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Jlenok [28]
3 years ago
7

Consider the following functions. f1(x) = 0, f2(x) = x, f3(x) = ex g(x) = c1f1(x) + c2f2(x) + c3f3(x) Solve for c1, c2, and c3 s

o that g(x) = 0 on the interval (−[infinity], [infinity]). If a nontrivial solution exists, state it. (If only the trivial solution exists, enter the trivial solution {0, 0, 0}.) {c1, c2, c3} =
Mathematics
1 answer:
Len [333]3 years ago
3 0

There are infinitely many solutions for any choice of c_1 and c_2=c_3=0.

x and e^x are linearly independent and the only way to get c_2f_2+c_3f_3=0 is to choose c_2=c_3=0.

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In right triangle abc shown below, AC=29 inches, AB=17 inches, and m
lapo4ka [179]
Need drawing not sure where m is
6 0
3 years ago
Please help me i only have few hours left i need as much as I can ge. ill get you brainliest and send you free mr.beast merch
masya89 [10]

Answer:

26cm²

Step-by-step explanation:

Perimeter

2x - 4 + 5x - 13 + x + 9 = 40

8x - 8 = 40

8x = 40 - 8

8x = 32

x = 4

Sides of the triangle

Height = 2x - 4 = 2 x 4 - 4 = 4cm

Hypotenuse = 5x - 13 = 5 x 4 - 13 = 7cm

Base = x + 9 = 4 + 9 = 13cm

Area of triangle

= 1/2 x base x height

= 1/2 x 13 x 4

= 26cm²

Hope this helps

7 0
3 years ago
Read 2 more answers
A survey of 35 individuals who passed the seven exams and obtained the rank of Fellow in the actuarial field finds the average s
arsen [322]

Answer:

150000-2.032\frac{15000}{\sqrt{35}}=144847.94    

150000+2.032\frac{15000}{\sqrt{35}}=155152.06

So on this case the 95% confidence interval would be given by (144847.94;155152.06)    

Step-by-step explanation:

Previous concepts

A confidence interval is "a range of values that’s likely to include a population value with a certain degree of confidence. It is often expressed a % whereby a population means lies between an upper and lower interval".

The margin of error is the range of values below and above the sample statistic in a confidence interval.

Normal distribution, is a "probability distribution that is symmetric about the mean, showing that data near the mean are more frequent in occurrence than data far from the mean".

\bar X=150000 represent the sample mean

\mu population mean (variable of interest)

s=15000 represent the sample standard deviation

n=35 represent the sample size  

Solution to the problem

The confidence interval for the mean is given by the following formula:

\bar X \pm t_{\alpha/2}\frac{s}{\sqrt{n}}   (1)

In order to calculate the critical value t_{\alpha/2} we need to find first the degrees of freedom, given by:

df=n-1=35-1=34

Since the Confidence is 0.95 or 95%, the value of \alpha=0.05 and \alpha/2 =0.025, and we can use excel, a calculator or a table to find the critical value. The excel command would be: "=-T.INV(0.025,34)".And we see that t_{\alpha/2}=2.032

Now we have everything in order to replace into formula (1):

150000-2.032\frac{15000}{\sqrt{35}}=144847.94    

150000+2.032\frac{15000}{\sqrt{35}}=155152.06

So on this case the 95% confidence interval would be given by (144847.94;155152.06)    

3 0
4 years ago
What is 1/12(3a-6b)=? And why
Ipatiy [6.2K]

   3a - 6b  =   3 • (a - 2b)<span> =
a-2b/4</span>
5 0
3 years ago
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Hello there. May somebody help me simplify this problem? It would be very kind.
marissa [1.9K]

Answer:

27 / (4x⁶y⁸)

Step-by-step explanation:

\frac{4(3x^{2}y^{4})^{3}}{(2x^{3}y^{5})^{4}}

When raised to a power, multiply the exponents.

\frac{4(3^{3}x^{6}y^{12})}{2^{4}x^{12}y^{20}} \\\\\frac{2^{2}3^{3}x^{6}y^{12}}{2^{4}x^{12}y^{20}}

When dividing, subtract the exponents.

2^{2-4}3^{3}x^{6-12}y^{12-20}\\2^{-2}3^{3}x^{-6}y^{-8}

Move negative exponents to the denominator.

\frac{3^{3}}{2^{2}x^{6}y^{8}}\\\frac{27}{4x^{6}y^{8}}

8 0
3 years ago
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