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rosijanka [135]
3 years ago
10

Which expressions represent the statement “three minus the product of seven and four”? Mark all that apply.

Mathematics
1 answer:
antiseptic1488 [7]3 years ago
4 0
The answer is A and D because as long as it is three minus you can still switch up the multiplication side
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Which statement accurately describes the relationship between mass and inertia?
Andreyy89
D mass is proportional to 1/inertia2
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Find the area of the circle if r = 5 meters. Leave the answer in terms of π.
olya-2409 [2.1K]

<em>Hope</em><em> </em><em>this</em><em> </em><em>will</em><em> </em><em>help</em><em> </em><em>u</em><em>.</em><em>.</em><em>.</em>

4 0
3 years ago
PLEASE PLEASE PLEASE HELP!!!!! 95 points!!!
zysi [14]

The "rule" being described here is nothing more than the input/output of a mathematical function.


For every input 'x' value supplied, you only need to subtract three to it. For every input 'y' value, you only need to add four to it.


Example: I'll use variable 'm' to represent this function. Variable 'p' will represent the current input point.


m(p) = p[x - 3, y + 4] = p[-7 - 3, 0 + 4] = p[-10, 4]. 'p[]" is just the point.

I tried my best hope that help!

5 0
4 years ago
Read 2 more answers
Suppose a rectangular pasture is to be constructed using 1 2 linear mile of fencing. The pasture will have one divider parallel
timama [110]

Answer:

\displaystyle A=\frac{1}{192}

Step-by-step explanation:

<u>Maximization With Derivatives</u>

Given a function of one variable A(x), we can find the maximum or minimum value of A by using the derivatives criterion. If A'(x)=0, then A has a probable maximum or minimum value.

We need to find a function for the area of the pasture. Let's assume the dimensions of the pasture are x and y, and one divider goes parallel to the sides named y, and two dividers go parallel to x.

The two divisions parallel to x have lengths y, thus the fencing will take 4x. The three dividers parallel to y have lengths x, thus the fencing will take 3y.

The amount of fence needed to enclose the external and the internal divisions is

P=4x+3y

We know the total fencing is 1/2 miles long, thus

\displaystyle 4x+3y=\frac{1}{2}

Solving for x

\displaystyle x=\frac{\frac{1}{2}-3y}{4}

The total area of the pasture is

A=x.y

Substituting x

\displaystyle A=\frac{\frac{1}{2}-3y}{4}.y

\displaystyle A=\frac{\frac{1}{2}y-3y^2}{4}

Differentiating with respect to y

\displaystyle A'=\frac{\frac{1}{2}-6y}{4}

Equate to 0

\displaystyle \frac{\frac{1}{2}-6y}{4}=0

Solving for y

\displaystyle y=\frac{1}{12}

And also

\displaystyle x=\frac{\frac{1}{2}-3\cdot \frac{1}{12}}{4}=\frac{1}{16}

Compute the second derivative

\displaystyle A''=-\frac{3}{2}.

Since it's always negative, the point is a maximum

Thus, the maximum area is

\displaystyle A=\frac{1}{12}\cdot \frac{1}{16}=\frac{1}{192}

6 0
3 years ago
Evaluate:<br> -48 + 9 - 18<br> Ik what the answer is <br> I will just give you hella points
IceJOKER [234]

Answer: -57

Step-by-step explanation: -48 + (9 - 18) -48 + -9 = -57

3 0
3 years ago
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