explanation:

Let us first analyze some principle theory. By definition we know that the velocity ( $v$ ) is a function of a distance ( $d$ ) covered in some time ( $t$ ), whilst acceleration ( $a$ ) is the velocity achieved in some time. These can also been expressed as:

$v = \frac{d}{t}\\$ and $a=\frac{v}{t}$

We also know that both velocity and acceleration are vectors (therefore they are characterized by both a magnitude and a direction). Finally we know that given a position vector we can find the velocity and the acceleration, by differentiating the vector with respect to time, once and twice, respectively.

Let us now solve our problem. Here we are givine the Position vector of a particle P (in two dimensional space of $i-j$ ) as:

$r=(3t^3-t+3)i+(2t^2+2t-1)j$ Eqn.(1)

Let us solve.

Part (a) Velocity: we need to differentiate Eqn.(1) with respect to time as:

$v(t)=\frac{dr}{dt}\\\\ v(t)=[(3)3t^2-1]i+[(2)2t+2]j\\\\v(t)=(9t^2-1)i+(4t+2)j$ Eqn.(2)

Part (b) Acceleration: we need to differentiate Eqn.(2) with respect to time as:

$a(t)=\frac{dv}{dt}\\ \\a(t)=[(2)9t]i+4j\\\\a(t)=(18t)i+4j$

Thus the expressions for the velocity and the acceleration of particle P in terms of t are

$v(t)=(9t^2-1)i+(4t+2)j$ and $a(t)=(18t)i+4j$

6 0

4 years ago