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3 years ago

the tigers won twice as many football games as they lost. they played 96 games. how many games did they win?

Mathematics

1 answer:

alukav5142 [94]3 years ago

7 0

Half of 96 is 48 and they won twice as many as they lost.. how can this math problem be real XD

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An hourglass consists of two sets of congruent composite figures on either end. Each composite figure is made up of a cone and a

belka [17]

Answer: third option 268.8

Explanation:

For this kind of proble it is very important that you attach the figure because it contains important information to understand the question.

I have attached the figure for better understanding.

1) The top portion (and the bottom is congruent but rotated 180°) of the hourglass is a figure equivalent to a cylinder on top and a cone on bottom.

So the total volume contained in the top portion is the volume of a cylinder + the volume of a cone.

This is how you calculate the volume of the top portion:

1) The height of the cylinder is 54 mm - 18 mm = 36 mm

2) The formula for the volume of a cylinder is V = π (radius)^2 * height

radius = 8 mm
height = 36 mm

=> V = π(8mm)^2 * 36 mm = 2304π (mm)^3

3) The formula for the volume of a cone is V = (1/3)π(radius)^2 * height

radius = 8 mm
height = 18 mm

V = (1/3)π(8mm)^2 * 18 mm = 384π (mm)^3

4) The total volume of the top portion is volumen of the cylindrical part + voume of the cone:

Total voluem = 2304π (mm)^3 + 384π (mm)^3 = 2688π (mm)^3

5) To find the number of seconds <span>it take until all of the sand has dripped to the bottom of the hourglass you have to divide the total volumen of sand by the rate:

time in seconds = total volume of sand / rate of dripping

time in seconds = [2688 π (mm)^3 ] / [10π (mm)^3 / s] = 268.8 s

That is the answer: 268.8 s
</span>

4 0

4 years ago

Read 2 more answers

Mr. Watson uses 10.5 gallons of gasoline to drive the first 313 miles of the trip. How many gallons of gasoline would he need to

adelina 88 [10]

450/313*10.5 = 15.09.... which is approx is 15

7 0

3 years ago

Hank is throwing a surprise party. The party costs $20 per person. Create an equation and graph to represent the relationship be

o-na [289]

Answer: Answer with Step-by-step explanation:

Since we have given that

Cost of party per person = $20.

Let the number of friends be 'f'.

Let the cost of party be 'p'.

According to question, our equation becomes

p= 2f

So, if p = 100, then f = p/2 = 100/2 = 50 if f = 10, then, p = 2f = 2 • 10 = 20 If p = 300 then, f = 300/2 = 150, If f = 20, then, p = 2f = 2 • 20 = 40 So, Table as follows:

Number of friends cost of party

50 100

10 20

150 300

20 40

3 0

3 years ago

I need help on this question

leva [86]

Answer:

15.7

Step-by-step explanation:

The equation for finding the circumference of a circle is

2 x pi x r(radius). Radius is half the diameter.

5/2=2.5

2 x 3.14 x 2.5

=15.7

7 0

3 years ago

The n term of a geometric sequence is denoted by Tn and the sum of the first n terms is denoted by Sn.Given T6-T4=5/2 and S5-S3=

Leno4ka [110]

1 step: $S_{5}=T_{1}+T_{2}+T_{3}+T_{4}+T_{5}$ , $S_{3}=T_{1}+T_{2}+T_{3}$ , then
$S_{5}-S_{3}=T_{4}+T_{5}=5$ .

2 step: $T_{n}=T_{1}*q^{n-1}$ , then
$T_{6}=T_{1}*q^{5}$
$T_{5}=T_{1}*q^{4}$
$T_{4}=T_{1}*q^{3}$
$T_{3}=T_{1}*q^{2}$
and $\left \{ {{T_{6}-T_{4}= \frac{5}{2} } \atop {T_{5}+T_{4}=5}} \right.$ will have form $\left \{ {{T_1*q^{5}-T_{1}*q^{3}= \frac{5}{2} } \atop {T_{1}*q^{4}+T_{1}*q^{3}=5} \right.$ .

3 step: Solve this system $\left \{ {{T_1*q^{3}*(q^{2}-1)= \frac{5}{2} } \atop {T_{1}*q^{3}*(q+1)=5} \right.$ and dividing first equation on second we obtain $\frac{q^{2}-1}{q+1}= \frac{ \frac{5}{2} }{5}$ . So, $\frac{(q-1)(q+1)}{q+1} = \frac{1}{2}$ and $q-1= \frac{1}{2}$ , $q= \frac{3}{2}$ - the common ratio.

4 step: Insert $q= \frac{3}{2}$ into equation $T_{1}*q^{3}*(q+1)=5$ and obtain $T_{1}* \frac{27}{8}*( \frac{3}{2}+1 ) =5$ , from where $T_{1}= \frac{16}{27}$ .

5 0

3 years ago