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2 years ago

Please help me with this..

Mathematics

1 answer:

ollegr [7]2 years ago

8 0

Answer:

Step-by-step explanation:

1.one solution

2.infinitely

3.one solution

4.one solution

5.no solution

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40 points! Find the distance UV between the points U(2,−2) and V(−5,−3). Round your answer to the nearest tenth, if necessary.

puteri [66]

Answer:

7.1

Step-by-step explanation:

Using the distance formula

d = sqrt ( (x2-x1)^2 + ( y2-y1)^2)

sqrt ( (-5-2)^2 + ( -3- -2)^2)

sqrt( ( -7)^2 + ( -3 +2)^2)

sqrt( ( 49 + 1)

sqrt( 50)

7.071067812

To the nearest tenth

7.1

8 0

3 years ago

Read 2 more answers

K over 1.5 =3 what is the value of k

Makovka662 [10]

4.5 is equal to k. This is the answer .

5 0

3 years ago

Read 2 more answers

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

2 years ago

suppose that the number of bacteria in a certain population increases according to a continuous exponential growth model. A samp

Tema [17]

Answer:

3.33% per hour

Step-by-step explanation:

Use the A=Pe^rt equation. A is the end amount, so it's 1892. P is the original amount, 1700. E is a constant, around 2.72. R is the growth constant. T is the time that passed, 3 hours. You can substitute the givens into the equation and get 1892=1700e^(3r). Divide by 1700 to isolate the e. This leaves you with 1892/1700=e^(3r). Do the natural log of each side cancel the e and bring the exponent down. This leaves you with ln(1892/1700)=3r. Divide by 3 to isolate r. ln(1892/1700) is .1. .1/3 is .03333. Multiply by 100 to get a percent. 3.33 percent is your final answer.

8 0

2 years ago

Find all sets of four consecutive integers whose sum is between 95 and 105

Nat2105 [25]

Hello,

Let n-2,n-1,n,n+1 the foru numbers
s=n-2+n-1+n+n+1=4n-2

95<=4n-2<=105
==>97<=7n<=107
==>24.25<=n<=26.75
==>25<=n<=26
if n=25 s=98
if n=26 s=102

7 0

3 years ago