All categories

3 years ago

Solve 2x + 4y = -8 and 4x + 8y = 5 by elimination. Please show work on how to get the answer.

Mathematics

1 answer:

goldfiish [28.3K]3 years ago

5 0

Answer:

No solution

Step-by-step explanation:

Given two equations are 2x+4y= -8 and 4x+8y= 5

These two lines are parallel They don't have a solution because they do not meet .

Multiply the first equation by 2 we get 4x+8y = -16

the other equation is 4x+8y= 5 clearly these two are parallel .

To prove by elimination method we have subtract those two equations we get 21≠0 which is not possible.

Therefore there is no solution to the given set of equation.

You might be interested in

I really need help rn T_T plz help me out!

pshichka [43]

Answers:

The lengths of sides PQ and RS are  13 
The lengths of sides QR and SP are 20

This is a 13 by 20 rectangle.

============================================================

Explanation:

Refer to the drawing below.

Let x be the length of side SP. Since we're dealing with a rectangle, the opposite side is the same length. Side QR is also x units long.

We're told that RS = SP - 7 which is the same as saying RS = x-7

We also know that PQ = x-7 as well because PQ is opposite side RS.

In short, we have these four sides in terms of x

PQ = x-7
QR = x
RS = x-7
SP = x

as shown in the drawing. The four sides add up to the perimeter of 66.

PQ+QR+RS+SP = perimeter

PQ+QR+RS+SP = 66

(x-7)+x+(x-7)+x = 66

4x-14 = 66

4x = 66+14

4x = 80

x = 80/4

x = 20

Use this x value to find the unknown side lengths.

PQ = x-7 = 20-7 = 13
QR = x = 20
RS = x-7 = 20-7 = 13
SP = x = 20

In short, this is a 13 by 20 rectangle.

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

Check:

perimeter = side1+side2+side3+side4

perimeter = PQ+QR+RS+SP

perimeter = 13+20+13+20

perimeter = 33+33

perimeter = 66

The answer is confirmed.

4 0

3 years ago

Suppose that E(θˆ1) = E(θˆ2) = θ, V(θˆ 1) = σ2 1 , and V(θˆ2) = σ2 2 . Consider the estimator θˆ 3 = aθˆ 1 + (1 − a)θˆ 2. a Show

katen-ka-za [31]

Answer:

Step-by-step explanation:

Given that:

$E( \hat \theta _1) = \theta \ \ \ \ E( \hat \theta _2) = \theta \ \ \ \ V( \hat \theta _1) = \sigma_1^2 \ \ \ \ V(\hat \theta_2) = \sigma_2^2$

If we are to consider the estimator $\hat \theta _3 = a \hat \theta_1 + (1-a) \hat \theta_2$

a. Then, for $\hat \theta_3$ to be an unbiased estimator ; Then:

$E ( \hat \theta_3) = E ( a \hat \theta_1+ (1-a) \hat \theta_2)$

$E ( \hat \theta_3) = aE ( \theta_1) + (1-a) E ( \hat \theta_2)$

$E ( \hat \theta_3) = a \theta + (1-a) \theta = \theta$

b) If $\hat \theta _1 \ \ and \ \ \hat \theta_2$ are independent

$V(\hat \theta _3) = V (a \hat \theta_1+ (1-a) \hat \theta_2)$

$V(\hat \theta _3) = a ^2 V ( \hat \theta_1) + (1-a)^2 V ( \hat \theta_2)$

Thus; in order to minimize the variance of $\hat \theta_3$ ; then constant a can be determined as :

$V( \hat \theta_3) = a^2 \sigma_1^2 + (1-a)^2 \sigma^2_2$

Using differentiation:

$\dfrac{d}{da}(V \ \hat \theta_3) = 0 \implies 2a \ \sigma_1^2 + 2(1-a)(-1) \sigma_2^2 = 0$

⇒

$a (\sigma_1^2 + \sigma_2^2) = \sigma^2_2$

$\hat a = \dfrac{\sigma^2_2}{\sigma^2_1+\sigma^2_2}$

This implies that

$\dfrac{d}{da}(V \ \hat \theta_3)|_{a = \hat a} = 2 \ \sigma_1^2 + 2 \ \sigma_2^2 > 0$

So, $V( \hat \theta_3)$ is minimum when $\hat a = \dfrac{\sigma_2^2}{\sigma_1^2+\sigma_2^2}$

As such; $a = \dfrac{1}{2}$ if $\sigma_1^2 \ \ = \ \ \sigma_2^2$

4 0

3 years ago

Which graph does not represent a function that is always increasing over the entire interval -2 < x < 2?

Novay_Z [31]

Answer:

The answer is C

Step-by-step explanation:

I got my answer from quizlet

8 0

3 years ago

Which statement is true of the determinant of a matrix? A. It can be either positive or zero, but never negative. B. It can have

Andre45 [30]

Answer: D

Step-by-step explanation:

A. Det can be negative.

B. Is gobbledygook

C. All components of the matrix are included.

D. The others have been eliminated, and it's true.

8 0

3 years ago

Please help me please will give brainliest to anyone

Schach [20]

Answer:

its C

Step-by-step explanation:

3 0

3 years ago