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dlinn [17]
3 years ago
7

Does anyone know how to do this. No one will help.

Mathematics
2 answers:
givi [52]3 years ago
8 0

Answer:

150.7

Step-by-step explanation:

Formula for volume of a cone = \pi r^2\frac{h}{3}

where r = radius and h = height

We are given the diameter ( 8yds ) and the height ( 9yds )

In order to find the volume we need to find the radius

We can do this my dividing the diameter by 2 ( this is because the radius is equal to half the length of the diameter)

8/2 =4

So the radius is equal to 4yds

Now we plug in the values into the formula

( remember it says use 3.14 for π )

V=(3.14)4^2\frac{9}{3} \\4^2=16\\16*3.14=50.24\\\frac{9}{3} =3\\3*50.24=150.72

Finally we round to the nearest tenth and get that the answer is 150.7yd²

BARSIC [14]3 years ago
6 0

Answer:

150.8yd^3

Step-by-step explanation:

I helped lol.

Have a nice day! :-)

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Instructions:Select the correct answer from each drop-down menu. ∆ABC has vertices at A(11, 6), B(5, 6), and C(5, 17). ∆XYZ has
Akimi4 [234]

to compare the triangles, first we will determine the distances of each side

<span>Distance = ((x2-x1)^2+(y2-y1)^2)^0.5
</span>Solving 

<span>∆ABC  A(11, 6), B(5, 6), and C(5, 17)</span>

<span>AB = 6 units   BC = 11 units AC = 12.53 units
</span><span>∆XYZ  X(-10, 5), Y(-12, -2), and Z(-4, 15)
</span><span>XY = 7.14 units   YZ = 18.79 units XZ = 11.66 units</span>

<span>∆MNO  M(-9, -4), N(-3, -4), and O(-3, -15).</span>

<span>MN = 6 units   NO = 11 units MO = 12.53 units
</span><span>∆JKL  J(17, -2), K(12, -2), and L(12, 7).
</span><span>JK = 5 units   KL = 9 units JL = 10.30 units
</span><span>∆PQR  P(12, 3), Q(12, -2), and R(3, -2)
</span><span>PQ = 5 units   QR = 9 units PR = 10.30 units</span> 
Therefore
<span>we have the <span>∆ABC   and the </span><span>∆MNO  </span><span> 
with all three sides equal</span> ---------> are congruent  
</span><span>we have the <span>∆JKL  </span>and the <span>∆PQR 
</span>with all three sides equal ---------> are congruent  </span>

 let's check

 Two plane figures are congruent if and only if one can be obtained from the other by a sequence of rigid motions (that is, by a sequence of reflections, translations, and/or rotations).

 1)     If ∆MNO   ---- by a sequence of reflections and translation --- It can be obtained ------->∆ABC 

<span> then </span>∆MNO<span> ≅</span> <span>∆ABC  </span> 

 a)      Reflexion (x axis)

The coordinate notation for the Reflexion is (x,y)---- >(x,-y)

<span>∆MNO  M(-9, -4), N(-3, -4), and O(-3, -15).</span>

<span>M(-9, -4)----------------->  M1(-9,4)</span>

N(-3, -4)------------------ > N1(-3,4)

O(-3,-15)----------------- > O1(-3,15)

 b)      Reflexion (y axis)

The coordinate notation for the Reflexion is (x,y)---- >(-x,y)

<span>∆M1N1O1  M1(-9, 4), N1(-3, 4), and O1(-3, 15).</span>

<span>M1(-9, -4)----------------->  M2(9,4)</span>

N1(-3, -4)------------------ > N2(3,4)

O1(-3,-15)----------------- > O2(3,15)

 c)   Translation

The coordinate notation for the Translation is (x,y)---- >(x+2,y+2)

<span>∆M2N2O2  M2(9,4), N2(3,4), and O2(3, 15).</span>

<span>M2(9, 4)----------------->  M3(11,6)=A</span>

N2(3,4)------------------ > N3(5,6)=B

O2(3,15)----------------- > O3(5,17)=C

<span>∆ABC  A(11, 6), B(5, 6), and C(5, 17)</span>

 ∆MNO  reflection------- >  ∆M1N1O1  reflection---- > ∆M2N2O2  translation -- --> ∆M3N3O3 

 The ∆M3N3O3=∆ABC 

<span>Therefore ∆MNO ≅ <span>∆ABC   - > </span>check list</span>

 2)     If ∆JKL  -- by a sequence of rotation and translation--- It can be obtained ----->∆PQR 

<span> then </span>∆JKL ≅ <span>∆PQR  </span> 

 d)     Rotation 90 degree anticlockwise

The coordinate notation for the Rotation is (x,y)---- >(-y, x)

<span>∆JKL  J(17, -2), K(12, -2), and L(12, 7).</span>

<span>J(17, -2)----------------->  J1(2,17)</span>

K(12, -2)------------------ > K1(2,12)

L(12,7)----------------- > L1(-7,12)

 e)      translation

The coordinate notation for the translation is (x,y)---- >(x+10,y-14)

<span>∆J1K1L1  J1(2, 17), K1(2, 12), and L1(-7, 12).</span>

<span>J1(2, 17)----------------->  J2(12,3)=P</span>

K1(2, 12)------------------ > K2(12,-2)=Q

L1(-7, 12)----------------- > L2(3,-2)=R

 ∆PQR  P(12, 3), Q(12, -2), and R(3, -2)

 ∆JKL  rotation------- >  ∆J1K1L1  translation -- --> ∆J2K2L2=∆PQR 

<span>Therefore ∆JKL ≅ <span>∆PQR   - > </span><span>check list</span></span>
6 0
3 years ago
Someone please help me in this question it is kind of confusing it is a question under pre calculus in first year calculus i wil
masya89 [10]
Neither P, nor A are on the sketch 
I guess P is the upper right corner of the rectangle 
and A=(0,1) 

P belongs to the line going through (1,0) and B(0,y) 
<span>but we don't know the y-coordinate of B </span>

<span>the triangle is right and isosceles, so pythagoras a²+a²=2² ... 2a²=4 ... a²=2 ... a=sqrt2 </span>
now look at the right triangle BOA 
<span>his hypotenuse is AB=sqrt2 and the <span>the kathete</span> OA is 1 </span>
so y²+1²=(sqrt2)² ... y²+1=2 ... y²=1.. y=1 
so the coordinates of B are (0,1) 

the line going through (1,0) and (0,1) is L(x)=-x+1 

P belongs to this line, so the coordinates of P are P(x,-x+1) (0<x<1) 

b) so if that's P, the height of the rectangle is -x+1 and the width=2x 
<span>so its area A(x)=2x*(-x+1)= -2x²+2x 

I hope my answer has come to your help. Thank you for posting your question here in Brainly.

</span>
5 0
3 years ago
Of the 21 children at a park, 5 of them are playing on the swings. What is the approximate
Bond [772]

The Answer Is B

Use proportions to find x% of the children playing on the swings.

\frac{21}{100}  =  \frac{5}{x\%}

x  = 23.80952380\%

Meaning B is correct.

3 0
3 years ago
Select the correct answer. What are the solutions of this quadratic equation?
ipn [44]

I believe your answer is C.

5 0
3 years ago
Read 2 more answers
An 18 ft. tree casts a 15 ft. long shadow as shown in the following figure. Which of the following is the angle formed by the su
BabaBlast [244]

Answer:

The angle is 50.2°.

Step-by-step explanation:

The tree and its shadow from a right triangle with the tree being the perpendicular and the shadow being the base as shown in the figure attached.

Therefore the angle θ formed by the sun's rays and the ground is:

\theta = tan^-1(\frac{18}{15} )=\boxed{50.2^o}

8 0
3 years ago
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