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3 years ago

How do you solve this? -5√112

Mathematics

1 answer:

3241004551 [841]3 years ago

3 0

Simplify the radicand by breaking the radicand up into a product of known factors.

Answer: -20√7

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I dont understand please help Which expression is equivalent to 8-(6r+2) A -6r+6 B 2r+2 C 6r+10 D -6r+10

Goryan [66]

Answer:

-6r+6

Step-by-step explanation:

Given data

We are given the expression

8-(6r+2)

let us expand it by opening the bracket

=8-6r-2

collect like terms

=8-2-6r

=6-6r

rearrange

=-6r+6

Hence option A is correct

8 0

2 years ago

What’s the answer 1/6+1/2

miskamm [114]

Answer:

4/6 or 2/3

Step-by-step explanation:

Using Common denominators:

3/6 = 1/2

1/6 + 3/6 = 4/6

This can reduce to 2/3

8 0

3 years ago

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PLZZ HELP I NEED HELP

Mkey [24]

Step-by-step explanation:

1) let the number=x

six times a number=6x

Condition:

6x+4=22

2) eleven times a number=11x

Condition:

11x-5=50

3) 9 times a number=9x

Condition:

9x-7=-16

Note:if you need to ask any question please let me know.

3 0

3 years ago

HI PLEASE HELP ME WITH MY CALCULUS 1 HW? I AM REALLY STUCK. I need help with parts d,e,g.

asambeis [7]

(d) The particle moves in the positive direction when its velocity has a positive sign. You know the particle is at rest when $t=0$ and $t=3$ , and because the velocity function is continuous, you need only check the sign of $v(t)$ for values on the intervals (0, 3) and (3, 6).

We have, for instance $v(1)\approx-0.91$ and $v(4)\approx0.91>0$ , which means the particle is moving the positive direction for $3$ , or the interval (3, 6).

(e) The total distance traveled is obtained by integrating the absolute value of the velocity function over the given interval:

$\displaystyle\int_0^6|v(t)|\,\mathrm dt=\int_0^3-v(t)\,\mathrm dt+\int_3^6v(t)\,\mathrm dt$

which follows from the definition of absolute value. In particular, if $x$ is negative, then $|x|=-x$ .

The total distance traveled is then 4 ft.

(g) Acceleration is the rate of change of velocity, so $a(t)$ is the derivative of $v(t)$ :

$a(t)=v'(t)=-\dfrac{\pi^2}9\cos\left(\dfrac{\pi t}3\right)$

Compute the acceleration at $t=4$ seconds:

$a(t)=\dfrac{\pi^2}{18}\dfrac{\rm ft}{\mathrm s^2}$

(In case you need to know, for part (i), the particle is speeding up when the acceleration is positive. So this is done the same way as part (d).)

6 0

3 years ago

What is the unit rate of 4 to 4.60

lapo4ka [179]

Answer:

Step-by-step explanation:

7 0

3 years ago

Read 2 more answers