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3 years ago

Evaluate each expression if x = −2, y = −3, and z = 5.

1 answer:

3 0

Basically what you have to do is substitute the variables for the values

-4+-9+5= -8
-2-3= -5
2= 2
6-4+1= 3
-9(-2+-4--3)= 27

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Help please 40 points

MrMuchimi

Part 1:

Step 1:

Find the value of G(16)

Using the equation for g(x) =√x, replace x with 16 and solve:

g(16) = √16 = 4

Step 2:

Now using the answer for g(16) replace x in the equation fir f(x) with 4 and solve:

f(x) = x +2

f(4) = 4+2 = 6

Part 2:

following the same steps as part 1:

f(g(x)) would be replacing x in the equation ofr f(x)with the equation for g(x)

f(g(x)) = √x +2

5 0

3 years ago

Read 2 more answers

Joe biked the following distances this past week to train for a triathlon. Joe biked 37 miles, 88 miles, 44 miles, 54 miles, 62

Artyom0805 [142]

Answer:

The median is 50.

Step-by-step explanation:

First you put all the numbers in order. (37, 44, 46, 54, 62, 88) then you look for the middle number. Since there is two middle numbers, you add them together then divide it by 2.

So 46+54= 100

100/2= 50

8 0

2 years ago

What is the 15th term of the sequence A(n) = -3n+50 ?

Lady bird [3.3K]

Answer:

hjjbvf for niinvolppllpp

3 0

3 years ago

A hollow metal sphere has 6 cm inner and 8cm outer radii. The surface charge densities on the exterior surface is +100 nC/m2 and

natulia [17]

Answer:

<h2>Outer Electric Field is 11250 N/C.</h2><h2>Inner Electric Field is -10000 N/C.</h2>

Step-by-step explanation:

First of all, we need to read carefully and analyse the problem. As you can see, is an electrical subject, and it's given surface charge densities and radius.

So, to calculate electric fields, we need to find the proper equation to do so: $E=k\frac{q}{r^{2} }$ ; as you can see, first we need to find the charges.

We can find all charges using the surface charge densities, because it has the next relation: $p=\frac{q}{A}$ ; which indicates that charge density is the amount of charge per area. But, there's a problem, we don't have areas, so we have to calculate them first with this relation: $S=4\pi r^{2}$ ; which gives us the surface of a sphere.

The inner surface: $Si=4\pi (0.06m)^{2} = 0.04 m^{2}$

The outer surface: $S=4\pi (0.08m)^{2}=0.08m^{2}$

Now we can calculate the charges,

Inner charge: $Qi=pA=(-100\frac{nC}{m^{2} } )(0.04m^{2} )=-4nC$

Outer charge: $Qo=pA=100\frac{nC}{m^{2} } )(0.08m^{2} )=8nC$

Then, we are able to calculate both fields:

Inner field: $Ei=k\frac{Qi}{r^{2} }=9x10^{9} \frac{Nm^{2} }{C^{2} }\frac{-4x10^{-9} }{0.06m^{2} }=-10000\frac{N}{C}$

Outer field: $Eo=k\frac{Qo}{r^{2} }=9x10^{9} \frac{Nm^{2} }{C^{2} }\frac{8x10^{-9} }{0.08m^{2} }=11250\frac{N}{C}$

The directions that field have is opposite each other, the inner one has an inside direction, and the outer electric field has an outside direction.

3 0

3 years ago

Help on algebra II pls asap

timama [110]

Because the inequality is less then or equal two the lines need to be solid, not dotted.

Also the less than would put the solution to the outside of the lines.

The correct graph is Stan.

Answer: the incorrect graph is Lee because he used dotted lines and has the solution on the wrong side of the lines

6 0

2 years ago