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3 years ago

1.5x - 2.5. x=3. (show work)

Mathematics

2 answers:

Oxana [17]3 years ago

7 0

Step-by-step explanation:

Simplifying

1.5x + -2.5x = 3

Combine like terms: 1.5x + -2.5x = -1x

-1x = 3

Solving

-1x = 3

Solving for variable 'x'.

Move all terms containing x to the left, all other terms to the right.

Divide each side by '-1'.

x = -3

Simplifying

x = -3

ELEN [110]3 years ago

4 0

Answer:

1.5(3)-2.5

4.5 - 2.5

Step-by-step explanation:

the answer is 2 ;)

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In the bronze medal game of the 2014 FIBA Asia Cup, and with his team down by two points, Philippines guard Paul Dalistan was fo

dsp73

Complete Question

Question: In the bronze medal game of the 2014 FIBA Asia Cup... In the bronze medal game of the 2014 FIBA Asia Cup, and with his team down by two points, Philippines guard Paul Dalistan was fouled while attempting a three-pointer at the buzzer. He missed the shot and headed to the line for three foul shots.

1. Watch the video of Dalistan preparing to take his first shot. Before he shoots, what is the probability that Philippines will win the game? Lose the game? Send it into overtime? Dalistan’s free throw percentage is 84.6%

Answer:

The probability that Philippines will win the game is $P(A) = 0.6055$

The probability that Philippines will lose the game is $P(B) = 0.00365$

The probability that the game will be sent to overtime is $P(C) = 0.06$

Step-by-step explanation:

From the question we are told that

The probability of Paul Dalistan making a shot is p = 0.846

The probability of Paul Dalistan not making the shot is $q = 1- p = 1- 0.846=0.154$

Paul Dalistan team was down by two points

This mean that for Paul Dalistan's team to win the game he must make the 3 free throws

So the probability that Philippines will win the game is mathematically represented as

$P(A) = p^3$

=> $P(A) = 0.846^3$

=> $P(A) = 0.6055$

Gnerally for Paul Dalistan's team to loss the game it means that Paul Dalistan did not make any of the free shot

Hence the probability that Philippines will loss the game is

$P(B) = q^3$

=> $P(B) = (0.154 )^3$

=> $P(B) = 0.00365$

Generally for the game to be sent to overtime then the game must be tied and for the game to be tied, Paul Dalistan have to make 2 out of the three shots

Hence the probability that game will go to overtime is

$P(C) = 3 * 0.846 * 0.154 * 0.154$

$P(C) = 0.06$

6 0

4 years ago

Use Lagrange multipliers to find the maximum and minimum values of the function subject to the given constraint. (If an answer d

aliya0001 [1]

The Lagrangian

$L(x,y,z,\lambda)=x^2+y^2+z^2+\lambda(x^4+y^4+z^4-13)$

has critical points where the first derivatives vanish:

$L_x=2x+4\lambda x^3=2x(1+2\lambda x^2)=0\implies x=0\text{ or }x^2=-\dfrac1{2\lambda}$

$L_y=2y+4\lambda y^3=2y(1+2\lambda y^2)=0\implies y=0\text{ or }y^2=-\dfrac1{2\lambda}$

$L_z=2z+4\lambda z^3=2z(1+2\lambda z^2)=0\implies z=0\text{ or }z^2=-\dfrac1{2\lambda}$

$L_\lambda=x^4+y^4+z^4-13=0$

We can't have $x=y=z=0$ , since that contradicts the last condition.

(0 critical points)

If two of them are zero, then the remaining variable has two possible values of $\pm\sqrt[4]{13}$ . For example, if $y=z=0$ , then $x^4=13\implies x=\pm\sqrt[4]{13}$ .

(6 critical points; 2 for each non-zero variable)

If only one of them is zero, then the squares of the remaining variables are equal and we would find $\lambda=-\frac1{\sqrt{26}}$ (taking the negative root because $x^2,y^2,z^2$ must be non-negative), and we can immediately find the critical points from there. For example, if $z=0$ , then $x^4+y^4=13$ . If both $x,y$ are non-zero, then $x^2=y^2=-\frac1{2\lambda}$ , and

$xL_x+yL_y=2(x^2+y^2)+52\lambda=-\dfrac2\lambda+52\lambda=0\implies\lambda=\pm\dfrac1{\sqrt{26}}$

$\implies x^2=\sqrt{\dfrac{13}2}\implies x=\pm\sqrt[4]{\dfrac{13}2}$

and for either choice of $x$ , we can independently choose from $y=\pm\sqrt[4]{\frac{13}2}$ .

(12 critical points; 3 ways of picking one variable to be zero, and 4 choices of sign for the remaining two variables)

If none of the variables are zero, then $x^2=y^2=z^2=-\frac1{2\lambda}$ . We have

$xL_x+yL_y+zL_z=2(x^2+y^2+z^2)+52\lambda=-\dfrac3\lambda+52\lambda=0\implies\lambda=\pm\dfrac{\sqrt{39}}{26}$

$\implies x^2=\sqrt{\dfrac{13}3}\implies x=\pm\sqrt[4]{\dfrac{13}3}$

and similary $y,z$ have the same solutions whose signs can be picked independently of one another.

(8 critical points)

Now evaluate $f$ at each critical point; you should end up with a maximum value of $\sqrt{39}$ and a minimum value of $\sqrt{13}$ (both occurring at various critical points).

Here's a comprehensive list of all the critical points we found:

$(\sqrt[4]{13},0,0)$

$(-\sqrt[4]{13},0,0)$

$(0,\sqrt[4]{13},0)$

$(0,-\sqrt[4]{13},0)$

$(0,0,\sqrt[4]{13})$

$(0,0,-\sqrt[4]{13})$

$\left(\sqrt[4]{\dfrac{13}2},\sqrt[4]{\dfrac{13}2},0\right)$

$\left(\sqrt[4]{\dfrac{13}2},-\sqrt[4]{\dfrac{13}2},0\right)$

$\left(-\sqrt[4]{\dfrac{13}2},\sqrt[4]{\dfrac{13}2},0\right)$

$\left(-\sqrt[4]{\dfrac{13}2},-\sqrt[4]{\dfrac{13}2},0\right)$

$\left(\sqrt[4]{\dfrac{13}2},0,\sqrt[4]{\dfrac{13}2}\right)$

$\left(\sqrt[4]{\dfrac{13}2},0,-\sqrt[4]{\dfrac{13}2}\right)$

$\left(-\sqrt[4]{\dfrac{13}2},0,\sqrt[4]{\dfrac{13}2}\right)$

$\left(-\sqrt[4]{\dfrac{13}2},0,-\sqrt[4]{\dfrac{13}2}\right)$

$\left(0,\sqrt[4]{\dfrac{13}2},\sqrt[4]{\dfrac{13}2}\right)$

$\left(0,\sqrt[4]{\dfrac{13}2},-\sqrt[4]{\dfrac{13}2}\right)$

$\left(0,-\sqrt[4]{\dfrac{13}2},\sqrt[4]{\dfrac{13}2}\right)$

$\left(0,-\sqrt[4]{\dfrac{13}2},-\sqrt[4]{\dfrac{13}2}\right)$

$\left(\sqrt[4]{\dfrac{13}3},\sqrt[4]{\dfrac{13}3},\sqrt[4]{\dfrac{13}3}\right)$

$\left(\sqrt[4]{\dfrac{13}3},\sqrt[4]{\dfrac{13}3},-\sqrt[4]{\dfrac{13}3}\right)$

$\left(\sqrt[4]{\dfrac{13}3},-\sqrt[4]{\dfrac{13}3},\sqrt[4]{\dfrac{13}3}\right)$

$\left(-\sqrt[4]{\dfrac{13}3},\sqrt[4]{\dfrac{13}3},\sqrt[4]{\dfrac{13}3}\right)$

$\left(\sqrt[4]{\dfrac{13}3},-\sqrt[4]{\dfrac{13}3},-\sqrt[4]{\dfrac{13}3}\right)$

$\left(-\sqrt[4]{\dfrac{13}3},\sqrt[4]{\dfrac{13}3},-\sqrt[4]{\dfrac{13}3}\right)$

$\left(-\sqrt[4]{\dfrac{13}3},-\sqrt[4]{\dfrac{13}3},\sqrt[4]{\dfrac{13}3}\right)$

$\left(-\sqrt[4]{\dfrac{13}3},-\sqrt[4]{\dfrac{13}3},-\sqrt[4]{\dfrac{13}3}\right)$

5 0

3 years ago

The area of a rectangular patio is 5 5/8 square yards, and it's length is 1 1/2 yards. What is the patio's width in yards?

MAXImum [283]

Hello there! Thank you for asking your question here at Brainly. I will be assisting you today with answering this problem, and will be teaching you how to deal with it on your own in the future.

First, let's take a look at our question, and evaluate it.

"The area of a rectangular patio is 5 5/8 square yards, and it's length is 1 1/2 yards. What is the patio's width in yards?"

To clarify this problem, we are looking for the patio's width.

Let's first understand what the area of a rectangular shape is.
The formula for the area of a rectangle is "Length times Width", or "L • W".

So this is how the equation should look like:
A = L • W
We have our area, 5 5/8, and we have our length, 1 1/2.

To make things more simple, let's convert our fractions to decimals. Now, to convert our fractions to decimals, let's set our denominators (the numbers on the bottom of a fraction) equal to another fraction, with x as the numerator (the numbers on the top of a fraction) and 100 as the denominator.

So we have 1/2 and x/100. Divide 100 by 2 to find x (as 1/2 of anything is dividing by 2).
100 / 2 = 50, so 1 1/2 = 1.50 as a decimal.

Now, let's try 5/8.
1/8 = 0.125, so multiply 0.125 by 5.
0.125 • 5 = 0.625.
5 5/8 = 5.625 as a decimal.

So, now we have our equation:
A = L • W
Plug in our numbers.
5.625 = 1.50x

To isolate and solve for x, we need to divide both sides by 1.50, so let's do that.
5.625 / 1.50 = 3.75
1.50x / 1.50 = x

We are now left with:
x = 3.75

Your answer is:
The patio's width is 3.75 yards.

I hope this helps!

8 0

3 years ago

7. The graph of part of linear function g is shown on the grid. Which inequality best represents the domain of the part shown?

PSYCHO15rus [73]

Answer:

Step-by-step explanation:

Domain means the change in x.

so the domain will be which values of x the fn goes through.

you can see the black line goes from -9 to 2 in x values.

the -9 isn't filled in, but the 2 is. filled in means it will be with an equals instead of without.

so x is between -9 and 2, 2 is with equals -9 is without,

so the first one.

3 0

3 years ago

How to find the fraction and decimal of 66 and 2/3?

puteri [66]

When converting to a fraction u multiply 66*3 getting 198 and u add the numerator getting 200 you u keep the demomenator which is 3 so 66and 2/3 is 200/3.

7 0

3 years ago