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gogolik [260]
1 year ago
9

The price of Stock A at 9 A.M. was ​$15.03. Since​ then, the price has been increasing at the rate of ​$0.05 each hour. At noon

the price of Stock B was ​$15.53. It begins to decrease at the rate of ​$
0.14 each hour. If the two rates​ continue, in how many hours will the prices of the two stocks be the​ same?
Mathematics
1 answer:
irina1246 [14]1 year ago
3 0

After 1.842 hours the price of Stock A and Stock B will be the same using a set of linear equations.

A grouping of one or more linear equations containing the same variables is known as a system of linear equations.

For Stock A :

The initial price = $15.03

The rate of increase = $0.05

For Stock B :

The rate of decrease = $15.53

The rate of decrease per hour = $0.13

Then the price of stock A at noon will be:

$ 15.03 + $ (0.05 × 3) = $ 15.03 + $ 0.15 = $15.18

The period of time during which the stock price will remain constant:

15.18 + 0.05t = 15.53 - 0.14t

Simplifying the terms we get,

0.05t + 0.14t = 15.53 - 15.18

0.19t = 0.35

t = (0.35 ÷ 0.19) hours

t = 1.842 hours

Therefore, the stocks will be of equal price after 1.842 hours.

Learn more about stocks here:

brainly.com/question/1193187

#SPJ9

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a girl was asked about her age and she said she is half her mothers age and she was asked how old was her mother and she said i
Vlad [161]

Answer:

The girl is 19 y/o, the mom is 38 y/o and the dad is 43 y/o.

Step-by-step explanation:

26+13+31= 70    NO  

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7 0
2 years ago
Read 2 more answers
Use the diagram to find lengths. BP is the perpendicular bisector of AC. QC is the
faust18 [17]

Answer:

The length of the side PC is 34 cm.

Step-by-step explanation:

We are given that BP is the perpendicular bisector of AC. QC is the perpendicular bisector of BD. AB = BC = CD.

Suppose BP = 16 cm and AD = 90 cm.

As, it is given that AD = 90 cm and the three sides AB = BC = CD.

From the figure it is clear that AD = AB + BC + CD

So, AB = \frac{90}{3} = 30 cm

BC = \frac{90}{3} = 30 cm

CD = \frac{90}{3} = 30 cm

Since the triangle, BPC is a right-angled triangle as \anglePBC = 90°, so we can use Pythagoras theorem in this triangle to find the length of the side PC.

Now, the Pythagoras theorem states that;

\text{Hypotenuse}^{2} = \text{Perpendicular}^{2} +\text{Base}^{2}

\text{PC}^{2} = \text{BP}^{2} +\text{BC}^{2}

\text{PC}^{2} = \text{16}^{2} +\text{30}^{2}

\text{PC}^{2} = 256+900 = 1156

\text{PC}=\sqrt{1156}

PC = 34 cm

Hence, the length of the side PC is 34 cm.

4 0
2 years ago
What is the slope of the line passing through the points (0, 4) and (6, 13)
skelet666 [1.2K]

Answer:

The slope is \frac{3}{2}

Step-by-step explanation:

Hi there!

We are given the points (0, 4) and (6, 13)

We want to find the slope of the line containing these two points

Slope (m) is the steepness of the line. When calculating from two points, you use the equation \frac{y_2-y_1}{x_2-x_1}, where (x_1, y_1) and (x_2, y_2) are points

Although we have what we need to find the slope, let's label the values of the points to avoid confusion and mistakes.

x_1=0\\y_1=4\\x_2=6\\y_2=13

Substitute these values into the equation

m=\frac{y_2-y_1}{x_2-x_1}

m=\frac{13-4}{6-0}

Subtract

m=\frac{9}{6}

Simplify

m=\frac{3}{2}

The slope of the line is 3/2

Hope this helps!

See more on calculating slopes from 2 points here: brainly.com/question/26738734

3 0
2 years ago
Find the coordinates of the missing endpoint if P is the midpoint of Line Segment NQ.
dimaraw [331]
  • Midpoint formula is (\frac{x_1+x_2}{2},\frac{y_1+y_2}{2}) .
<h3>19.</h3>

So starting with this one, we will be solving for the coordinates of the unknown endpoint separately. Starting with the x-coordinate, since we know that the midpoint x-coordinate is 5 and the x-coordinate of N is 2, our equation is set up as such: \frac{x+2}{2}=5 From here we can solve for the x-coordinate of Q.

Firstly, multiply both sides by 2: x+2=10

Next, subtract both sides by 2 and your x-coordinate is x=8

With finding the y-coordinate, it's a similar process as with the x-coordinate except that we are using the y-coordinates of the midpoint and endpoint N.

\frac{y+0}{2}=2\\ y=4

<u>Putting it together, the missing endpoint is (8,4).</u>

<em>(The process is pretty much the same with the other problems, so I'll go through them real quickly.)</em>

<h3>20.</h3>

\frac{x+5}{2}=6\\ x+5=12\\ x=7

\frac{y+4}{2}=3\\ y+4=6\\ y=2

<u>The missing endpoint is (7,2).</u>

<h3>21.</h3>

\frac{x+3}{2}=-1\\ x+3=-2\\ x=-5

\frac{y+9}{2}=5\\ y+9=10\\y=1

<u>The missing endpoint is (-5,1).</u>

8 0
3 years ago
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