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Tanya [424]
3 years ago
11

Which value of n makes the following equation true? √n=4 A) 2 B) 4 C) 8 D) 16

Mathematics
2 answers:
antoniya [11.8K]3 years ago
6 0

Answer:

Hence, the value of n such that the given equation \sqrt{n}=4 holds true is: n=16.

Step-by-step explanation:

We have to find the value of n  such that it make the following statement or equation true:

\sqrt{n}=4

As we are given a square root on the variable term 'n' on left side of the equation hence , on squaring both side of the equation we obtain the value of n as:

n=(4)^2

This means:

n=16

Hence, the value of n such that the given equation \sqrt{n}=4 holds true is 16.

Len [333]3 years ago
4 0

Answer:

16

Step-by-step explanation:

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Solve for x plz. Ens
Digiron [165]

Answer:

9

Step-by-step explanation:

triangles always = 180 degrees

we know there is a 90 degree angle so 180-90=90

we know the other angle is 20 degrees so you do 90-20 which equals 70

now we know that the angle with x equals 70

so then you do 70-16 which equals 54

now just divide 54 by 6 and you get 9

so x=9

hope this helped ;)

6 0
3 years ago
Is 22/32 equivalent to 11/16?<br><br>PLEASE I NEED HELP!!!!!!!!!!!!
saveliy_v [14]

Hello! :)

Yes they are: 11/16 = 11(2)/16(2)= 22/32

So, they are equivalent.

I hope I answered in time!

Good luck!

~ Destiny ^_^

5 0
3 years ago
Read 2 more answers
Bev earns $2 a week for taking out her neighbor’s trash cans. Complete the table of values. Then state the domain and range.
Eduardwww [97]

Domain : \{1,2,3,4\}

Range: \{2,4,6,8\}

Explanation:

It is given that Bev earns $2 a week.

Thus, for n number of weeks, we shall multiply $2 with n.

Where n=1,2,3,4 is the number of weeks.

For 1 week, Bev earns 2n=2(1)=2 ⇒ $2

For 2 weeks, Bev earns 2n=2(2)=4 ⇒ $4

For 3 weeks, Bev earns 2n=2(3)=6 ⇒ $6

For 4 weeks, Bev earns 2n=2(4)=8 ⇒ $8

Thus, we can see that the domain is the number of weeks.

Hence, Domain : \{1,2,3,4\}

Also, the money earned is the range.

Hence, Range: \{2,4,6,8\}

6 0
3 years ago
Graph the inequality below on the number line.
marshall27 [118]

Answer:

You start at -9

-Open circle

-shade towards the right (where all the positive numbers are)

Open circle - ones that don't have a greater than or equal to. If it does then it's a closed one.

Shade towards the right cause y is less than 9

3 0
3 years ago
Read 2 more answers
A projectile is fired from a cliff 190 feet above the water at an inclination of 45 to the horizontal, with a muzzle velocity of
jenyasd209 [6]

Answer:

Step-by-step explanation:

I'm not sure if this problem is physics based or calculus based.  So I used calculus because it just makes more sense to do so.

If the projectile is fired from 190 feet in the air, then h0 in our quadratic will be 190.  Since we are being asked to find both the displacement in the x-dimension and in the y-dimension, we need a bit of physics here as well.  The velocity in the y-dimension is found in

v_{0y}=v_{0}sin\theta  and the velocity in the x-dimension is found in

v_{0x}=v_{0}cos\theta

Since the sin and the cos of 45 is the same, it's made a bit simpler for us.  The velocity in both the x and the y dimension is 35.35533906 feet per second.

We can use that now to write the quadratic we need to start solving these rather tedious problems.

The position function for the projectile is

s(t)=-16t^2+35.35533906t+190

I'm going to kind of mix things up a bit, because in order to find the distance in the horizontal dimension that the object is when it's at its max height, we need to first find out how long it takes to get to its max height.  We will first take the derivative of the position function to get the velocity function of the projectile.  The first derivative of the position function is

v(t)=-32t+35.35533906

Remember that the first derivative is the velocity function of the projectile.  You should know from either physics or calculus that at its max height, the velocity of an object is 0 (because it has to stop in the air in order to turn around and come back down).  Setting the velocity function equal to 0 and solving for time will give us the time that the object is at the max height.

0=-32t+35.35533906

I'm going to factor out the -32 to make things easier:

0=-32(t-1.104854346) which gives us that at approximately 1.10485 seconds the object is at its max height.  

Moving over to the horizontal distance question now.  The displacement the object experiences in the horizontal dimension is found in d = rt.  We know the horizontal velocity and now we know how long it takes to get to its max height, so the horizontal distance is found in

d = (35.35533906(1.10485) so

d = 39.06 feet  When the object is at its max height the object is a horizontal distance of 39.06 feet from the face of the cliff.  That's a.

Now to find the max height, we will use again how long it took to get to the max height and sub it in for t in the position function.

s(1.10485) = -16(1.10485)^2 + 35.35533906(1.10485) + 190 to get that the max height is

209 feet.  That's b.

Now for c.  We are asked when the object will hit the water.  We know that when the object hits the water it is no longer in the air and has a height of 0 above the water.  Sub in a 0 for s(t) in the original position function and factor to solve for t:

0=-16t^2+35.35533906t + 190 and solve for t by factoring however you find to be the easiest.  Quadratic formula works great!

We find that the times are -2.51394 seconds and 4.72365 seconds.  Since time will NEVER be negative, we know that the time it takes to hit the water is 4.72365 seconds.

Whew!!!

8 0
4 years ago
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