What is the additive inverse of -7?<br> O A. -7<br> O B. 7<br> C. -|-7|<br> D. -(-(-7))

Please help!!!!<br><br>what is the least common denominator of 1/6, 10/11, 5/12?<br>

Vesna [10]

<h3>Answer: LCM = 132</h3>

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Work Shown:

LCM = least common denominator

List out the prime factorization of each denominator

6 = 2*3
11 = 1*11
12 = 2*2*3

So we have the list of primes 2,3, and 11 that help form the denominators when we multiply some of them together.

The prime 2 shows up at most twice, so 2*2 = 4 is a factor of the LCM

The prime 3 shows up at most one time, meaning 3 is also a factor

The prime 11 shows up at most one time, so 11 is another factor

Multiply these factors to get 4*3*11 = 12*11 = 132

The LCM is 132

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Another Approach:

Focus on 1/6 and 10/11 for now. The LCM is 66 because 6*11 = 66. We simply multiply the denominators together. Then we divide over the GCF 1 to get 66/1 = 66.

The LCM of 1/6 and 10/11 is 66

The fractions 1/6 and 10/11 are equivalent to 11/66 and 60/66 respectively

The original list of fractions updates to 11/66, 60/66, 5/12

We've gone from 3 different denominators to now 2 different denominators.

Repeat the steps of multiplying the denominators and dividing by the GCF

66*12 = 792

792/(gcf of 66 and 12) = 792/6 = 132

So the LCM of all the fractions is 132.

3 0

3 years ago

Waht is the value of ((2/3)^0)^-3

mestny [16]

Answer:

$\left(\left(\frac{2}{3}\right)^0\right)^{-3}=1$

Step-by-step explanation:

$\mathrm{Apply\:exponent\:rule}:\quad \left(a^b\right)^c=a^{bc},\:\quad \:a\ge 0\\\left(\left(\frac{2}{3}\right)^0\right)^{-3}=\left(\frac{2}{3}\right)^{0\cdot \left(-3\right)}\\=\left(\frac{2}{3}\right)^0\\=1$

<3

Red

6 0

3 years ago

Michael keeps track of how much time he uses his

Valentin [98]

Answer:

we have to see the table

Step-by-step explanation:

say that y is the calculated time

and x is the weeks

4 0

3 years ago

Please help!! ASAP!!

ser-zykov [4K]

Answer:

The scientific notation will be Letter D

6 0

3 years ago

Read 2 more answers

Questions attached as screenshot below:Please help me I need good explanations before final testI pay attention

Nikitich [7]

The acceleration of the particle is given by the formula mentioned below:

$a=\frac{d^2s}{dt^2}$

Differentiate the position vector with respect to t.

$\begin{gathered} \frac{ds(t)}{dt}=\frac{d}{dt}\sqrt[]{\mleft(t^3+1\mright)} \\ =-\frac{1}{2}(t^3+1)^{-\frac{1}{2}}\times3t^2 \\ =\frac{3}{2}\frac{t^2}{\sqrt{(t^3+1)}} \end{gathered}$

Differentiate both sides of the obtained equation with respect to t.

$\begin{gathered} \frac{d^2s(t)}{dx^2}=\frac{3}{2}(\frac{2t}{\sqrt[]{(t^3+1)}}+t^2(-\frac{3}{2})\times\frac{1}{(t^3+1)^{\frac{3}{2}}}) \\ =\frac{3t}{\sqrt[]{(t^3+1)}}-\frac{9}{4}\frac{t^2}{(t^3+1)^{\frac{3}{2}}} \end{gathered}$

Substitute t=2 in the above equation to obtain the acceleration of the particle at 2 seconds.

$\begin{gathered} a(t=1)=\frac{3}{\sqrt[]{2}}-\frac{9}{4\times2^{\frac{3}{2}}} \\ =1.32ft/sec^2 \end{gathered}$

The initial position is obtained at t=0. Substitute t=0 in the given position function.

$\begin{gathered} s(0)=-23\times0+65 \\ =65 \end{gathered}$

8 0

1 year ago

What is the additive inverse of -7? O A. -7 O B. 7 C. -|-7| D. -(-(-7))