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3 years ago

What is x to the -6 power

Mathematics

1 answer:

Allisa [31]3 years ago

6 0

X^-6

You can find the derivative:

-6x^ -7

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If f(x)=-9x-2, find f(7)

Elza [17]

Answer:

x= -1/5

Let me know if you need an explanation :)

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3 years ago

Tan(x)= 3/2 <br> Find X <br> Plz help

Novosadov [1.4K]

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7 0

3 years ago

Please help idk any of this

AnnyKZ [126]

m<1=152

m<2=28

m<3=152

m<4=28

m<5=152

m<6=28

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not completly sure but good luck

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3 years ago

BRAINLIEST IF CORRECT!!!

goblinko [34]

Answer:

Multiply a fraction by its reciprocal to show that their product is always 1

Step-by-step explanation:

5 0

3 years ago

Read 2 more answers

The right end of a relaxed standard spring is at the origin; the left end is clamped at some point on the negative x-axis. Holdi

ella [17]

The work required to stretch the spring is 350J.

<h3>Hooke's Law</h3>

This law can be written by the formula: F=kx, where:

F= elastic force (N)

k= spring constant (N/m)

x= linear displacement (m)

<h3>Spring Work</h3>

For finding the spring work in J, you should apply the formula $W=\frac{k*(\Delta x)^2}{2}$ , where:

W= work (J)

k= spring constant (N/m)

x= difference between the linear displacements (m)

The question gives:

x=5 cm=0.05m requires a force of 1.4N

x=8 cm =0.08m

Step 1 - First, you should find the spring constant from Hooke's law, for x=5 cm and F=1.4N.

$F=kx\\ 1.4=k*0.05\\ \\ k=\frac{1.4}{0.05} =28\frac{N}{m}$

Step 2 - Now you can apply the formula for spring work.

$W=\frac{k*(\Delta x)^2}{2}\\ \\ W=\frac{28*(8-3)^2}{2} \\ \\ W=\frac{28*(5)^2}{2}\\ \\ W=\frac{28*25}{2}\\ \\ W=\frac{700}{2}=350J$

Read more about the spring work here:

brainly.com/question/3317535

5 0

3 years ago