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3 years ago

7

what is the value of the x variable in the solution to the following system of equations? 2x 2y = 6 x 3y = −1 a. −2 b. 2 c. 5 d.

−5

1 answer:

tatuchka [14]3 years ago

7 0

2x + 2y = 6 . . . . . (1)
x + 3y = -1 . . . . . (2)

(1) x 3 => 6x + 6y = 18 . . . . . (3)
(2) x 2 => 2x + 6y = -2 . . . . . (4)

(3) - (4) => 4x = 20
x = 20/4 = 5

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403, 892 in expanded form

eimsori [14]

Hey there!

<h3>"Expanded form" is basically, setting it up in a addition form to give you that result</h3>

$4 = 400,000 \\ 0 = 0 \\ 3 = 3,000 \\ 8 = 800 \\ 9 = 90 \\ 2 =2 \\ \\ \\ \\ Answer: 400,000 + 0 + 3,000 + 800 + 90 + 2$

Good luck on your assignment and enjoy your day!

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4 0

2 years ago

Last year, Josh had $20,000 to invest. He invested some of it in an account that paid 7% simple interest per year, and he invest

Ilya [14]

Answer: $12000 was invested at 7% while $8000 was invested at 9%.

Step-by-step explanation:

Based on the information given in the question, let the amount that was invested at 7% be n.

Therefore,

0.07n + 0.09(20000 - n) = 1560

0.07n + 1800 - 0.09n = 1560

-0.02n = 1560 - 1800

-0.02n = -240

n = 240/0.02

n = 120

Therefore, $12000 was invested at 7% while $8000 was invested at 9%.

3 0

2 years ago

Find the dimensions of the open rectangular box of maximum volume that can be made from a sheet of cardboard 11 in. by 7 in. by

11111nata11111 [884]

Answer:

The dimension of the open rectangular box is $8.216\times 4.216\times 1.392$ .

The volume of the box is 8.217 cubic inches.

Step-by-step explanation:

Given : The open rectangular box of maximum volume that can be made from a sheet of cardboard 11 in. by 7 in. by cutting congruent squares from the corners and folding up the sides.

To find : The dimensions and the volume of the open rectangular box ?

Solution :

Let the height be 'x'.

The length of the box is '11-2x'.

The breadth of the box is '7-2x'.

The volume of the box is $V=l\times b\times h$

$V=(11-2x)\times (7-2x)\times x$

$V=4x^3-36x^2+77x$

Derivate w.r.t x,

$V'(x)=4(3x^2)-2(36x)+77$

$V'(x)=12x^2-72x+77$

The critical point when V'(x)=0

$12x^2-72x+77=0$

Solve by quadratic formula,

$x=\frac{18+\sqrt{93}}{6},\frac{18-\sqrt{93}}{6}$

$x=4.607,1.392$

Derivate again w.r.t x,

$V''(x)=24x-72$

Now, $V''(4.607)=24(4.607)-72=38.568>0$ (+ve)

$V''(1.392)=24(1.392)-72=-38.592$ (-ve)

So, there is maximum at x=1.392.

The length of the box is $l=11-2x$

$l=11-2(1.392)=8.216$

The breadth of the box is $b=7-2x$

$b=7-2(1.392)=4.216$

The height of the box is h=1.392.

The dimension of the open rectangular box is $8.216\times 4.216\times 1.392$ .

The volume of the box is $V=l\times b\times h$

$V=8.216\times 4.216\times 1.392$

$V=48.217\ in.^3$

The volume of the box is 8.217 cubic inches.

5 0

2 years ago

Suppose I ask you to pick any four cards at random from a deck of 52, without replacement, and bet you one dollar that at least

Tatiana [17]

Answer:

a) No, because you have only 33.8% of chances of winning the bet.

b) No, because you have only 44.7% of chances of winning the bet.

Step-by-step explanation:

a) Of the total amount of cards (n=52 cards) there are 12 face cards (3 face cards: Jack, Queen, or King for everyone of the 4 suits: clubs, diamonds, hearts and spades).

The probabiility of losing this bet is the sum of:

- The probability of having a face card in the first turn

- The probability of having a face card in the second turn, having a non-face card in the first turn.

- The probability of having a face card in the third turn, having a non-face card in the previous turns.

- The probability of having a face card in the fourth turn, having a non-face card in the previous turns.

<u><em>1) The probability of having a face card in the first turn</em></u>

In this case, the chances are 12 in 52:

$P_1=P(face\, card)=12/52=0.231$

<u><em>2) The probability of having a face card in the second turn, having a non-face card in the first turn.</em></u>

In this case, first we have to get a non-face card (there are 40 in the dech of 52), and then, with the rest of the cards (there are 51 left now), getting a face card:

$P_2=P(non\,face\,card)*P(face\,card)=(40/52)*(12/51)=0.769*0.235=0.181$

<u><em>3) The probability of having a face card in the third turn, having a non-face card in the first and second turn.</em></u>

In this case, first we have to get two consecutive non-face card, and then, with the rest of the cards, getting a face card:

$P_3=(40/52)*(39/51)*(12/50)\\\\P_3=0.769*0.765*0.240=0.141$

<u><em>4) The probability of having a face card in the fourth turn, having a non-face card in the previous turns.</em></u>

In this case, first we have to get three consecutive non-face card, and then, with the rest of the cards, getting a face card:

$P_4=(40/52)*(39/51)*(38/50)*(12/49)\\\\P_4=0.769*0.765*0.76*0.245=0.109$

With these four probabilities we can calculate the probability of losing this bet:

$P=P_1+P_2+P_3+P_4=0.231+0.181+0.141+0.109=0.662$

The probability of losing is 66.2%, which is the same as saying you have (1-0.662)=0.338 or 33.8% of winning chances. Losing is more probable than winning, so you should not take the bet.

b) If the bet involves 3 cards, the only difference with a) is that there is no probability of getting the face card in the fourth turn.

We can calculate the probability of losing as the sum of the first probabilities already calculated:

$P=P_1+P_2+P_3=0.231+0.181+0.141=0.553$

There is 55.3% of losing (or 44.7% of winning), so it is still not convenient to bet.

5 0

3 years ago

If the sum of six consecutive even integers is 390 what is the smallest of the six integers

saveliy_v [14]

The answer is 60 but im going to explain it down below if you want to know how to do it :)

first because we dont know the values of the 6 numbers we have to lay it out like this.
x
x+2
x+4
x+6
x+8
x+10
-------- (then we add it together)
6x+30 = 390 (the sum of the 6 numbers)

using this equation we find out what x is

so we do 390-30 to find what 6x is, which is 360
because now we know that 6x=360, we divide 360 by 6 to find out what x is
360 divided by 6 =60..... x = 60

so now we know that the smallest integer is 60
to check the answer we could add the 6 numbers which are:
60+62+64+66+68+70=390 :)

so the smallest integer is 60

5 0

2 years ago