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3 years ago

You’re five neighboring restaurants charge $9.95, $12.95, 8.50, 8.95, and $10.54 hamburger special. What is the mean price

Mathematics

1 answer:

Maslowich3 years ago

7 0

$10.18 Add all the numbers then divide by the amount in this case 5

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Pooja's plant began sprouting 2days before Pooja bought it, and she had it for 98 days until it died. At its tallest, the plant

masya89 [10]

98 for x and 30 for y

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4 years ago

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The original price of a set of bunk beds is $70. How much will Ronnie pay if he buys it during the sale?

qaws [65]

The answer to this question would be 40$

8 0

4 years ago

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Which of the following is a valid objective function for a linear programming problem

OLga [1]

The objective function of a linear programming problem is a linear function.

A valid objective function is (b) Min 4x+3y+(2/3)z

The form of the objective function of a linear programming problem is:

$\mathbf{ax + by + cz +.....}$

The above function means that:

The objective function cannot contain the product of the variables
The objective function cannot contain the quotient of the variables

The above highlights mean that:

The objective function can contain the sum of the variables
The objective function can contain the difference of the variables

Hence, the valid objective function is (b) Min 4x+3y+(2/3)z

Read more about objective functions at:

brainly.com/question/11206462

8 0

3 years ago

There were 240 students at a primary school. After some new students were transferred into the school, there were 282 students i

stepan [7]

Answer:

Numbers of students at a primary school before= 240

Number of increased students =42

Again,

Increased number of students in percent(%)= 42/240×100%= 17.5%

Therefore 17.5% students are increased this year.

3 0

3 years ago

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Find the derivative.

Aleksandr [31]

Answer:

Using either method, we obtain: $t^\frac{3}{8}$

Step-by-step explanation:

a) By evaluating the integral:

$\frac{d}{dt} \int\limits^t_0 {\sqrt[8]{u^3} } \, du$

The integral itself can be evaluated by writing the root and exponent of the variable u as: $\sqrt[8]{u^3} =u^{\frac{3}{8}$

Then, an antiderivative of this is: $\frac{8}{11} u^\frac{3+8}{8} =\frac{8}{11} u^\frac{11}{8}$

which evaluated between the limits of integration gives:

$\frac{8}{11} t^\frac{11}{8}-\frac{8}{11} 0^\frac{11}{8}=\frac{8}{11} t^\frac{11}{8}$

and now the derivative of this expression with respect to "t" is:

$\frac{d}{dt} (\frac{8}{11} t^\frac{11}{8})=\frac{8}{11}\,*\,\frac{11}{8}\,t^\frac{3}{8}=t^\frac{3}{8}$

b) by differentiating the integral directly: We use Part 1 of the Fundamental Theorem of Calculus which states:

"If f is continuous on [a,b] then

$g(x)=\int\limits^x_a {f(t)} \, dt$

is continuous on [a,b], differentiable on (a,b) and $g'(x)=f(x)$

Since this this function $u^{\frac{3}{8}$ is continuous starting at zero, and differentiable on values larger than zero, then we can apply the theorem. That means:

$\frac{d}{dt} \int\limits^t_0 {u^\frac{3}{8} } } \, du=t^\frac{3}{8}$

5 0

3 years ago