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4 years ago

9

A wheel with a diameter of 12 inches will move 337 when rolled five times. How far will the same wheel move to the nearest inch,

in 16 rolls.

1 answer:

charle [14.2K]4 years ago

6 0

You will have to add 337+12= 349

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Neko [114]

Expansions and compressions are transformations that change the length or width of the graph of a function.

To graph y = a*f (x)
if a> 1, the graph of y = f (x) is expanded vertically by a factor a.
We have then:
f (x) = 2/5 x ^ 2-2
The function g (x) is a vertical stretch of f (x) by a factor of 2:
g (x) = 2f (x)
g (x) = 2 (2/5 x ^ 2-2)
g (x) = 4/5 x ^ 2-4

Answer:
The equation of g (x) is:
g (x) = 4/5 x ^ 2-4

8 0

3 years ago

Use the table and equation to answer the following question.

ale4655 [162]

go ask ur mom for help not me ty

8 0

2 years ago

Find the GCF of the given polynomial.<br> 16a 4 b 4 + 32a 3 b 5 - 48a 2 b 6

astra-53 [7]

Best Answer: <span> 8x^6y^5 - 3x^8y^3

GCF of 8 and 3 is 1
GCF of x^6 and x^8 is x^6 the lowest exponent of x.
GCF of y^5 and y^3 is y^3 the lowest exponent of y.
Total GCF = 1 × x^6 × y^3
= x^6y^3

8x^6y^5 - 3x^8y^3
= x^6y^3(8y^2 – 3x^2)
---- </span><span>This is the correct answer!!!</span>

3 0

3 years ago

A jeweler had a fixed amount of gold to make bracelets and necklaces. The amount c

kvv77 [185]

The jeweler made 7 bracelets and 9 necklaces .

In the question ,

it is given that

amount of gold in each bracelet is $=$ 6 grams

amount of gold in each necklace is $=$ 24 grams .

let the number of bracelets $=$ "b" .

let the number of necklaces $=$ "n" .

since the total number of necklace and bracelet is 16 .

the equation is b + n = 16

so , b = 16 - n

also given that total gold used is 258 grams .

so the equation is 6b + 24n = 258

Substituting b= 16 - n in the equation ,

we get , 6*(16 - n) + 24n = 258

96 - 6n + 24n = 258

96 + 18n = 258

18n = 258 - 96

18n = 162

n = 162/18

n = 9

and b = 16 - 9 = 7

Therefore , The jeweler made 7 bracelets and 9 necklaces .

Learn more about Equations here

brainly.com/question/10994155

#SPJ1

3 0

2 years ago

A graduate school plans to increase its enrollment capacity by developing its facilities and the programs it offers. Their enrol

kap26 [50]

Answer:

4.4 years

Step-by-step explanation:

Next time, please share the answer choices.

In this case, the beginning enrollment capacity was 120. The rate of increase was 3.00, the terminal enrollment cap 3,240. We let t represent the number of years:

3,240 students = (120 students)(3.00)^t

We need to solve this for t, the number of years required before enrollment will hit 3,240 students.

Solve 3240 = 120(3)^t for t:

Divide both sides by 120, obtaining 27.

Then we have

127 = 3^t

Logarithms are the best tool to use here to find t. Take the natural log of both sides:

log 127 = t log 3. Solving for t: t = (log 127) / (log 3), or t = 2.104 / 0.477 = 4.41

Thus, starting at 120 students, enrollment reached 3,240 students in 4.4 years.

7 0

3 years ago