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3 years ago

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On Tuesday the value of a stock decreased by 1/2 point. On Wednesday, the value increased by 3/4 point. On Thursday the value de

creased by 3/8 point. What integer represents the change in stock value after three days?

1 answer:

Tpy6a [65]3 years ago

3 0

The change is -(1/2 + 3/4 + 3/8).
If you add the fractions, you get -(13/8).

You might be interested in

Arte-miy333 [17]

Answer:

We know that for a line:

y = a*x + b

where a is the slope and b is the y-intercept.

Any line with a slope equal to -(1/a) will be perpendicular to the one above.

So here we start with the line:

3x + 4y + 5 = 0

let's rewrite this as:

4y = -3x - 5

y = -(3/4)*x - (5/4)

So a line perpendicular to this one, has a slope equal to:

- (-4/3) = (4/3)

So the perpendicular line will be something like:

y = (4/3)*x + c

We know that this line passes through the point (a, 3)

this means that, when x = a, y must be equal to 3.

Replacing these in the above line equation, we get:

3 = (4/3)*a + c

c = 3 - (4/3)*a

Then the equation for our line is:

y = (4/3)*x + 3 - (4/3)*a

We can rewrite this as:

y = (4/3)*(x -a) + 3

now we need to find the point where this line ( y = -(3/4)*x - (5/4)) and the original line intersect.

We can find this by solving:

(4/3)*(x -a) + 3 = y = -(3/4)*x - (5/4)

(4/3)*(x -a) + 3 = -(3/4)*x - (5/4)

(4/3)*x - (3/4)*x = -(4/3)*a - 3 - (5/4)

(16/12)*x - (9/12)*x = -(4/3)*a - 12/4 - 5/4

(7/12)*x = -(4/13)*a - 17/4

x = (-(4/13)*a - 17/4)*(12/7) = - (48/91)*a - 51/7

And the y-value is given by inputin this in any of the two lines, for example with the first one we get:

y = -(3/4)*(- (48/91)*a - 51/7) - (5/4)

= (36/91)*a + (153/28) - 5/4

Then the intersection point is:

( - (48/91)*a - 51/7, (36/91)*a + (153/28) - 5/4)

And we want that the distance between this point, and our original point (3, a) to be equal to 4.

Remember that the distance between two points (a, b) and (c, d) is:

distance = √( (a - c)^2 + (b - d)^2)

So here, the distance between (a, 3) and ( - (48/91)*a - 51/7, (36/91)*a + (153/28) - 5/4) is 4

4 = √( (a + (48/91)*a + 51/7)^2 + (3 - (36/91)*a + (153/28) - 5/4 )^2)

If we square both sides, we get:

4^2 = 16 = (a + (48/91)*a + 51/7)^2 + (3 - (36/91)*a - (153/28) + 5/4 )^2)

Now we need to solve this for a.

16 = (a*(1 + 48/91) + 51/7)^2 + ( -(36/91)*a + 3 - 5/4 + (153/28) )^2

16 = ( a*(139/91) + 51/7)^2 + ( -(36/91)*a - (43/28) )^2

16 = a^2*(139/91)^2 + 2*a*(139/91)*51/7 + (51/7)^2 + a^2*(36/91)^2 + 2*(36/91)*a*(43/28) + (43/28)^2

16 = a^2*( (139/91)^2 + (36/91)^2) + a*( 2*(139/91)*51/7 + 2*(36/91)*(43/28)) + (51/7)^2 + (43/28)^2

At this point we can see that this is really messy, so let's start solving these fractions.

16 = (2.49)*a^2 + a*(23.47) + 55.44

0 = (2.49)*a^2 + a*(23.47) + 55.44 - 16

0 = (2.49)*a^2 + a*(23.47) + 39.44

Now we can use the Bhaskara's formula for quadratic equations, the two solutions will be:

$a = \frac{-23.47 \pm \sqrt{23.47^2 - 4*2.49*39.4} }{2*2.49} \\\\a = \frac{-23.47 \pm 12.57 }{4.98}$

Then the two possible values of a are:

a = (-23.47 + 12.57)/4.98 = -2.19

a = (-23.47 - 12.57)/4.98 = -7.23

4 0

2 years ago

27.3% of as fraction

Kruka [31]

Answer:

273/1000

Step-by-step explanation:

7 0

2 years ago

Read 2 more answers

If you add 12 to my number and then multiply the result by 3, you will get 64 more than two-thirds of my number. Find my number.

anastassius [24]

Answer:

The number is 12

Step-by-step explanation:

[] First, let's turn all these words into something mathematical. N will equal "my number"

-><u> If you add 12 to my number</u> and then multiply the result by 3, you will get 64 more than two-thirds of my number.

-> n + 12 <u>and then multiply the result by 3</u>, you will get 64 more than two-thirds of my number.

-> 3(n + 12), you will get 64 more than <u>two-thirds of my number</u>.

-> 3(n + 12) = <u>64 more than</u> $\frac{2n}{3}$

-> 3(n + 12) = 64 + $\frac{2n}{3}$

[] Phew, okay. Now it is something we can solve and less scary;

[Given]

3(n + 12) = 64 + $\frac{2n}{3}$

[Distribute]

3n + 36 = 64 + $\frac{2n}{3}$

[Multiply both sides by 3]

9n + 108 = 192 + 2n

[Subtract 108 and 2n from both sides]

7n = 84

[Divide both sides by 4]

n = 12

Have a nice day!

I hope this is what you are looking for, but if not - comment! I will edit and update my answer accordingly. (ノ^∇^)

- Heather

8 0

2 years ago

2x^3-x^2-3x=210<br> the answer is 5 but I want to know why.

mel-nik [20]

Answer:

$x=5,\frac{-9+\sqrt{255}i }{4} ,\frac{-9-\sqrt{255}i }{4}$

Step-by-step explanation:

1) Move all terms to one side.

$2x^{3} -x^{2} -3x-210=0$

2) Factor $2{x}^{3}-{x}^{2}-3x-210$ using Polynomial Division.

1 - Factor the following.

$2x^{3} -x^{2} -3x-210$

2 - First, find all factors of the constant term 210.

$1,2,3,4,5,6,7,10,14,15,21,30,35,42,70,105,210$

3) Try each factor above using the Remainder Theorem.

Substitute 1 into x. Since the result is not 0, x-1 is not a factor..

$2*1^{3} -1^{2} -3*1-210=-212$

Substitute -1 into x. Since the result is not 0, x+1 is not a factor..

$2(-1)^{3} -(-1)^{2} -3*-1-210=-210$

Substitute 2 into x. Since the result is not 0, x-2 is not a factor..

$2*2^{3} -2^{2} -3*2-210=-204$

Substitute -2 into x. Since the result is not 0, x+2 is not a factor..

$2{(-2)}^{3}-{(-2)}^{2}-3\times -2-210 = -224$

Substitute 3 into x. Since the result is not 0, x-3 is not a factor..

$2\times {3}^{3}-{3}^{2}-3\times 3-210 = -174$

Substitute -3 into x. Since the result is not 0, x+3 is not a factor..

$2{(-3)}^{3}-{(-3)}^{2}-3\times -3-210 = -264$

Substitute 5 into x. Since the result is 0, x-5 is a factor..

$2\times {5}^{3}-{5}^{2}-3\times 5-210 =0$

------------------------------------------------------------------------------------------

⇒ $x-5$

4) Polynomial Division: Divide $2{x}^{3}-{x}^{2}-3x-210$ by $x-5$ .

$2x^{2}$ $9x$ $42$

-------------------------------------------------------------------------

$x-5$ | $2x^{3}$ $-x^{2}$ $-3x$ $-210$

$2x^{3}$ $-10x^{2}$

-----------------------------------------------------------------------

$9x^{2}$ $-3x$ $-210$

--------------------------------------------------------------------------

$42x$ $-210$

$42x$ $-210$

-------------------------------------------------------------------------

5) Rewrite the expression using the above.

$2x^2+9x+42$

$(2x^2+9x+42)(x-5)=0$

3) Solve for $x.$

$x=5$

4) Use the Quadratic Formula.

1 - In general, given $a{x}^{2}+bx+c=0$ , there exists two solutions where:

$x=\frac{-b+\sqrt{b^{2} -4ac} }{2a} ,\frac{-b-\sqrt{b^2-4ac} }{2a}$

2 - In this case, $a=2,b=9$ and $c = 42.$

$x=\frac{-9+\sqrt{9^2*-4*2*42} }{2*2} ,\frac{-9-\sqrt{9^2-4*2*42} }{2*2}$

3 - Simplify.

$x=\frac{-9+\sqrt{255}i }{4} ,\frac{-9-\sqrt{255}i }{4}$

5) Collect all solutions from the previous steps.

$x=5,\frac{-9+\sqrt{255}i }{4} ,\frac{-9-\sqrt{255}i }{4}$

4 0

2 years ago

Read 2 more answers

Henry has 2 accounts with a credit limit of $6000. He has $500 in monthly debts, including a $200 car payment. He has never made

Volgvan

I think it's D as an answer but I could be wrong

8 0

3 years ago