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3 years ago

Roberts photo service charges $15 to print 20 photos what is the price per photo

Mathematics

1 answer:

andriy [413]3 years ago

3 0

15÷20=0.75

he charges $0.75 per photo

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527,519 in expanded form using exponets

Ahat [919]

527,519 = (5 x 10^6) + (2 x 10^5) +
(7 x 10^4) + (5 x 10^2) + (1 x 10^1) +
(9 x 10^0)

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3 years ago

A recipe requires sugar, flour and cocoa in the ratio 5:4:1

WARRIOR [948]

Let x be the amount of cocoa you use in the recipe.

then 4x is the amount of flour used, and 5x is the amount sugar used.

x+4x+5x= 550g
10x=550g
x=55g

So, you use 55g of cocoa, 220g of flour, and 275g of sugar.

hope this helps! :)

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3 years ago

Brandon bought a baseball on for $15.00. He had a coupon for a 20% discount. How much did Brandon pay for the baseball?

IRINA_888 [86]

Brandon payed $12.00. So C.

8 0

3 years ago

Read 2 more answers

What is the rate of change between (-3.7, 1.4) and (-3.5,1)?

Anarel [89]

Answer:

-2

Step-by-step explanation:

the rate of change or slope (M) is -2

Hope this helps:P

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2 years ago

When a bactericide is added to a nutrient broth in which bacteria are growing, the bacteria population continues to grow for a

baherus [9]

Answer:

a) 1296 bacteria per hour

b) 0 bacteria per hour

c) -1296 bacteria per hour

Step-by-step explanation:

We are given the following information in the question:

The size of the population at time t is given by:

$b(t) = 9^6 + 6^4t-6^3t^2$

We differentiate the given function.

Thus, the growth rate is given by:

$\displaystyle\frac{db(t)}{dt} = \frac{d}{dt}(9^6 + 6^4t-6^3t^2)\\\\= 6^4-2(6^3)t$

a) Growth rates at t = 0 hours

$\displaystyle\frac{db(t)}{dt} \bigg|_{t=0}= 6^4-2(6^3)(0) = 1296\text{ bacteria per hour}$

b) Growth rates at t = 3 hours

$\displaystyle\frac{db(t)}{dt} \bigg|_{t=3}= 6^4-2(6^3)(3) = 0\text{ bacteria per hour}$

c) Growth rates at t = 6 hours

$\displaystyle\frac{db(t)}{dt} \bigg|_{t=6}= 6^4-2(6^3)(6) = -1296\text{ bacteria per hour}$

3 0

3 years ago