Ask question

Ask question

All categories

3 years ago

14

Alex bought 9 2/3 pounds of hamburger meat for a cookout. He plans to make 1/3 pound hamburgers.What is the maximum number of 1/

3 pound hamburgers Alex can make?

2 answers:

sladkih [1.3K]3 years ago

8 0

29 you divide 1/3 by 9 and add the extra 2

Alchen [17]3 years ago

7 0

It’s 29 hamburgers because 1 hamburger is 1/3 of beef then 9 pounds equals 27 burgers and then you have the left over 2. So it is 29

You might be interested in

Suppose measure of angle ABC equals 110, what is the measure of angle DBC?

shtirl [24]

There is no way of solving that unless we are given more information. Are ABC and DBC supplementary angles? If so, then m<DBC equals 70.

8 0

3 years ago

Round your answer to the nearest hundredth LA

m_a_m_a [10]

Answer:

Step-by-step explanation:

Since this is a right triangle, we will use a right triangle trig identity to solve our problem. The side opposite the reference angle A is given as 5, and the side adjacent to the reference angle is given as 7. The trig ratio that uses the sides opposite and adjacent is the tangent ratio:

$tanA=\frac{5}{7}$

Hit the 2nd button on your calculator then the tan button which displays:

$tan^{-1}($

Enter 5/7 after that open parenthesis and hit = to get that

A = 35.54°

8 0

3 years ago

A population of insects increases at a rate 230 + 8t + 0.9t2 insects per day (t in days). Find the insect population after 6 day

____ [38]

Answer:

The insect population after 6 days is of 1639 insects.

Step-by-step explanation:

A population of insects increases at a rate 230 + 8t + 0.9t2 insects per day

This means that $r(t) = 230 + 8t + 0.9t^2$

The population of insects after x days is given by:

$P(t) = \int_{0}^{x}r(t)dt$

So

$P(x) = \int_{0}^{x} (230 + 8t + 0.9t^2)$

$P(x) = 230t + 4t^2 + 0.3t^3 + K|_{0}^{x}$

$P(x) = 230x + 4x^2 + 0.3x^3 + K$

In which K is the initital population(which is 50). So

$P(x) = 230x + 4x^2 + 0.3x^3 + 50$

After 6 days:

$P(6) = 230*6 + 4*6^2 + 0.3*6^3 + 50 = 1638.8$

Rounding to the nearest insect, 1639

The insect population after 6 days is of 1639 insects.

7 0

3 years ago

Use the Law of Cosines to find the specified missing value. Approximate your answer to the nearest tenth.

Charra [1.4K]

<h3>Answer: 176.6</h3>

Work Shown:

$b^2 = a^2 + c^2 - 2*a*c*\cos(B)\\\\b^2 = 150^2 + 30^2 - 2*150*30*\cos(150)\\\\b^2 \approx 31194.22863406\\\\b \approx \sqrt{31194.22863406}\\\\b \approx 176.618879608211\\\\b \approx 176.6\\\\$

4 0

2 years ago

Write 720,080 in expanded form with exponents

disa [49]

Why are you having us do your homework

7 0

3 years ago

Read 2 more answers