The formula to find<span> a </span>circle's area<span> (radius)</span>2<span> usually expressed as π ⋅ r 2 where r is the radius of a </span>circle<span>. </span>Area<span> of </span>Circle<span> Concept. The </span>area of a circle<span> is all the space inside a </span>circle's<span> circumference.</span>
Answer:
5^8
Step-by-step explanation:
Diego added the exponents. This was an error. If he was simplifying
5^2 × 5^4, then he could add the exponents and get a correct answer. But his problem had a power raised to a power. In this case, you multiply the exponents to simplify.
(5^2)^4 means
5^2×5^2×5^2×5^2
which is
5×5×5×5×5×5×5×5
which is 5^8.
9514 1404 393
Answer:
2√30 ∠-120°
Step-by-step explanation:
The modulus is ...
√((-√30)² +(-3√10)²) = √(30 +90) = √120 = 2√30
The argument is ...
arctan(-3√10/-√30) = arctan(√3) = -120° . . . . a 3rd-quadrant angle
The polar form of the number can be written as ...
(2√30)∠-120°
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<em>Additional comments</em>
Any of a number of other formats can be used, including ...
(2√30)cis(-120°)
(2√30; -120°)
(2√30; -2π/3)
2√30·e^(i4π/3)
Of course, the angle -120° (-2π/3 radians) is the same as 240° (4π/3 radians).
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At least one app I use differentiates between (x, y) and (r; θ) by the use of a semicolon to separate the modulus and argument of polar form coordinates. I find that useful, as a pair of numbers (10.95, 4.19) by itself does not convey the fact that it represents polar coordinates. As you may have guessed, my personal preference is for the notation 10.95∠4.19. (The lack of a ° symbol indicates the angle is in radians.)
Making assumptions about where parentheses should be,
<span>Let u = -7x </span>
<span>du = -7dx </span>
<span>dv = e^(2x) dx </span>
<span>v = e^(2x)/2 </span>
<span>∫ -7xe^(2x) dx = </span>
<span>-7xe^(2x)/2 - ∫ e^(2x)/2 (-7) dx = </span>
<span>-7xe^(2x)/2 + 7e^(2x)/4 + c</span>
The computation shows that the placw on the hill where the cannonball land is 3.75m.
<h3>How to illustrate the information?</h3>
To find where on the hill the cannonball lands
So 0.15x = 2 + 0.12x - 0.002x²
Taking the LHS expression to the right and rearranging we have:
-0.002x² + 0.12x -.0.15x + 2 = 0.
So we have -0.002x²- 0.03x + 2 = 0
I'll multiply through by -1 so we have
0.002x² + 0.03x -2 = 0.
This is a quadratic equation with two solutions x1 = 25 and x2 = -40 since x cannot be negative x = 25.
The second solution y = 0.15 * 25 = 3.75
Learn more about computations on:
brainly.com/question/4658834
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Complete question:
The flight of a cannonball toward a hill is described by the parabola y = 2 + 0.12x - 0.002x 2 . the hill slopes upward along a path given by y = 0.15x. where on the hill does the cannonball land?
