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kakasveta [241]
3 years ago
13

-2x+5c+6c-3x= Simplify and write the answer.

Mathematics
2 answers:
Irina18 [472]3 years ago
8 0
-2x+5c+6c-3x
-5x+11c=
Shalnov [3]3 years ago
7 0

Answer:

-5x+11c

Step-by-step explanation:

add the alike terms

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Subtract (8x + 9) - (6x - 1)
il63 [147K]

Answer:

=2x+10

Step-by-step explanation:

Let's simplify step-by-step.

8x+9−(6x−1)

Distribute the Negative Sign:

=8x+9+−1(6x−1)

=8x+9+−1(6x)+(−1)(−1)

=8x+9+−6x+1

Combine Like Terms:

=8x+9+−6x+1

=(8x+−6x)+(9+1)

=2x+10

Answer:

=2x+10

7 0
3 years ago
Evaluate the surface integral ModifyingBelow Integral from nothing to nothing Integral from nothing to nothing With Upper S f (x
kykrilka [37]

Parameterize S by

\vec s(u,v)=6\cos u\sin v\,\vec\imath+6\sin u\sin v\,\vec\jmath+6\cos v\,\vec k

with 0\le u\le2\pi and 0\le v\le\frac\pi2. Take a normal vector to S,

\dfrac{\partial\vec s}{\partial v}\times\dfrac{\partial\vec s}{\partial u}=36\cos u\sin^2v\,\vec\imath+36\sin u\sin^2v\,\vec\jmath+36\cos v\sin v\,\vec k

which has norm

\left\|\dfrac{\partial\vec s}{\partial v}\times\dfrac{\partial\vec s}{\partial u}\right\|=36\sin v

Then the integral of f(x,y,z)=x^2+y^2 over S is

\displaystyle\iint_Sx^2+y^2\,\mathrm dS=\iint_S\left((6\cos u\sin v)^2+(6\sin u\sin v)^2\right)\left\|\frac{\partial\vec s}{\partial v}\times\frac{\partial\vec s}{\partial u}\right\|\,\mathrm du\,\mathrm dv

=\displaystyle36^2\int_0^{\pi/2}\int_0^{2\pi}\sin^3v\,\mathrm du\,\mathrm dv=\boxed{1728\pi}

6 0
3 years ago
How do I solve this?
Tju [1.3M]

Answer:  A:  3x^2y^(3/2)

Step-by-step explanation:

This can be written as

(81*x^8*y^6)^(1/4)

Then multiply each exponent by (1/4):

81^(1/4)*x^(8(1/4))y^6(1/4))

81^(1/4) = 3

x^(8(1/4)) = x^2

y^6(1/4)) = y^(3/2)

The result:  3x^2y^(3/2)

7 0
2 years ago
A dairy farm uses the somatic cell count (SCC) report on the milk it provides to a processor as one way to monitor the health of
Eva8 [605]

Answer:

t=\frac{25000-210000}{\frac{37500}{\sqrt{5}}}=2.37  

p_v =P(t_{4}>2.37)=0.038  

If we compare the p value and a significance level assumed \alpha=0.1 we see that p_v so we can conclude that we reject the null hypothesis, and the actual true mean is higher than 210250 at 5% of significance.  

Step-by-step explanation:

Data given and notation

\bar X=250000 represent the sample mean  

s=37500 represent the standard deviation for the sample

n=5 sample size  

\mu_o =210250 represent the value that we want to test  

\alpha=0.1 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 to be tested  

We need to conduct a hypothesis in order to determine if the mean is higher than 210250, the system of hypothesis would be:  

Null hypothesis:\mu \leq 210250  

Alternative hypothesis:\mu > 210250  

Compute the test statistic  

We don't know the population deviation, so for this case 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".  

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

t=\frac{25000-210000}{\frac{37500}{\sqrt{5}}}=2.37  

Now we need to find the degrees of freedom for the t distirbution given by:

df=n-1=5-1=4

Conclusion

Since is a one right tailed test the p value would be:  

p_v =P(t_{4}>2.37)=0.038  

If we compare the p value and a significance level assumed \alpha=0.1 we see that p_v so we can conclude that we reject the null hypothesis, and the actual true mean is higher than 210250 at 5% of significance.  

6 0
3 years ago
A conical container can hold 120π cubic centimeters of water. The diameter of the base of the container is 12 centimeters.
Tanzania [10]

Answer: The height of the container is 10 centimeters. If its diameter and height were both doubled, the container's capacity would be 8 times its original capacity.

Step-by-step explanation:

The volume of a cone can be calculated with this formula:

V=\frac{\pi r^2h}{3}

Where "r" is the radius and "h" is the height.

We know that the radius is half the diameter. Then:

r=\frac{12cm}{2}=6cm

We know the volume and the radius of the conical container, then we can find "h":

120\pi cm^3=\frac{\pi (6cm)^2h}{3}\\\\(3)(120\pi cm^3)=\pi (6cm)^2h\\\\h=\frac{3(120\pi cm^3)}{\pi (6cm)^2}\\\\h=10cm

The diameter and height doubled are:

d=12cm*2=24cm\\h=10cm*2=20cm

Now the radius is:

r=\frac{24cm}{2}=12cm

And the container capacity is

V=\frac{\pi (12cm)^2(20cm)}{3}=960\pi cm^3

Then, to compare the capacities, we can divide this new capacity by the original:

 \frac{960\pi cm^3}{120\pi cm^3}=8

Therefore,  the container's capacity would be 8 times its original capacity.

6 0
2 years ago
Read 2 more answers
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