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zmey [24]
3 years ago
13

Determine the next step for solving the quadratic equation by completing the square. 0 = –2x2 + 2x + 3 –3 = –2x2 + 2x –3 = –2(x2

– x) –3 + = –2(x2 – x + ) = –2(x – )2 = (x – )2 The two solutions are
Mathematics
2 answers:
VLD [36.1K]3 years ago
8 0
We have that
0 = –2x²<span>+ 2x + 3
</span>

Group terms that contain the same variable, and move the constant to the opposite side of the equation

(–2x²+ 2x)=-3

Factor the leading coefficient 

-2*(x²-x)=-3

Complete the square. Remember to balance the equation by adding the same constants to each side

-2*(x²-x+0.25)=-3-0.50

Rewrite as perfect squares

-2*(x-0.5)²=-3.50

(x-0.5)²=-3.50/-2

(x-0.5)²=1.75------> square root both sides

(+/-)(x-0.5)=√7/2

(+)(x-0.5)=√7/2----> (√7/2)+0.5----> x=1.823

(-)(x-0.5)=√5/2-----> x=0.5-(√7/2)----> x=-0.823

the solutions are

x=1.823

x=-0.823

guapka [62]3 years ago
5 0

Answer:

-1/2, 1/4, 1/2

Step-by-step explanation:

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A rectangular room is 12.5 feet wide and 14 feet long. How many linear feet of baseboards are needed to trim the room?
QveST [7]

If you do width times length you will get 30. Good luck

8 0
3 years ago
Which of the following is the most appropriate unit to describe the rate at which people are entering a theme park?
ad-work [718]

The <u><em>correct answer</em></u> is:

d) People per hour, because the dependent quantity is the people

Explanation:

In this situation, the two quantities are people and hours. These are the two things in this problem we can count or measure.

The independent variable is the one that causes a change, while the dependent variable is the one that <em>gets</em> changed. In this situation, the number of people change every hour; this means the number of people <em>gets</em> changed, which makes it the dependent variable. This means that the independent variable must be time.

Since people is dependent and time is independent, "people per hour" would be the best form of this statement.

6 0
3 years ago
If you have 7x^2 + 3 if the input is 4 what is the output
Usimov [2.4K]
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7(4)^2 + 3
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112 + 3
115.
5 0
3 years ago
Read 2 more answers
How do I solve these questions?
stich3 [128]
Y = -4x - 5.....(0,?)...so we know x = 0, so sub in 0 for x and solve for y
y = -4(0) - 5
y = -5....so one point is (0,-5)
do the same for the other points and u get (1,-9) , (-1,-1)

y = -4x - 5.....slope is -4 and ur y int (where ur line crosses the y axes) is (0,-5).
to find the x int (where the line crosses the x axes), sub in 0 for y and solve for x.
0 = -4x - 5
4x = -5
x = -5/4.....or - 1 1/4(for graphing purposes)...x int is (-5/4,0)
so plot all ur points and connect the dots
============================
4x + 9y = 0...(-9,?)..so x = -9..so sub in -9 for x and solve for y
4(-9) + 9y = 0
-36 + 9y = 0
9y = 36
y = 36/9
y = 4...so ur point is (-9,4)

do the same for the others and ur points are : (0,0) , (9,-4)

4x + 9y = 0
9y = -4x....so ur slope is -4 and ur y int is (0,0)
ur x int is : (0,0)....so basically u have a line going through the origin (0,0)
plot all ur points and connect the dots
6 0
3 years ago
5. A tank initially contains 20 lb of salt dissolved in 200-gallon of water. Starting at time t = 0, a solution containing 0.5 l
sladkih [1.3K]

Answer:

(a) Amount of salt as a function of time

A(t)=100-80e^{-0.02t}

(b) The time at which the amount of salt in the tank reaches 50 lb is 23.5 minutes.

(c) The amount of salt when t approaches to +inf is 100 lb.

Step-by-step explanation:

The rate of change of the amount of salt can be written as

\frac{dA}{dt} =rate\,in\,-\,rate\,out\\\\\frac{dA}{dt}=C_i*q_i-C_o*q_o=C_i*q_i-\frac{A(t)}{V}*q_o\\\\\frac{dA}{dt}=0.5*4-\frac{A(t)}{200}*4\\\\\frac{dA}{dt} =\frac{100-A(t)}{50}=-(\frac{A(t)-100}{50})

Then we can rearrange and integrate

\frac{dA}{dt}= -(\frac{A(t)-100}{50})\\\\\int \frac{dA}{A-100}=-\frac{1}{50}  \int dt\\\\ln(A-100)=-\frac{t}{50}+C_0\\\\ A-100=Ce^{-0.02*t}\\\\\\A(0)=20 \rightarrow 20-100=Ce^0=C\\C=-80\\\\A=100-80e^{-0.02t}

Then we have the model of A(t) like

A(t)=100-80e^{-0.02t}

(b) The time at which the amount of salt reaches 50 lb is

A(t)=100-80e^{-0.02t}=50\\\\e^{-0.02t}=(50-100)/(-80)=0.625\\\\-0.02*t=ln(0.625)=-0.47\\\\t=(-0.47)/(-0.02)=23.5

(c) When t approaches to +infinit, the term e^(-0.02t) approaches to zero, so the amount of salt in the solution approaches to 100 lb.

7 0
3 years ago
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