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Mashutka [201]
1 year ago
7

Write a system of linear inequalities so the points (1, 2) and (4, −3) are solutions of the system, but the point (−2, 8) is not

a solution of the system. PLEASE SOMEONE HELP AND SOLVE IT AS A SYSTEM OF LINEAR EQUATIONS
Mathematics
2 answers:
dalvyx [7]1 year ago
7 0
I can’t see it where is it it’s blurry
Travka [436]1 year ago
4 0

Answer:

<u>The sample system is given below:</u>

  • y < 5
  • y > x - 8

<em>see the attached for the graph</em>

The points (1, 2) and (4, -3) are in the intersected region but the point (-2, 8) is outside

  • <em>The easy way to get the inequalities to plot the points and add lines to divide the regions where desired point are inside and not required point is outside. </em>
  • <em>I put y = 5 and y = x - 8 lines and shaded the region below y = 5 line and above the other line.</em>

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A consumer is considering two different purchasing options for the car of their choice. The first option, which is leasing, is d
AVprozaik [17]

Answer:

  • y=250x+4000
  • y=400x+400

Step-by-step explanation:

Given that:

  • x represents the number of months of ownership; and
  • y represents the total paid for the car after ‘x' months.

<u>First Option (Leasing)</u>

250x - y + 4000 = 0

Expressing the equation in the Slope-Intercept Form y=mx+b, we have:

y=250x+4000

<u>Second Option (Financing)</u>

$400 for 0 months of ownership, (0,400), and $4400 for 10 months of ownership, (10, 4400).

First, we determine the slope of the line joining (0,400) and (10,4400)

Slope, m= \dfrac{4400-400}{10-0}= \dfrac{4000}{10}=400

We have:

y=400x+b

When y=400, x=0

400=400(0)+b

b=400

Therefore, the Slope-Intercept Form of the second option is:

y=400x+400

<u>Significance</u>

  • In the first option, there is a down payment of $4000 and a monthly payment of $250.
  • In the second option, there is a down payment of $400 and a monthly payment of $400.

<u>Part B</u>

We notice from the graph that after 24 months, the cost for leasing and financing becomes the same ($10,000). Therefore, a consumer will be better off financing since the downpayment for leasing is higher.

<u>i.e </u>

  • When x=0, y=$4000 for leasing
  • When x=0, y=$400 for financing

4 0
2 years ago
1. What is m∠CAD?
bulgar [2K]
< CAD = 100....if u add < ACB + < CBA u get < CAD
================
< DAB = 125 and < ACB = 30

if DAB = 125.....then BAC = 180 - 125 = 55
and all 3 interior angles of a triangle = 180
< BAC + < ACB + < ABC = 180
55 + 30 + < ABC = 180
85 + < ABC = 180
< ABC = 180 - 85
< ABC = 95 <===


6 0
2 years ago
Read 2 more answers
Consider the initial value problem: 2ty′=8y, y(−1)=1. Find the value of the constant C and the exponent r so that y=Ctr is the s
VikaD [51]

The correct question is:

Consider the initial value problem

2ty' = 8y, y(-1) = 1

(a) Find the value of the constant C and the exponent r such that y = Ct^r is the solution of this initial value problem.

b) Determine the largest interval of the form a < t < b on which the existence and uniqueness theorem for first order linear differential equations guarantees the existence of a unique solution.

c) What is the actual interval of existence for the solution obtained in part (a) ?

Step-by-step explanation:

Given the differential equation

2ty' = 8y

a) We need to find the value of the constant C and r, such that y = Ct^r is a solution to the differential equation together with the initial condition y(-1) = 1.

Since Ct^r is a solution to the initial value problem, it means that y = Ct^r satisfies the said problem. That is

2tdy/dt - 8y = 0

Implies

2td(Ct^r)/dt - 8(Ct^r) = 0

2tCrt^(r - 1) - 8Ct^r = 0

2Crt^r - 8Ct^r = 0

(2r - 8)Ct^r = 0

But Ct^r ≠ 0

=> 2r - 8 = 0 or r = 8/2 = 4

Now, we have r = 4, which implies that

y = Ct^4

Applying the initial condition y(-1) = 1, we put y = 1 when t = -1

1 = C(-1)^4

C = 1

So, y = t^4

b) Let y = F(x,y)................(1)

Suppose F(x, y) is continuous on some region, R = {(x, y) : x_0 − δ < x < x_0 + δ, y_0 −ę < y < y_0 + ę} containing the point (x_0, y_0). Then there exists a number δ1 (possibly smaller than δ) so that a solution y = f(x) to (1) is defined for x_0 − δ1 < x < x_0 + δ1.

Now, suppose that both F(x, y)

and ∂F/∂y are continuous functions defined on a region R. Then there exists a number δ2

(possibly smaller than δ1) so that the solution y = f(x) to (1) is

the unique solution to (1) for x_0 − δ2 < x < x_0 + δ2.

c) Firstly, we write the differential equation 2ty' = 8y in standard form as

y' - (4/t)y = 0

0 is always continuous, but -4/t has discontinuity at t = 0

So, the solution to differential equation exists everywhere, apart from t = 0.

The interval is (-infinity, 0) n (0, infinity)

n - means intersection.

7 0
1 year ago
A random sample of 65 high school seniors was selected from all high school seniors at a certain high school. The following scat
Over [174]

Step-by-step explanation:

The line of best fit is y = 2.599x + 105.08.  At x = 20, the estimated height is y = 157.06.  The actual height is 160 cm, so the residual is:

160 cm − 157.06 cm = 2.94 cm

3 0
1 year ago
We suspect that automobile insurance premiums (in dollars) may be steadily decreasing with the driver's driving experience (in y
ANTONII [103]

Answer:

D) a chi square test for independence.

Step-by-step explanation:

Given that we  suspect that automobile insurance premiums (in dollars) may be steadily decreasing with the driver's driving experience (in years), so we choose a random sample of drivers who have similar automobile insurance coverage and collect data about their ages and insurance premiums.

We are to check whether two variables insurance premiums and driving experience are associated.

Two categorical variables are compared for different ages and insurance premiums.

Hence a proper test would be

D) a chi square test for independence.

8 0
1 year ago
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