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Rom4ik [11]
3 years ago
10

How do you work this out? We have to make one positive and the other negative then put it on a graph correctly

Mathematics
1 answer:
ANEK [815]3 years ago
7 0
The answer to Q1 along with an explanation is shown in the picture;
Having a look at all the questions, they are all essentially the same type so can all be solved similarly;
So by following the method and explanation, you should be able to do the rest of the questions.

You might be interested in
Limit (sin4x-4sinx)/x^3 when x close to zero
BartSMP [9]

\Large \boxed{\sf \bf \ \ \lim_{x\rightarrow0} \ {\dfrac{sin(4x)-4sin(x)}{x^3}}=-10 \ \  }

Step-by-step explanation:

Hello, please consider the following.

Using Maclaurin series expansion, we can find an equivalent of sin(x) in the neighbourhood of 0.

sin(x) \sim  \left(x-\dfrac{x^3}{3!}\right)\\\\\text{So, in the neighbourhood of 0}\\\\\begin{aligned}(sin(4x)-4sin(x)) &\sim \left( 4x-\dfrac{(4x)^3}{3!}-4x+\dfrac{4x^3}{3!}\right)\\\\&\sim \left(\dfrac{x^3*4*(1-4^2)}{3*2}\right)\\\\&\sim \left(\dfrac{x^3*2*(-15)}{3}\right)\\\\&\sim \left(x^3*2*(-5)\right)\\\\&\sim \left(x^3*(-10)\right)\\\end{aligned}

Then,

\displaystyle \lim_{x\rightarrow0} \ {\dfrac{sin(4x)-4sin(x)}{x^3}}\\\\= \lim_{x\rightarrow0} \ {\dfrac{-10*x^3}{x^3}}\\\\=-10

Thank you

4 0
3 years ago
Read 2 more answers
The null hypothesis for an ANOVA is that all treatments/samples come from populations with the same mean. The alternative hypoth
LenKa [72]

Answer:

Alternative hypothesis: "AT LEAST ONE" of the population means is different from the others

Step-by-step explanation:

Analysis of variance (ANOVA) "is used to analyze the differences among group means in a sample".  

The sum of squares "is the sum of the square of variation, where variation is defined as the spread between each individual value and the grand mean"  

If we assume that we have n groups and we want to check if the population means are equal, th best way to check this it's with an ANOVA test.

The hypothesis for this case are:

Null hypothesis: \mu_{1}=\mu_{2}=\dots =\mu_{n}

Or in words:

Null hypothesis: All treatments/samples come from populations with the same mean

Alternative hypothesis: Not all the means are equal \mu_{i}\neq \mu_{j}, i,j=1.2,\dots ,n

Or we can say:

Alternative hypothesis: "AT LEAST ONE" of the population means is different from the others

8 0
3 years ago
When the level of confidence and sample standard deviation remain the same, a confidence interval for a population mean based on
miskamm [114]

Answer:

A narrower than

Step-by-step explanation:

6 0
3 years ago
What is the value of xin the diagram?<br><br> A. 9<br> B. 8<br> C. 7<br> D. 6
RoseWind [281]

Answer:

7

Step-by-step explanation:

The ratio of the similar triangles is

14       49

---- = ------

2          x

Using cross products

14x = 2*49

14x = 98

Divide by 14

14x/14 = 98/7

x =7

7 0
3 years ago
#54 Simplify and write the answers using positive exponents only.
kramer
Gg easy

remember
(x^m)/(x^n)=x^(m-n)
and
(x^m)^n=x^(mn)
and
(xyz)^m=(x^m)(y^m)(z^m)
and
(m/n)^a=(m^a)/(n^a)
and
x^-n=1/(x^n)

so
( \frac{6mn^{-2}}{3m^{-1}n^2} )^{-3}=
( \frac{6}{3} )^{-3}( \frac{m}{m^{-1}} )^{-3}( \frac{n^{-2}}{n^2} )^{-3}=
(2^{-3})((m^2)^{-3})((n^{-4})^{-3})=
( \frac{1}{2^3})(m^{-6})(n^{12})=
( \frac{1}{8} )( \frac{1}{m^6})(n^{12})=
\frac{n^{12}}{8m^6}

8 0
3 years ago
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