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egoroff_w [7]
3 years ago
12

Anyone know this one I’m kinda bad at math lol

Mathematics
2 answers:
swat323 years ago
8 0

Answer:

1/ 9^5

Step-by-step explanation:

When dividing exponents with the same base, subtract the exponents

9^2 / 9^7

9 ^ (2-7)

9^ -5

We know that a^ -b = 1/ a^b

1/ 9^5

jenyasd209 [6]3 years ago
8 0

You can rewrite it as:  (using the law of dividing exponents)

9^(2-7) = 9^(-5) = 1/(9^5)

The fourth one

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Write a two step equation involving division and addition that has a solution of x = 25
Tamiku [17]
Start with the solution
x=-25
use the reverse of what they asked
they wanted division and addition
we use multiplication and subtract

ok
x=-25
we can use subtraction and multiplication in any order
minus 10 from both sides
x-10=-35
multiply both sides by 4
4(x-10)=-140
4x-40=-140 is a possible equation
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3 years ago
Simplify 1.25 (-x -4)
statuscvo [17]

Answer:

1.25(-xsquare - 8x + 16)

Step-by-step explanation:

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2 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

=\displaystyle\frac12\int_{-\pi/4}^{\pi/4}(36-18\sec^2\theta)\,\mathrm d\theta=9\pi-18

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

6 0
3 years ago
Scenario for 300=60t
earnstyle [38]
Do you need to find what t is? If so, do 300 divided by 60 to get the t isolated. which is, 5.

you can check by doing 60*5 which is 300
4 0
3 years ago
Read 2 more answers
What is the approximate area of a semicircle with a radius of 8 ft? Use π=3.14 . Enter your answer in the box. The area of the s
yan [13]

Answer:

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Step-by-step explanation:

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3 years ago
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