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galina1969 [7]
2 years ago
13

Solve the simultaneous equations by substitution x-3y=-7 x=5-y

Mathematics
1 answer:
WINSTONCH [101]2 years ago
5 0

Answer:

x = 2, y = 3

Step-by-step explanation:

x - 3y = - 7 → (1)

x = 5 - y → (2)

Substitute x = 5 - y into (1)

5 - y - 3y = - 7

5 - 4y = - 7 ( subtract 5 from both sides )

- 4y = - 12 ( divide both sides by - 4 )

y = 3

Substitute y = 3 into (2)

x = 5 - 3 = 2

solution is (2, 3 )

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A line is drawn through (–7, 11) and (8, –9). The equation y – 11 = y minus 11 equals StartFraction negative 4 Over 3 EndFractio
yaroslaw [1]

⇒Given equation of line Passing through  (–7, 11) and (8, –9) is given by

       y-11= \frac{-4}{3}(x+7)

⇒Equation of line Passing through  (–7, 11) and (8, –9) is given by

   \rightarrow \frac{y-y_{1}}{x-x_{1}}=\frac{y_{2}-y_{1}}{x_{2}-x_{1}}\\\\\rightarrow \frac{y-11}{x-(-7)}=\frac{-9-11}{8-(-7)}\\\\\rightarrow \frac{y-11}{x+7}=\frac{-20}{15}\\\\\rightarrow \frac{y-11}{x+7}=\frac{-4}{3}\\\\\rightarrow 3y-33=-4x-28\\\\\rightarrow 4x+3y=33-28\\\\ \rightarrow 4x+3y=5

Option C

          4x+3y=5

6 0
3 years ago
Read 2 more answers
Which function has the greater rate of change?
Gnoma [55]

Answer:

Function 1

Step-by-step explanation:

<u>Function 1:</u>

y=\dfrac{5}{6}x+10

The rate of change of the function 1 is:

\dfrac{5}{6}\approx 0.83

<u>Function 2:</u>

The rate of change of the function 2 is:

\dfrac{15.75-14.25}{1-(1)}=\dfrac{1.5}{2}=0.75

Hence, the greater rate of change has function 1.

3 0
3 years ago
Maani lai<br>1.8x = 4.68​
Neporo4naja [7]

Answer:2.6

Step-by-step explanation:

8 0
3 years ago
An evolutionary biologist examined the relative fitness of Escherichia coli bacteria grown for 2000 generations, about 300 days,
Varvara68 [4.7K]

Answer:

Step-by-step explanation:

Hello!

The variable of interest is:

X: fitness of a line of E. coli grown on an acidic environment.

n= 6 E. coli lines

Recorded fitness for each line: 1.24, 1.22, 1.23, 1.24, 1.18, 1.09

The relative fitness of 1 indicates that both bacteria types are equally fit.

A relative fitness larger than 1 indicates that the acid-evolved line is more fit than the parental line kept at neutral pH when both are grown in acidic conditions.

Meaning that if the average fitness of the E. coli lines grown on an acidic environment is greater than 1 then they are better adjusted to live in acidic conditions, symbolically: μ > 1

The statistic hypotheses are:

H₀: μ ≤ 1

H₁: μ > 1

α: 0.05

Assuming that the variable has a normal distribution you have to apply a one-sample t-test:

t= \frac{X[bar]-Mu}{\frac{S}{\sqrt{n} } } ~~t_{n-1}

X[bar]= 1.20

S= 0.06

t_{H_0}= \frac{1.20-1}{\frac{0.06}{\sqrt{6} } } = 8.40

The p-value for this test is 0.0002

Since the p-value= 0.0002 is less than α:0.05 the decision is to reject the null hypothesis.

Then at a 5% significance level, there is significant evidence to conclude that the bacteria evolved in acidic pH are better adapted to acidic conditions.

I hope you have a SUPER day!

7 0
3 years ago
Please Help me with number two and three
pentagon [3]

Answer:

number 2 is C and number 3 is D

Step-by-step explanation:

5 0
3 years ago
Read 2 more answers
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