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shtirl [24]
3 years ago
7

A 25 lb. bag Balkfoods Small Red Beans cost $52.92. What is the unit price per Ib?

Mathematics
1 answer:
lozanna [386]3 years ago
8 0

Answer:

$2.1 per pound

Step-by-step explanation:

price/pounds

52.92 / 25 = 2.1

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James had a peach that was 98 mm in diameter. One day
irinina [24]

Answer:

1927 times

Step-by-step explanation:

To do this, you divide the second value by the first value.

This gives us 188,869 divided by 98, which equals around 1927.

Hope this helps.

7 0
3 years ago
Find the area of the following figure below
abruzzese [7]

Answer:

30

Step-by-step explanation:

A=(a+b)/2xh

A=(6+9)/2x4

A=15/2x4

A=7.5x4

A=30

7 0
3 years ago
I don't get this question can someone help?
dedylja [7]

Answer:

I dont know

Step-by-step explanation:

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6 0
3 years ago
MCR3U1 Culminating 2021.pdf
11111nata11111 [884]

Answer:

(a) y = 350,000 \times (1 + 0.07132)^t

(b) (i) The population after 8 hours is 607,325

(ii) The population after 24 hours is 1,828,643

(c) The rate of increase of the population as a percentage per hour is 7.132%

(d) The doubling time of the population is approximately, 10.06 hours

Step-by-step explanation:

(a) The initial population of the bacteria, y₁ = a = 350,000

The time the colony grows, t = 12 hours

The final population of bacteria in the colony, y₂ = 800,000

The exponential growth model, can be written as follows;

y = a \cdot (1 + r)^t

Plugging in the values, we get;

800,000 = 350,000 \times (1 + r)^{12}

Therefore;

(1 + r)¹² = 800,000/350,000 = 16/7

12·㏑(1 + r) = ㏑(16/7)

㏑(1 + r) = (㏑(16/7))/12

r = e^((㏑(16/7))/12) - 1 ≈ 0.07132

The  model is therefore;

y = 350,000 \times (1 + 0.07132)^t

(b) (i) The population after 8 hours is given as follows;

y = 350,000 × (1 + 0.07132)⁸ ≈ 607,325.82

By rounding down, we have;

The population after 8 hours, y = 607,325

(ii) The population after 24 hours is given as follows;

y = 350,000 × (1 + 0.07132)²⁴ ≈ 1,828,643.92571

By rounding down, we have;

The population after 24 hours, y = 1,828,643

(c) The rate of increase of the population as a percentage per hour =  r × 100

∴   The rate of increase of the population as a percentage = 0.07132 × 100 = 7.132%

(d) The doubling time of the population is the time it takes the population to double, which is given as follows;

Initial population = y

Final population = 2·y

The doubling time of the population is therefore;

2 \cdot y = y \times (1 + 0.07132)^t

Therefore, we have;

2·y/y =2 = (1 + 0.07132)^t

t = ln2/(ln(1 + 0.07132)) ≈ 10.06

The doubling time of the population is approximately, 10.06 hours.

8 0
3 years ago
Find the intersection points using substitution or elimination for each system of equations:
natka813 [3]

Answer:

Step-by-step explanation:

Okay, so I think I know what the equations are, but I might have misinterpreted them because of the syntax- I think when you ask a question you can use the symbols tool to input it in a more clear way, otherwise you can use parentheses and such.

Problem 1:

(x²)/4 +y²= 1

y= x+1

*substitute for y*

Now we have a one-variable equation we can solve-

x²/4 + (x+1)² = 1

x²/4 + (x+1)(x+1)= 1

x²/4 + x²+2x+1= 1

*subtract 1 from both sides to set equal to 0*

x²/4 +x^2+2x=0

x²/4 can also be 1/4 * x²

1/4 * x² +1*x² +2x = 0

*combine like terms*

5/4 * x^2+2x+ 0 =0

now, you can use the quadratic equation to solve for x

a= 5/4

b= 2

c=0

the syntax on this will be rough, but I'll do my best...

x= (-b ± √(b²-4ac))/(2a)

x= (-2 ±√(2²-4*(5/4)*(0))/(2*(5/4))

x= (-2 ±√(4-0))/(2.5)

x= (-2±2)/2.5

x will have 2 answers because of ±

x= 0 or x= 1.6

now plug that back into one of the equations and solve.

y= 0+1 = 1

y= 1.6+1= 2.6

Hopefully this explanation was enough to help you solve problem 2.

Problem 2:

x² + y² -16y +39= 0

y²- x² -9= 0

6 0
3 years ago
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