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3 years ago

15

And amusement parks his child and adult tickets on a ratio of 8 to 1 on Saturday they sold 147 more child tickets then adult tic

kets how many tickets that the amusement park sell on Saturday

2 answers:

BARSIC [14]3 years ago

6 0

Dosen't make sense.. 8-1?

Contact [7]3 years ago

3 0

Answer:

they sold 3 tickets

Step-by-step explanation:

they multiply by 8 divided 147

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List all the prime numbers between 50 and 60

leva [86]

Answer:

53, 57, 59

Step-by-step explanation:

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3 years ago

How many solutions does the following equation have? −4(x+5)=−4x−20

Nady [450]

Answer:

Infinite Solution!

Step-by-step explanation:

First, We simplify the right side.

Distribute -4, -4x-20=-4x-20

Now we add +4x to both sides, now the equation stands as -20=-20

We know when the solution is same #= same #. We have infinite solution!

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3 years ago

Calculate the area of the shaded region between the regular polygons. Round to the nearest hundredth.

wolverine [178]

Answer:

I hope I'll help you with this but I am not sure of it all...

The area of the shaded region is 46,77

Step-by-step explanation:

To determine the area of the shaded hexagon, we must first calculate te area of the whole shaded hexagon and minus that by the white little hexagon.

The formula to calculate the area of a hexagon is (3√3 s2)/ 2

(where s = one side of the hexagon)

if one side a the little white hexagon equals 6 ft;

3√3 6ft.2/2 = 18√3/2 = 9√3

now for the bigger hexagon where one side is 12 ft;

3√3 12 ft.2/2 = 72√3/2 = 36√3

now the area of the bigger one minus the area of the smaller one;

36√3 - 9√3 = 46,77

3 0

3 years ago

The isotope samarium-151 decays into europium-151, with a half-life of around 96.6 years. A rock contains 5 grams of samarium-15

Olegator [25]

Part A:

The amount of substance left of a radioactive substance with a half life $t_{ \frac{1}{2} }$ and amount of initial substance $N_0$ is given by

$N(t)=N_0\left( \frac{1}{2} \right)^{ \frac{t}{t_{ \frac{1}{2} }} }$

Given that th<span>e isotope samarium-151 has a half-life of around 96.6 years and the rock contains 5 grams of samarium-151 when it reaches its closure temperature, and it contains 0.625 grams when it is discovered.

</span><span>The time since the rock reached its closure temperature is obtained as follows:

$0.625=5\left( \frac{1}{2} \right)^{ \frac{t}{96.6} } \\ \\ \Rightarrow\left( \frac{1}{2} \right)^{ \frac{t}{96.6} }= \frac{0.625}{5} =0.125 \\ \\ \Rightarrow\frac{t}{96.6}\log{ \frac{1}{2} }=\log0.125 \\ \\ \Rightarrow\frac{t}{96.6}= \frac{\log0.125}{\log{ \frac{1}{2} }} =3 \\ \\ \Rightarrow t=3(96.6)=289.8$

Therefore, </span>t<span>he time since the rock reached its closure temperature is 289.8 years.

Part B:

Given that the </span><span>rock contains 5 grams of samarium-151 when it reaches its closure temperature, and it contains 0.625 grams when it is discovered. The number of grams of europium-151 when the rock was discovered is 5 - 0.625 = 4.375</span>

7 0

4 years ago

What is the measure of EG?

IrinaVladis [17]

The first thing you should know in this case is that a circumference has a total measure in 360 degrees.
We have then that the formula to find EG in this case is:
EG + GF + FE = 360
We cleared EG:
EG = 360-GF-FE
We substitute the values:
EG = 360-83-66
EG = 211
Answer
EG = 211 degrees.

7 0

3 years ago