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jenyasd209 [6]
3 years ago
8

What is the solution of 4−6x≥−8 A.) x≥2 B.) x≤23 C.) x≤2 D.) x≤−2

Mathematics
2 answers:
Alenkasestr [34]3 years ago
5 0

Answer: I cant tell which one is a or b or c or d but ill say its b

Drupady [299]3 years ago
5 0
The correct answer is A, x is greater than or equal to 2.

Work: subtract -4 from both sides. So -6 is greater than it equal to -12. Then divide both sides by -6 to get x is greater than or equal to 2.
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The expression
swat32

Answer:

The expression

1

÷

1/4

is given. Give a real-life application to explain this expression and then simplify it.

Given a life expression to solve this, when one divide 100 pieces of pencil by 25 which is 1/4 of the entire pencil.

1/1/4=1 x 4/1

=4

Step-by-step explanation:

divide 1 by 1 over 4, then using cross multiply

then we have 1 multiply 4 divided by 1

4 is the final answer

6 0
3 years ago
If f(x) = 3x - 9 and g(x) = x², what is (gºf)(5)?​
Xelga [282]

Answer:

36

Step-by-step explanation:

Here, f(x) is the input to g(x), forming a composite function.

To evaluate  (gºf)(5),

1) find the value of f(5).  It is f(5) = 6.

2) Use this result as the input to g(x):  g(6) = (6)^2 = 36

3 0
3 years ago
In a basketball game, the two teams scored a total of 210 points. the home team won by 20 points how many points did the home te
guapka [62]

Answer:

105

Step-by-step explanation:


7 0
3 years ago
Read 2 more answers
A company wishes to manufacture some boxes out of card. The boxes will have 6 sides (i.e. they covered at the top). They wish th
Serhud [2]

Answer:

The dimensions are, base b=\sqrt[3]{200}, depth d=\sqrt[3]{200} and height h=\sqrt[3]{200}.

Step-by-step explanation:

First we have to understand the problem, we have a box of unknown dimensions (base b, depth d and height h), and we want to optimize the used material in the box. We know the volume V we want, how we want to optimize the card used in the box we need to minimize the Area A of the box.

The equations are then, for Volume

V=200cm^3 = b.h.d

For Area

A=2.b.h+2.d.h+2.b.d

From the Volume equation we clear the variable b to get,

b=\frac{200}{d.h}

And we replace this value into the Area equation to get,

A=2.(\frac{200}{d.h} ).h+2.d.h+2.(\frac{200}{d.h} ).d

A=2.(\frac{200}{d} )+2.d.h+2.(\frac{200}{h} )

So, we have our function f(x,y)=A(d,h), which we have to minimize. We apply the first partial derivative and equalize to zero to know the optimum point of the function, getting

\frac{\partial A}{\partial d} =-\frac{400}{d^2}+2h=0

\frac{\partial A}{\partial h} =-\frac{400}{h^2}+2d=0

After solving the system of equations, we get that the optimum point value is d=\sqrt[3]{200} and  h=\sqrt[3]{200}, replacing this values into the equation of variable b we get b=\sqrt[3]{200}.

Now, we have to check with the hessian matrix if the value is a minimum,

The hessian matrix is defined as,

H=\left[\begin{array}{ccc}\frac{\partial^2 A}{\partial d^2} &\frac{\partial^2 A}{\partial d \partial h}\\\frac{\partial^2 A}{\partial h \partial d}&\frac{\partial^2 A}{\partial p^2}\end{array}\right]

we know that,

\frac{\partial^2 A}{\partial d^2}=\frac{\partial}{\partial d}(-\frac{400}{d^2}+2h )=\frac{800}{d^3}

\frac{\partial^2 A}{\partial h^2}=\frac{\partial}{\partial h}(-\frac{400}{h^2}+2d )=\frac{800}{h^3}

\frac{\partial^2 A}{\partial d \partial h}=\frac{\partial^2 A}{\partial h \partial d}=\frac{\partial}{\partial h}(-\frac{400}{d^2}+2h )=2

Then, our matrix is

H=\left[\begin{array}{ccc}4&2\\2&4\end{array}\right]

Now, we found the eigenvalues of the matrix as follow

det(H-\lambda I)=det(\left[\begin{array}{ccc}4-\lambda&2\\2&4-\lambda\end{array}\right] )=(4-\lambda)^2-4=0

Solving for\lambda, we get that the eigenvalues are:  \lambda_1=2 and \lambda_2=6, how both are positive the Hessian matrix is positive definite which means that the functionA(d,h) is minimum at that point.

4 0
3 years ago
Mr. Sullivan places tiles numbered 1 through 18 in a bag and assigns a number to represent each of the 18 students in his class.
Marat540 [252]

Answer:

Mr. Sullivan <u>is</u> being fair because one tile has <u>an equal</u> probability of being selected over another tile.

Step-by-step explanation:

The process is fair because every tile has the same probability of being selected, and as the selection goes on, the selected students are removed.

5 0
3 years ago
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