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3 years ago

I need help on both of those 4 problems

Mathematics

1 answer:

Masja [62]3 years ago

7 0

Are u trying to see if these r true and false?

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A clown had 10 balloons that he sold at a carnival dor 6 cents each. if he sold all 10 balloons, how mush money did he make

saveliy_v [14]

Answer:

Step-by-step explanation:

You would do 10*0.06 to get 0.6

So her sold 60 cents worth of balloons

6 0

3 years ago

Read 2 more answers

What is 12.25 as a fraction

Finger [1]

Answer:

$\large\boxed{12.25=\dfrac{49}{4}}$

Step-by-step explanation:

$12.\underbrace{25}_{2}=\dfrac{1225}{1\underbrace{00}_2}=\dfrac{1225:25}{100:25}=\dfrac{49}{4}\\\\12.25=12+0.25=12+\dfrac{25}{100}=12+\dfrac{25:25}{100:25}=12+\dfrac{1}{4}\\\\=\dfrac{12\cdot4}{4}+\dfrac{1}{4}=\dfrac{48+1}{4}=\dfrac{49}{4}$

8 0

3 years ago

Mr. Collins ordered 8,000 oz of stone how many tons of stone did he order?

padilas [110]

Answer:

0.25.

Step-by-step explanation:

7 0

3 years ago

find the formula for a rational function that satisfies all of the following. its x-intercept is (-3,0). it has a vertical asymp

Norma-Jean [14]

Let that be

$\\ \rm\Rightarrow y=\dfrac{a(f(x))}{g(x)}$

Two vertical asymptotes at -1 and 0

$\\ \rm\Rightarrow y=\dfrac{a(f(x)}{x(x+1)}$

If we simply

$\\ \rm\Rightarrow y=\dfrac{a(f(x)}{x^2+1}$

Denominator has degree 2
Numerator should have degree as 2 and coefficient as 3 inorder to get horizontal asymptote y=3 means the quadratic equation should contain 3x²
But there should be a x intercept at -3 so one zeros should be -3

Find a equation

3x²+9x

Find zeros

3x²+9x=0
3x(x+3)=0
x=0,-3

Horizontal asymptote

3x²/x²
3

So our equation is

$\\ \rm\Rightarrow y=\dfrac{3x^2+9x}{x(x+1)}$

Graph attached

4 0

2 years ago

Please help it would be much appreciated!

sergiy2304 [10]

Answer:

Part a) Option d

Part b) Option a

Step-by-step explanation:

Part a

if we look at the options given and the data available

Option a) x^4+9

Putting x= 2 we get (2^4) + 9 =25

Putting x= 3 we get (3^4) + 9 =90 but f(x) =125 so not correct option

Option b) (4^x)+9

Putting x= 2 we get (4^2) + 9 =25

Putting x= 3 we get (4^3) + 9 =73 but f(x) =125 so not correct option

Option c) x^5

Putting x= 2 we get (2^5) =32 but f(x) =25 so not correct option

Option d) 5^x

Putting x= 2 we get (5^2) =25

Putting x= 3 we get (5^3) =125

Putting x= 4 we get (5^4) =625

So Option d is correct.

Part (b)

3(2)^3x

can be solved as:

=3(2^3)^x

=3(8)^x

So, correct option is a

8 0

3 years ago