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nydimaria [60]
3 years ago
9

How do i solve this?Y=-10/3x-4y=-2/3x+4

Mathematics
1 answer:
barxatty [35]3 years ago
6 0
Follow diagram. Remember to graph both functions on the graph. They should intersect at (-3,6).

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The total resistance in a circuit with two parallel resistors is 2 ohms and R1 is 6 ohms. Using the equation for R2, in terms of
ivanzaharov [21]
<h2>Answer:</h2>

R_2  is 3 ohms

<h2>Step-by-step explanation:</h2>

In a circuit containing two resistors R_1 and R_2 connected together in parallel, the total resistance R_T is given by;

\frac{1}{R_T} = \frac{1}{R_1}  + \frac{1}{R_2}              ---------(i)

<em>Make </em>R_2<em> subject of the formula;</em>

=> \frac{1}{R_2} = \frac{1}{R_T}  - \frac{1}{R_1}  

=> \frac{1}{R_2} = \frac{R_1 - R_T}{R_TR_1}

=>   {R_2} = \frac{R_TR_1}{R_1 - R_T}       ---------------(ii)

From the question,

R_1 = 6Ω

R_T = 2Ω

Substitute these values into equation (ii) as follows;

=> {R_2} = \frac{2*6}{6 - 2}

{R_2} = \frac{12}{4}

R_2 = 3Ω

Therefore, the value of R_2 = 3 ohms or R_2 = 3Ω

3 0
2 years ago
Find the distance between each pair of points E(-1, 0),F(12, 0)
Olegator [25]

Answer:

The answer is

<h2>13 \:  \: units</h2>

Step-by-step explanation:

The distance between two points can be found by using the formula

<h3>d =  \sqrt{ ({x1 - x2})^{2}  +  ({y1 - y2})^{2} }  \\ </h3>

where

(x1 , y1) and (x2 , y2) are the points

From the question the points are

E(-1, 0) and F(12, 0)

The distance between them is

<h3>|EF|  =  \sqrt{ ({ - 1 - 12})^{2}  +  ({0 - 0})^{2} }  \\  =  \sqrt{ ({ - 13})^{2} }  \\  =  \sqrt{169}</h3>

We have the final answer as

<h3>|EF| = 13 \:  \: units</h3>

Hope this helps you

5 0
3 years ago
A survey of 76 commercial airline flights of under 2 hours resulted in a sample average late time for a flight of 2.55 minutes.
nika2105 [10]

Answer:

The best point of estimate for the true mean is:

\hat \mu = \bar X = 2.55

2.55-1.96\frac{12}{\sqrt{76}}=-0.148    

2.55+1.96\frac{12}{\sqrt{76}}=5.248    

Since the time can't be negative a good approximation for the confidence interval would be (0,5.248) minutes. The interval are tellling to us that at 95% of confidence the average late time is lower than 5.248 minutes.

Step-by-step explanation:

Information given

\bar X=2.55 represent the sample mean for the late time for a flight

\mu population mean

\sigma=12 represent the population deviation

n=76 represent the sample size  

Confidence interval

The best point of estimate for the true mean is:

\hat \mu = \bar X = 2.55

The confidence interval for the true mean is given by:

\bar X \pm z_{\alpha/2}\frac{\sigma}{\sqrt{n}}   (1)

The Confidence level given is 0.95 or 95%, th significance would be \alpha=0.05 and \alpha/2 =0.025. If we look in the normal distribution a quantile that accumulates 0.025 of the area on each tail we got z_{\alpha/2}=1.96

Replacing we got:

2.55-1.96\frac{12}{\sqrt{76}}=-0.148    

2.55+1.96\frac{12}{\sqrt{76}}=5.248    

Since the time can't be negative a good approximation for the confidence interval would be (0,5.248) minutes. The interval are tellling to us that at 95% of confidence the average late time is lower than 5.248 minutes.

7 0
3 years ago
Neeed a lot of help pls answer this pls
daser333 [38]

Answer:

32

Step-by-step explanation:

100: 25

4:1

x : 8

(8x 4) : 8

32: 8

ans 32

7 0
3 years ago
Read 2 more answers
The vertices of figure BCDE have coordinates B(4, 4) , C(−4, 4), D(−4, −4) , and E(4, −4) . The vertices of figure B'C'D'E' have
Vedmedyk [2.9K]
<span>a translation 2 units left and 2 units up</span>
3 0
3 years ago
Read 2 more answers
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