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3 years ago

Solve the following System Of Equations Using Substitution or Elimination Methods. Show Work.

Mathematics

1 answer:

Ostrovityanka [42]3 years ago

8 0

Answer:

There are two pairs of solutions: (2,7) and (-1,4)

Step-by-step explanation:

We will use substitution.

y = x^2 + 3

y = x +5

Since the second equation is equal to y, replace y in the first equation with the second equation.

y = x^2 + 3

x + 5 = x^2 + 3

Rearrange so that one side is equal to 0.

5 - 3 = x^2 - x

2 = x^2 - x

0 = x^2 - x - 2

You may use quadratic formula or any form of factoring to find the zeros (x values that make the equation equal to 0).

a = 1, b = -1, c = -2

Zeros = $\frac{-b + \sqrt{b^{2}-4ac } }{2a}$ and $\frac{-b - \sqrt{b^{2}-4ac } }{2a}$

Zeros = 2 and -1

Now that you have your x values, plug them into the equations to find their corresponding y values.

y = x^2 + 3

y = (2)^2 + 3

y = 7

Pair #1: (2,7)

y = x^2 + 3

y = (-1)^2 + 3

y = 4

Pair #2: (-1,4)

Therefore, there are two pairs of solutions: (2,7) and (-1,4).

You might be interested in

A quiz-show contestant is presented with two questions, question 1 and question 2, and she can choose which question to answer f

Mrrafil [7]

Answer:

The contestant should try and answer question 2 first to maximize the expected reward.

Step-by-step explanation:

Let the probability of getting question 1 right = P(A) = 0.60

Probability of not getting question 1 = P(A') = 1 - P(A) = 1 - 0.60 = 0.40

Let the probability of getting question 2 right be = P(B) = 0.80

Probability of not getting question 2 = P(B') = 1 - P(B) = 1 - 0.80 = 0.20

To obtain the better option using the expected value method.

E(X) = Σ xᵢpᵢ

where pᵢ = each probability.

xᵢ = cash reward for each probability.

There are two ways to go about this.

Approach 1

If the contestant attempts question 1 first.

The possible probabilities include

1) The contestant misses the question 1 and cannot answer question 2 = P(A') = 0.40; cash reward associated = $0

2) The contestant gets the question 1 and misses question 2 = P(A n B') = P(A) × P(B') = 0.6 × 0.2 = 0.12; cash reward associated with this probability = $200

3) The contestant gets the question 1 and gets the question 2 too = P(A n B) = P(A) × P(B) = 0.6 × 0.8 = 0.48; cash reward associated with this probability = $300

Expected reward for this approach

E(X) = (0.4×0) + (0.12×200) + (0.48×300) = $168

Approach 2

If the contestant attempts question 2 first.

The possible probabilities include

1) The contestant misses the question 2 and cannot answer question 1 = P(B') = 0.20; cash reward associated = $0

2) The contestant gets the question 2 and misses question 1 = P(A' n B) = P(A') × P(B) = 0.4 × 0.8 = 0.32; cash reward associated with this probability = $100

3) The contestant gets the question 2 and gets the question 1 too = P(A n B) = P(A) × P(B) = 0.6 × 0.8 = 0.48; cash reward associated with this probability = $300

Expected reward for this approach

E(X) = (0.2×0) + (0.32×100) + (0.48×300) = $176

Approach 2 is the better approach to follow as it has a higher expected reward.

The contestant should try and answer question 2 first to maximize the expected reward.

Hope this helps!!!

3 0

3 years ago

A company receives shipments of a component used in the manufacture of a component for a high-end acoustic speaker system. When

Tanya [424]

Answer:

$ME= 2.33*\sqrt{\frac{0.096*(1-0.096)}{250}}= 0.0434$

Step-by-step explanation:

For this case we have a sample size of n = 250 units and in this sample they found that 24 units failed one or more of the tests.

We are interested in the proportion of units that fail to meet the company's specifications, and we can estimate this with:

$\hat p = \frac{24}{250}= 0.096$

The margin of error is the range of values below and above the sample statistic in a confidence interval.

Normal distribution, is a "probability distribution that is symmetric about the mean, showing that data near the mean are more frequent in occurrence than data far from the mean".

The confidence interval for a proportion is given by this formula

$\hat p \pm z_{\alpha/2} \sqrt{\frac{\hat p(1-\hat p)}{n}}$

For the 98% confidence interval the value of $\alpha=1-0.98=0.02$ and $\alpha/2=0.01$ , with that value we can find the quantile required for the interval in the normal standard distribution.

$z_{\alpha/2}=2.33$

And the margin of error would be:

$ME= 2.33*\sqrt{\frac{0.096*(1-0.096)}{250}}= 0.0434$

4 0

4 years ago

What is 8(3 + a) = 64 a =

ahrayia [7]

Answer: a = 3

Step-by-step explanation:

3 0

3 years ago

kareem took 5 hours to drive 470km from Sudbury to Brantford. for part of the trip he drove at 100km/h. for the rest of the trip

alexira [117]

Answer:

He drove 200 km at 100 km/h and 270 km at 90 km/h

Step-by-step explanation:

Distance = Speed × Time

Assume that Kareem drove at 100 km/h for x hour

∵ Kareem drove at 100 km/h for x hours

∴ He drove a distance = 100 × x

∴ He drove a distance = 100 x km

∵ He took 5 hours in whole trip

- Subtract x from 5 to find the time that he took in the rest

of the trip

∴ The time of the rest of the trip = 5 - x hours

∵ For the rest of the trip he drove at 90 km/h

∴ He drove a distance = 90 × (5 - x)

∴ He drove a distance = 450 - 90 x

∵ He drove 470 km in whole trip

- Add the two distance and equate the sum by 470

∴ 100 x + 450 - 90 x = 470

- Add the like terms in the left hand side

∴ 10 x + 450 = 470

- Subtract 450 from both sides

∴ 10 x = 20

- Divide both sides by 10

∴ x = 2

Substitute the value of x in each distance to find the distance of each part

∵ The distance of the first part = 100 x

∴ The distance of the first part = 100(2)

∴ The distance of the first part = 200 km

∵ The distance of the second part = 450 - 90 x

∴ The distance of the second part = 450 - 90(2) = 450 - 180

∴ The distance of the second part = 270 km

He drove 200 km at 100 km/h and 270 km at 90 km/h

4 0

3 years ago

Read the ques

Dafna11 [192]

Answer:

0.25feet

Step-by-step explanation:

The equation is not well written. Let the equation of the height be modelled as;

h = -16d²+8d+4

The velocity of the ball is zero at its maximum height.

Velocity = change in displacement/time

v = dh/dd

Differentiate

v = -32d+8

Since dh/dd = v = 0

0 = -32d+8

Add 32d to both sides

0+32d = 8

32d = 8

Divide both sides by 32

32d/32 = 8/32

d = 1/4

d = 0.25feet

Hence the maximum height of the tennis ball is 0.25feet

Note that the modeled equation was assumed. You can apply the same calculation to any equation given

3 0

3 years ago