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3 years ago

Solve the linear system of equations 8x−4y=20 5x−8y=62 x= y=

Mathematics

1 answer:

LUCKY_DIMON [66]3 years ago

3 0

Answer:

x = -2

x = -9

Step-by-step explanation:

The linear system is:

8x - 4y = 20 ⇒ (1)

5x - 8y = 62 ⇒ (2)

To solve the system multiply equation (1) by -2 to make the coefficients of y equal in values and different in signs, then add the resulting equation with equation (2)

∵ -2(8x) + -2(-4y) = -2(20)

∴ -16x + 8y = -40 ⇒ (3)

- Add equation (2) and (3)

∴ -11x = 22

- Divide both sides by -11

∴ x = -2

Substitute the value of x in equation (1) or (2)

∵ 8(-2) - 4y = 20

∴ -16 - 4y = 20

- Add 16 to both sides

∴ - 4y = 36

- Divide both sides by -4

∴ y = -9

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xenn [34]

Answer:

y = -2x + 3

Step-by-step explanation:

4x + 2y = 6

⇔ 2y = $\frac{6 - 4x}{2} = \frac{6}{2} - \frac{4x}{2} = 3 - 2x = -2x + 3$

5 0

3 years ago

Which of the following is equal to 393,500? 0.3935 × 10 -6 39.35 × 10 -4 3.935 × 10 -5 3.935 × 10 5

SCORPION-xisa [38]

Answer:

No answer is equal to 393,500

Step-by-step explanation:

You can try to solve each question.

0.3935 × 10 -6 = -2.065

39.35 × 10 -4 = 389.5

3.935 × 10 -5 = 34.35

3.935 × 10 +5 = 44.35

None of these equal to 393,500

(Maybe I am getting things wrong I am not in a smart mood today)

4 0

4 years ago

P(x)equals=R(x)minus−C(x). Given R(x)equals=59 x minus 0.3 x squared59x−0.3x2 and Upper C left parenthesis x right parenthesi

Deffense [45]

Answer:

$P(x)=-0.3x^2+56x-14$

Step-by-step explanation:

The given functions are

$R(x)=59x-0.3x^2$

$C(x)=3x+14$

It is given that

$P(x)=R(x)-C(x)$

Substitute the values of given functions in the above equation.

$P(x)=59x-0.3x^2-(3x+14)$

$P(x)=59x-0.3x^2-3x-14$

Combine like terms.

$P(x)=-0.3x^2+(59x-3x)-14$

$P(x)=-0.3x^2+56x-14$

Therefore, the required function is $P(x)=-0.3x^2+56x-14$ .

4 0

3 years ago

Please help me !!!!!!!!

Neko [114]

Hi :)
the correct answer is J
hope this helps!

6 0

3 years ago

A researcher compares the effectiveness of two different instructional methods for teaching electronics. A sample of 138 student

yan [13]

Answer:

The 98% confidence interval for the true difference between testing averages for students using Method 1 and students using Method 2 is (-8.04, 0.84).

Step-by-step explanation:

The (1 - α)% confidence interval for the difference between population means is:

$CI=(\bar x_{1}-\bar x_{2})\pm z_{\alpha/2}\times \sqrt{\frac{\sigma^{2}_{1}}{n_{1}}+\frac{\sigma^{2}_{2}}{n_{2}}}$

The information provided is as follows:

$n_{1}= 138\\n_{2}=156\\\bar x_{1}=61\\\bar x_{2}=64.6\\\sigma_{1}=18.53\\\sigma_{2}=13.43$

The critical value of z for 98% confidence level is,

$z_{\alpha/2}=z_{0.02/2}=2.326$

Compute the 98% confidence interval for the true difference between testing averages for students using Method 1 and students using Method 2 as follows:

$CI=(\bar x_{1}-\bar x_{2})\pm z_{\alpha/2}\times \sqrt{\frac{\sigma^{2}_{1}}{n_{1}}+\frac{\sigma^{2}_{2}}{n_{2}}}$

$=(61-64.6)\pm 2.326\times\sqrt{\frac{(18.53)^{2}}{138}+\frac{(13.43)^{2}}{156}}\\\\=-3.6\pm 4.4404\\\\=(-8.0404, 0.8404)\\\\\approx (-8.04, 0.84)$

Thus, the 98% confidence interval for the true difference between testing averages for students using Method 1 and students using Method 2 is (-8.04, 0.84).

5 0

3 years ago