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makkiz [27]
2 years ago
8

16. Solve -3x2 - 9x + 12 = 0.

Mathematics
1 answer:
AURORKA [14]2 years ago
7 0

Given :

  • -3x² - 9x + 12 = 0

To find :-

  • Value of x .

Solution :-

Given equation ,

  • -( 3x² + 9x -12 ) = 0
  • 3x² + 9x - 12 = 0
  • 3x² + 12x - 3x - 12 = 0
  • 3x ( x +4) -3( x +4) = 0
  • (3x -3)(x +4) = 0
  • x = 3/3 , -4
  • x = 1 , -4
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Is 4/3 and 8/6 a proportion
Alexus [3.1K]

Answer:

yes

Step-by-step explanation:

If you divide one into the other and the answer is 1 then they are of the same ratio.

6 0
3 years ago
Read 2 more answers
PLEASE HELP ME SOLVE this
Alisiya [41]

Answer:

y=-12/11

Step-by-step explanation:

y-(-12/11)=0(x-(-12/13))

y+12/11=0(x+12/13)

y+12/11=0

y=-12/11

Since the slope equals 0, then the line is horizontal.

3 0
3 years ago
A manufacturer of potato chips would like to know whether its bag filling machine works correctly at the 420 gram setting. It is
anastassius [24]

Answer:

t=\frac{424-420}{\frac{26}{\sqrt{61}}}=1.202    

p_v =2*P(t_{(60)}>1.202)=0.234  

If we compare the p value and the significance level given \alpha=0.01 we see that p_v>\alpha so we can conclude that we have enough evidence to fail reject the null hypothesis, so we can conclude that the true mean is NOT different from  420. So the specification is satisfied.

Step-by-step explanation:

Data given and notation  

\bar X=424 represent the sample mean

s=26 represent the sample standard deviation

n=61 sample size  

\mu_o =420 represent the value that we want to test

\alpha=0.01 represent the significance level for the hypothesis test.  

t would represent the statistic (variable of interest)  

p_v represent the p value for the test (variable of interest)  

State the null and alternative hypotheses.  

We need to conduct a hypothesis in order to check if the true mean is different from 420 or not, the system of hypothesis would be:  

Null hypothesis:\mu = 420  

Alternative hypothesis:\mu \neq 420  

If we analyze the size for the sample is > 30 but we don't know the population deviation so is better apply a t test to compare the actual mean to the reference value, and the statistic is given by:  

t=\frac{\bar X-\mu_o}{\frac{s}{\sqrt{n}}}  (1)  

t-test: "Is used to compare group means. Is one of the most common tests and is used to determine if the mean is (higher, less or not equal) to an specified value".  

Calculate the statistic

We can replace in formula (1) the info given like this:  

t=\frac{424-420}{\frac{26}{\sqrt{61}}}=1.202    

P-value

The first step is calculate the degrees of freedom, on this case:  

df=n-1=61-1=60  

Since is a two sided test the p value would be:  

p_v =2*P(t_{(60)}>1.202)=0.234  

Conclusion  

If we compare the p value and the significance level given \alpha=0.01 we see that p_v>\alpha so we can conclude that we have enough evidence to fail reject the null hypothesis, so we can conclude that the true mean is NOT different from  420. So the specification is satisfied.

4 0
3 years ago
Find the Laplace transformation of each of the following functions. In each case, specify the values of s for which the integral
MAVERICK [17]

Answer:

a. \frac {2} {s-1} converges to s> 1.

b. \frac{3}{e^3 \left(s-5 \right)} converges to s> 5.

c. - \frac {2}{s + 3} converges to s> - 3.

d. \frac {s}{s^2 + 25} converges to s> 0.

e. \frac {10} {s^2 + 1} converges even s> 0.

f. \frac {12}{s^2 + 4} converges to s> 0.

g. -\frac {5\left(\cos\left (1\right) s-2 \sin\left(1\right)\right)}{s^2 + 4} converges to s> 0.

h. \frac {1} {s ^ 2 + 4} converges to s> 0.

Step-by-step explanation:

a. L \left\{2e^t \right\} = 2L \left\{e^t \right\} = 2 \cdot \frac {1} {s-1} = \frac {2} {s-1} converges to s> 1.

b. L \left\{3e^{5t-3} \right\} = 3e^{-3} L \left\{e^{5t} \right\} = 3e^{-3} L \left\{e^{5t} \right\} = \frac{3}{e^3 \left(s-5 \right)} converges to s> 5.

c. L \left\{-2e^{-3t} \right\} = -2L \left\{e^{-3t} \right\} = - \frac {2}{s + 3} converges to s> - 3.

d. L \left\{\cos\left (5t \right)\right\} = \frac {s}{s^2 + 25} converges to s> 0.

e. L \left\{10 \sin\left(t\right)\right\} = 10L\left\{\sin\left(t\right)\right\} = \frac {10} {s^2 + 1} converges even s> 0.

f. L \left\{6\sin \left(2t \right) \right\} = 6L\left\{\sin\left (2t\right)\right\} = \frac {12}{s^2 + 4} converges to s> 0.

g. L \left\{-5\cos\left(2t + 1\right) \right\} = -5L\left\{\cos\left(2t + 1 \right)\right\} = -\frac {5\left(\cos\left (1\right) s-2 \sin\left(1\right)\right)}{s^2 + 4} converges to s> 0.

h. L\left\{\sin \left(t\right)\cos \left(t\right)\right\} = L\left\{\sin\left(2t\right)\frac{1}{2}\right\} =\frac{1}{2}\cdot \frac{2}{s^2+4} = \frac {1} {s ^ 2 + 4} converges to s> 0.

7 0
3 years ago
Jasmine is in a bicycle race. She has already biked 10 miles and is now biking at a rate of 18 miles per hour. Write and equatio
Dafna1 [17]

Answer:

y = 18x + 10

slope = 18

y-intercept = 10

Step-by-step explanation:

The situation given in the problem is linear and can be represented using slope-intercept form of a linear equation: y = mx + b, where 'm' is equal to the slope, 'b' is equal to the y-intercept, 'x' is the number of hours and 'y' is the total number of miles.  

Given that Jasmine has already biked 10 miles, her initial value, or y-intercept would be 10.  Since she is now biking at a rate of 18 miles per hour, her rate of change, or slope, would be 18.  

Using these values for 'm' and 'b': y = 18x + 10

6 0
3 years ago
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