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malfutka [58]
3 years ago
9

C = 7a^4+ + 5a²b2 – 3b^4 D=5a^4 + 7a²b^2 + 3b^4 C - D =

Mathematics
2 answers:
levacccp [35]3 years ago
8 0

Answer:

C-D is: \mathbf{4b^4 -2a^b^2 -6b^4}

Step-by-step explanation:

We are given:

C = 7a^4+ + 5a^2b2 - 3b^4\\D=5a^4 + 7a^2b^2 + 3b^4

We need to find C-D.

Finding C-D we actually have to subtract D from C

So, finding C-D

C-D\\= 7a^4+ + 5a^2b2 - 3b^4-(5a^4 + 7a^2b^2 + 3b^4)\\=7a^4+ + 5a^2b2 - 3b^4-5a^4-7a^2b^2-3b^4

Now, combining the like terms.

Like terms are those that have same variables and exponents.

In our case, 7a^4\: and\: 3b^4, 5a^2b^2\: and\: -7a^b^2, -3b^4\:and\:-3b^4

=7a^4 - 3b^4+5a^2b^2 -7a^b^2 -3b^4-3b^4\\=4b^4 -2a^b^2 -6b^4

So, we get C-D is: \mathbf{4b^4 -2a^b^2 -6b^4}

Gekata [30.6K]3 years ago
5 0

Answer:

2a^4 +12a

2 {a}^{4}  + 12 { a}^{2}  \\  \times  \times

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Announcements for 84 upcoming engineering conferences were randomly picked from a stack of IEEE Spectrum magazines. The mean len
Marina86 [1]

Answer:

Step-by-step explanation:

Hello!

You need to construct a 95% CI for the population mean of the length of engineering conferences.

The variable has a normal distribution.

The information given is:

n= 84

x[bar]= 3.94

δ= 1.28

The formula for the Confidence interval is:

x[bar]±Z_{1-\alpha/2}*(δ/n)

Lower bound(Lb): 3.698

Upper bound(Ub): 4.182

Error bound: (Ub - Lb)/2 = (4.182-3.698)/2 = 0.242

I hope it helps!

6 0
4 years ago
What is the equation in point-slope form of the line that passes through the points (7, 5)(7, 5) and (−4, −1)(−4, −1) ? y+4=116(
Alex Ar [27]
(7,5)(-4,-1)
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y - y1 = m(x - x1)
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now we sub...pay close attention to ur signs
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3 years ago
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Find −1/5+17/10. Please help me!
alina1380 [7]
I got 1.5, but I could be wrong.
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4 years ago
Please help w this! Its a calculus question! look at the picture for the problem,
neonofarm [45]

Since you mentioned calculus, perhaps you're supposed to find the area by integration.

The square is circumscribed by a circle of radius 6, so its diagonal (equal to the diameter) has length 12. The lengths of a square's side and its diagonal occur in a ratio of 1 to sqrt(2), so the square has side length 6sqrt(2). This means its sides occur on the lines x=\pm3\sqrt2 and y=\pm3\sqrt2.

Let R be the region bounded by the line x=3\sqrt2 and the circle x^2+y^2=36 (the rightmost blue region). The right side of the circle can be expressed in terms of x as a function of y:

x^2+y^2=36\implies x=\sqrt{36-y^2}

Then the area of this circular segment is

\displaystyle\iint_R\mathrm dA=\int_{-3\sqrt2}^{3\sqrt2}\int_{3\sqrt2}^{\sqrt{36-y^2}}\,\mathrm dx\,\mathrm dy

=\displaystyle\int_{-3\sqrt2}^{3\sqrt2}(\sqrt{36-y^2}-3\sqrt2)\,\mathrm dy

Substitute y=6\sin t, so that \mathrm dy=6\cos t\,\mathrm dt

=\displaystyle\int_{-\pi/4}^{\pi/4}6\cos t(\sqrt{36-(6\sin t)^2}-3\sqrt2)\,\mathrm dt

=\displaystyle\int_{-\pi/4}^{\pi/4}(36\cos^2t-18\sqrt2\cos t)\,\mathrm dt=9\pi-18

Then the area of the entire blue region is 4 times this, a total of \boxed{36\pi-72}.

Alternatively, you can compute the area of R in polar coordinates. The line x=3\sqrt2 becomes r=3\sqrt2\sec\theta, while the circle is given by r=6. The two curves intersect at \theta=\pm\dfrac\pi4, so that

\displaystyle\iint_R\mathrm dA=\int_{-\pi/4}^{\pi/4}\int_{3\sqrt2\sec\theta}^6r\,\mathrm dr\,\mathrm d\theta

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so again the total area would be 36\pi-72.

Or you can omit using calculus altogether and rely on some basic geometric facts. The region R is a circular segment subtended by a central angle of \dfrac\pi2 radians. Then its area is

\dfrac{6^2\left(\frac\pi2-\sin\frac\pi2\right)}2=9\pi-18

so the total area is, once again, 36\pi-72.

An even simpler way is to subtract the area of the square from the area of the circle.

\pi6^2-(6\sqrt2)^2=36\pi-72

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3 years ago
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saul85 [17]
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b(0)=50(1.4)^{0}
b(0)=50(1)
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b) To find if the function is growing or decaying, we are going to convert its decimal part to a fraction. Then, we will compare the numerator and the denominator of the fraction. If the numerator is grater than the denominator, the function is growing; if the opposite is true, the function is decaying.
Remember that to convert a decimal into a fraction we are going to add the denominator 1 to our decimal and then we'll multiply both of them by a power of ten for each number after the decimal point:
\frac{1.4}{1} . \frac{10}{10} = \frac{14}{10} = \frac{7}{5}
Now we can rewrite our exponential function:
b(t)=50( \frac{7}{5})^{t}
Since the numerator is grater than the denominator, it is growing faster than the denominator; therefore the function is growing.

c) The only thing we need to do here is evaluate the function at t=5:
b(t)=50(1.4)^{t}
b(5)=50(1.4)^{5}
b(5)=50(5.37824)
b(5)=268.912

We can conclude that after 5 hours <span>Dr. Silas began her study will be 268.9 bacteria in the sample.</span>
4 0
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