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3 years ago

Add 28, 362, and 104. Choose the appropriate number for each place value in the sum. Hundreds

Mathematics

2 answers:

Tanya [424]3 years ago

7 0

494

Hundreds 4. Tens 9. Ones 4

Molodets [167]3 years ago

4 0

Answer:

28+362+104=494

hundreds: 4

tens: 9

ones: 4

hope this helps :)

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Find the derivative of ((t^4-6)^3)*((t^3+6)^4)

rjkz [21]

=d/dx((t^4-6)^3) * (t^3+6)^4 + d/dx((t^3+6)^4) * (t^4-6)^3
=3*(t^4-6)^2 * (t^3+6)^4 * d/dx(t^4-6) + 4*(t^3+6)^3 * (t^4-6)^3 * d/dx(t^3+6)
=3*(t^4-6)^2 * (t^3+6)^4 * 4t^3 + 4*(t^3+6)^3 * (t^4-6)^3 * 3t^2
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Step-by-step explanation:

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3 years ago

15*x+3 = x-6 solve equation

Ber [7]

Answer:

Step-by-step explanation:

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3 years ago

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What is 0.36∞ written as a fraction in simplest form? PLZ HELP MEH

LenKa [72]

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Sep 5, 2016

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3 years ago

The position of a particle moving in a straight line at any time t is x(t) = 2t^2 + 6t + 5. What is the acceleration of the part

grandymaker [24]

Answer:

The acceleration of the particle at any time t

$a = 4\ m/s^{2}$

Step-by-step explanation:

Given:

The position of a particle moving in a straight line at any time t is.

$x(t) = 2t^{2} + 6t + 5$

The velocity of the particle $v = \frac{d(x)}{d(t)}$

$v = \frac{d}{d(t)}(2t^{2} + 6t + 5)$

$v= 2\times 2t + 6$

$v= 4t + 6$

So the velocity of the particle $v= (4t + 6)\ m/s$

The acceleration of the particle $a = \frac{d(v)}{d(t)}$

$a = \frac{d}{d(t)}(4t + 6)$

$a = 4\ m/s^{2}$

In this condition the acceleration does not depending upon the time, so the acceleration is constant

Therefore the acceleration of the particle at any time t $a = 4\ m/s^{2}$

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3 years ago