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hram777 [196]
3 years ago
9

Which expression is equivalent ?

Mathematics
1 answer:
Natali [406]3 years ago
8 0
ANSWER
A ..... I hope this helped
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Which shows the correct substitution of the values a, b, and c from the equation 1 = -2x + 3x² + 1 into the quadratic
antoniya [11.8K]
The coefficient 'a' for the quadratic term is a = 3
The coefficient 'b' for the linear term is b = -2
The coefficient 'c' is 0
Step-by-step explanation :
The given expression is,




To solve this problem we are using quadratic formula.
The general quadratic equation is,

Formula used :

Now we a have to solve the above equation and we get the value of 'x'.

a = 3, b = -2, c = 0






The coefficient 'a' for the quadratic term is a = 3
The coefficient 'b' for the linear term is b = -2
The coefficient 'c' is 0
8 0
2 years ago
Find the product of<br> (-4x2+2x)(9x3)
RSB [31]

Answer:

Can't guess whether you have written 'x' or you have put a sign of multiplication!!

6 0
2 years ago
Solve by factoring<br> x^2-10=9x
BabaBlast [244]

Answer:

x=10

Step-by-step explanation:

x^2 - 10 - (9x) = 0

x^2 - 9x - 10 = 0

(x - 10) + (x + 1) = 0

x= 10

x = -1

4 0
2 years ago
Cho tập hợp A=(2;7)và tập hợp B=[7;10]tìm tập hợp a∩B
ahrayia [7]

sorry po di ko ma intindihan

Step-by-step explanation:

sorry po

6 0
2 years ago
The following data gives the speeds (in mph), as measured by radar, of 10 cars traveling north on I-15. 76 72 80 68 76 74 71 78
____ [38]

Answer:

71.123 mph ≤ μ ≤ 77.277 mph

Step-by-step explanation:

Taking into account that the speed of all cars traveling on this highway have a normal distribution and we can only know the mean and the standard deviation of the sample, the confidence interval for the mean is calculated as:

m-t_{a/2,n-1}\frac{s}{\sqrt{n} } ≤ μ ≤ m+t_{a/2,n-1}\frac{s}{\sqrt{n} }

Where m is the mean of the sample, s is the standard deviation of the sample, n is the size of the sample, μ is the mean speed of all cars, and t_{a/2,n-1} is the number for t-student distribution where a/2 is the amount of area in one tail and n-1 are the degrees of freedom.

the mean and the standard deviation of the sample are equal to 74.2 and 5.3083 respectively, the size of the sample is 10, the distribution t- student has 9 degrees of freedom and the value of a is 10%.

So, if we replace m by 74.2, s by 5.3083, n by 10 and t_{0.05,9} by 1.8331, we get that the 90% confidence interval for the mean speed is:

74.2-(1.8331)\frac{5.3083}{\sqrt{10} } ≤ μ ≤ 74.2+(1.8331)\frac{5.3083}{\sqrt{10} }

74.2 - 3.077 ≤ μ ≤ 74.2 + 3.077

71.123 ≤ μ ≤ 77.277

8 0
2 years ago
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