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irina [24]
3 years ago
9

Need help please show how you did it.

Mathematics
1 answer:
jolli1 [7]3 years ago
5 0

Answer:

(0, 1).

Method 1 (Substitution):

Substituting our two y's, we get the following:

x^2 + 5x + 1 = x^2 + 2x + 1 \Rightarrow 3x = 0 \Rightarrow x = 0

Thus, the only set of solutions is (0, 1). A quick sketch (either by hand or on Desmos) can confirm this.

Method 2 (Elimination):

We have two equations. We'll let the top one be equation 1 and the bottom one be equation 2. Eliminating as many variables as we can, we subtract (2) from (1) to get:

0 = 3x => x = 0.

So the only set of solutions is (0, 1).

Method 3 (Gaussian elimination):

We can place this in an augmented matrix and row reduce.

\left[\begin{array}{cccc}1&5&1 & 1\\1&2&1 & 1\end{array}\right]

Row reducing this gives us:

\left[\begin{array}{cccc}1&5&1 & 1\\0&3&0 & 0\end{array}\right]

This tells us that the only solution for x is x = 0 (since we read this as "3x = 0") and thus, the only solution we get is (0, 1).

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Solve by substitution<br> -4x+y=6<br> -5x-y=21
inn [45]

Answer:

x=-3, y=-6. (-3, -6).

Step-by-step explanation:

-4x+y=6

-5x-y=21

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

-9x=27

x=27/-9

x=-3

-5(-3)-y=21

15-y=21

y=15-21

y=-6

7 0
3 years ago
A string passing over a smooth pulley carries a stone at one end. While its other end is attached to a vibrating tuning fork and
nasty-shy [4]

Answer:

correct option is C)  2.8

Step-by-step explanation:

given data

string vibrates form =  8 loops

in water loop formed =  10 loops

solution

we consider  mass of stone = m

string length = l

frequency of tuning = f

volume = v

density of stone = \rho

case (1)  

when 8 loop form with 2 adjacent node is \frac{\lambda }{2}

so here

l = \frac{8 \lambda _1}{2}      ..............1

l = 4 \lambda_1\\\\\lambda_1 = \frac{l}{4}

and we know velocity is express as

velocity = frequency × wavelength   .....................2

\sqrt{\frac{Tension}{mass\ per\ unit \length }}   =   f × \lambda_1

here tension = mg

so

\sqrt{\frac{mg}{\mu}}   =   f × \lambda_1     ..........................3

and

case (2)  

when 8 loop form with 2 adjacent node is \frac{\lambda }{2}

l = \frac{10 \lambda _1}{2}      ..............4

l = 5 \lambda_1\\\\\lambda_1 = \frac{l}{5}

when block is immersed

equilibrium  eq will be

Tenion + force of buoyancy = mg

T + v × \rho × g = mg

and

T = v × \rho - v × \rho × g    

from equation 2

f × \lambda_2 = f  × \frac{1}{5}  

\sqrt{\frac{v\rho _{stone} g - v\rho _{water} g}{\mu}} = f \times \frac{1}{5}     .......................5

now we divide eq 5 by the eq 3

\sqrt{\frac{vg (\rho _{stone} - \rho _{water})}{\mu vg \times \rho _{stone}}} = \frac{fl}{5} \times \frac{4}{fl}

solve irt we get

1 - \frac{\rho _{stone}}{\rho _{water}}  = \frac{16}{25}

so

relative density \frac{\rho _{stone}}{\rho _{water}} = \frac{25}{9}

relative density = 2.78 ≈ 2.8

so correct option is C)  2.8

3 0
3 years ago
g If a hypothesis test with a significance level of α rejects H0: μ1=μ2 in favor of Ha: μ1≠μ2, then thecorresponding (1- α)% con
Effectus [21]

Answer:

The statement, (1- <em>α</em>)% confidence interval for (μ₁ - μ₂) does not contain zero is TRUE.

Step-by-step explanation:

The hypothesis for a test is defined as follows:

<em>H</em>₀: μ₁ = μ₂ vs. <em>H</em>ₐ: μ₁ ≠ μ₂

It is provided that the test was rejected st the significance level <em>α</em>%.

If a decision is to made using the confidence interval the conditions are:

If the null hypothesis value is not included in the (1 - <em>α</em>)% confidence interval then the null hypothesis will be rejected and vice versa.

In this case the null hypothesis value is:

<em>H</em>₀: μ₁ - μ₂ = 0.

If the value 0 is not included in the (1 - <em>α</em>)% confidence interval for the difference between two means, then the null hypothesis will be rejected.

Thus the statement, (1- <em>α</em>)% confidence interval for (μ1- μ2) does not contain zero is TRUE.

5 0
3 years ago
Solve for x 21 27 x-3 x-1
Nataly_w [17]

Answer

27x^22-3x-1

hope this is right!!

Step-by-step explanation:

7 0
2 years ago
You make a scale drawing of a tree using the scale 5 in. = 27 ft. if the tree is 67.5 ft. tall, how tall is the scale drawing?
evablogger [386]
You would get the equivalent amount to 5in = ?ft then divide that by 67.5
8 0
3 years ago
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