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irakobra [83]
3 years ago
15

I’m having doubts about my answer can someone help me?

Mathematics
1 answer:
Lena [83]3 years ago
8 0
Yes it it correct

1/3p - 2
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joel has 24 sports trophies. of the trophies, 1/8 are soccer trophies. how many soccer trophies does joel have?
hichkok12 [17]

Answer:

3 soccer trophies does Joel have

Step-by-step explanation:

As per the statement:

Joel has 24 sports trophies.

⇒Total sport trophies Joel have= 24

It is also given: of the trophies, 1/8 are soccer trophies

We have to find how many soccer trophies does Joel have.

\text{Joel have soccer trophies} =\frac{1}{8} \cdot 24

Simplify:

\text{Joel have soccer trophies} =3

Therefore, 3 soccer trophies does Joel have

3 0
4 years ago
Read 2 more answers
Thank you so much, my friend
ss7ja [257]

Answer:

Step-by-step explanation:

This is quite a doozy, my friend. We will set up a d = rt table, fill it in...and pray.

The table will look like this before we even fill anything in:

            d        =        r        *        t

SUV

sedan

Ok now we start to pick apart the problem. Motion problems are the hardest of all story problems ever. This is because there are about 100 ways a motion problem can be presented. So far what we KNOW for an indisputable fact is that the distance from Georgetown to Greenville is 120 km. So we fill that in, making the table:

             d      =      r      *      t

SUV     120

sedan  120

The next part is derived from the sentence "After an hour, the SUV was 24 km ahead of the sedan." This tells us the rate of the SUV in terms of the sedan. If the SUV is 24 km ahead of the sedan in 1 hour, that tells us that the rate of the sedan is r and the rate of the SUV is r + 24 km/hr. BUT we have other times in this problem, one of them being 25 minutes. We have a problem here because the times either have to be in hours or minutes, but not both. So we will change that rate to km/min. Doing that:

24 \frac{km}{hr} × \frac{1hr}{60min}=.4\frac{km}{min} So now we can fill in the rates in the table:

            d      =      r      *      t

SUV    120    =   r + .4

sedan 120    =     r

They left at the same time, so now the table looks like this:

             d      =      r      *      t

SUV    120     =   r + .4  *      t

sedan  120    =      r      *      t

We will put in the time difference of 25 minutes in just a sec.

If d = rt, then the equation for each row is as follows:

SUV:   120 = (r + .4)t

sedan:   120 = rt

Since the times are the same (because they left at the same time, we will set the equations each equal to t. The distances are the same, too, I know that, but if we set the distances equal to each other and then solve the equations for a variable, the distances cancel each other out, leaving us with nowhere to go. Trust me, I tried that first! Didn't work.

Solving the first equation for time:

sedan:  \frac{120}{r}=t  That's the easy one. Now the SUV. This is where that time difference of 25 minutes comes in from the last sentence. Let's think about what that sentence means in terms of the times of each of these vehicles. If the sedan arrived 25 minutes after the SUV, then the sedan was driving 25 minutes longer; conversely, if the sedan arrived 25 minutes after the SUV, then the SUV was driving 25 minutes less than the sedan. The latter explanation is the one I used in the equation. Again, if the SUV was driving 25 minutes less than the sedan, and the equations are solved for time, then the equation for the SUV in terms of time is

\frac{120}{r+.4}=t-25 and we solve that for t:

\frac{120}{r+.4}+25=t

Again, going off the fact that times they both leave are the same, we set the equations equal to one another and solve for r:

\frac{120}{r+.4}+25=\frac{120}{r}

I began by first multiplying everything through by (r + .4) to get rid of it in the denominator. Doing that:

[r+.4](\frac{120}{r+.4}) +[r+.4](25)=[r+.4](\frac{120}{r}) which simplifies very nicely to

120+25(r+.4)=\frac{120}{r}(r+.4)  So maybe it's not so nice. Let's keep going:

120+25r+10=\frac{120r}{r}+\frac{48}{r} and keep going some more:

130+25r=120+\frac{48}{r} and now we multiply everything through by r to get rid of THAT denominator:

r(130)+r(25r)=r(120)+r(\frac{48}{r}) giving us:

130r+25r^2=120r+48 Now we have a second degree polynomial we have to solve by factoring. Get everything on one side and factor using the quadratic formula.

25r^2+10r-48=0

That factors to

r = 1.2 and r = -1.6 and both of those rates are in km/minute. First of all, we cannot have a negative rate (this is not physics where we are dealing with velocity which CAN be negative) so we throw out the -1.6 and convert the rate of 1.2 km/minute back to km/hr:

1.2\frac{km}{min} × \frac{60min}{1hr} and we get

r = 72 km/h, choice B.

Wow...what a pain THAT was, right?!

5 0
3 years ago
How do you get an equation to solve for all 3 sides of a triangle?
slavikrds [6]

In your solving toolbox (along with your pen, paper and calculator) you have these 3 equations:

1. The angles always add to 180°:

A + B + C = 180°

When you know two angles you can find the third.

 

2. Law of Sines (the Sine Rule):

Law of Sines

When there is an angle opposite a side, this equation comes to the rescue.

Note: angle A is opposite side a, B is opposite b, and C is opposite c.

 

3. Law of Cosines (the Cosine Rule):

Law of Cosines

This is the hardest to use (and remember) but it is sometimes needed  

to get you out of difficult situations.

It is an enhanced version of the Pythagoras Theorem that works  

on any triangle.

With those three equations you can solve any triangle (if it can be solved at all).

Six Different Types (More Detail)

There are SIX different types of puzzles you may need to solve. Get familiar with them:

1. AAA:

This means we are given all three angles of a triangle, but no sides.

AAA Triangle

AAA triangles are impossible to solve further since there are is nothing to show us size ... we know the shape but not how big it is.

We need to know at least one side to go further. See Solving "AAA" Triangles .

 

2. AAS

This mean we are given two angles of a triangle and one side, which is not the side adjacent to the two given angles.

AAS Triangle

Such a triangle can be solved by using Angles of a Triangle to find the other angle, and The Law of Sines to find each of the other two sides. See Solving "AAS" Triangles.

 

3. ASA

This means we are given two angles of a triangle and one side, which is the side adjacent to the two given angles.

ASA Triangle

In this case we find the third angle by using Angles of a Triangle, then use The Law of Sines to find each of the other two sides. See Solving "ASA" Triangles .

 

4. SAS

This means we are given two sides and the included angle.

SAS Triangle

For this type of triangle, we must use The Law of Cosines first to calculate the third side of the triangle; then we can use The Law of Sines to find one of the other two angles, and finally use Angles of a Triangle to find the last angle. See Solving "SAS" Triangles .

 

5. SSA

This means we are given two sides and one angle that is not the included angle.

SSA Triangle

In this case, use The Law of Sines first to find either one of the other two angles, then use Angles of a Triangle to find the third angle, then The Law of Sines again to find the final side. See Solving "SSA" Triangles .

 

6. SSS

This means we are given all three sides of a triangle, but no angles.

SSS Triangle

In this case, we have no choice. We must use The Law of Cosines first to find any one of the three angles, then we can use The Law of Sines (or use The Law of Cosines again) to find a second angle, and finally Angles of a Triangle to find the third angle.

7 0
4 years ago
Read 2 more answers
Evaluate the surface integral. s x2 + y2 + z2 ds s is the part of the cylinder x2 + y2 = 4 that lies between the planes z = 0 an
Leya [2.2K]
Parameterize the lateral face T_1 of the cylinder by

\mathbf r_1(u,v)=(x(u,v),y(u,v),z(u,v))=(2\cos u,2\sin u,v

where 0\le u\le2\pi and 0\le v\le3, and parameterize the disks T_2,T_3 as

\mathbf r_2(r,\theta)=(x(r,\theta),y(r,\theta),z(r,\theta))=(r\cos\theta,r\sin\theta,0)
\mathbf r_3(r,\theta)=(r\cos\theta,r\sin\theta,3)

where 0\le r\le2 and 0\le\theta\le2\pi.

The integral along the surface of the cylinder (with outward/positive orientation) is then

\displaystyle\iint_S(x^2+y^2+z^2)\,\mathrm dS=\left\{\iint_{T_1}+\iint_{T_2}+\iint_{T_3}\right\}(x^2+y^2+z^2)\,\mathrm dS
=\displaystyle\int_{u=0}^{u=2\pi}\int_{v=0}^{v=3}((2\cos u)^2+(2\sin u)^2+v^2)\left\|{{\mathbf r}_1}_u\times{{\mathbf r}_2}_v\right\|\,\mathrm dv\,\mathrm du+\int_{r=0}^{r=2}\int_{\theta=0}^{\theta=2\pi}((r\cos\theta)^2+(r\sin\theta)^2+0^2)\left\|{{\mathbf r}_2}_r\times{{\mathbf r}_2}_\theta\right\|\,\mathrm d\theta\,\mathrm dr+\int_{r=0}^{r=2}\int_{\theta=0}^{\theta=2\pi}((r\cos\theta)^2+(r\sin\theta)^2+3^2)\left\|{{\mathbf r}_3}_r\times{{\mathbf r}_3}_\theta\right\|\,\mathrm d\theta\,\mathrm dr
=\displaystyle2\int_{u=0}^{u=2\pi}\int_{v=0}^{v=3}(v^2+4)\,\mathrm dv\,\mathrm du+\int_{r=0}^{r=2}\int_{\theta=0}^{\theta=2\pi}r^3\,\mathrm d\theta\,\mathrm dr+\int_{r=0}^{r=2}\int_{\theta=0}^{\theta=2\pi}r(r^2+9)\,\mathrm d\theta\,\mathrm dr
=\displaystyle4\pi\int_{v=0}^{v=3}(v^2+4)\,\mathrm dv+2\pi\int_{r=0}^{r=2}r^3\,\mathrm dr+2\pi\int_{r=0}^{r=2}r(r^2+9)\,\mathrm dr
=136\pi
7 0
4 years ago
Please and thank you to whoever helps
Zepler [3.9K]

Answer:

  1131.0 m³

Step-by-step explanation:

Using the formula for the volume of a cylinder, fill in the appropriate values and do the arithmetic.

  V = πr²h

Since this formula needs the radius of the cylinder and you are given the diameter, you must divide the diameter by 2 to get the radius:

  r = d/2 = (12 m)/2 = 6 m

Then the volume is ...

  V = π(6 m)²(10 m) = 360π m³ ≈ 1131.0 m³ . . . . . rounded to tenths

8 0
4 years ago
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