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nataly862011 [7]
3 years ago
11

Hannah paid $15.79 for a dress that was

Mathematics
1 answer:
vichka [17]3 years ago
7 0

Answer:D.37%

Step-by-step explanation:*24.99x0.37=9.24

24.99-9.24=15.79

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A scalene triangle ______ has symmetry with respect to a line containing a median of the triangle.
Maru [420]
Hey there!

The definition of a scalene triangle is a triangle that has no equal sides, meaning that there cannot be a line of symmetry. A line of symmetry would be contained in a triangle that's an equilateral triangle, for example, which has three even sides. But since there are no even sides, there's no way for the triangle to be split in half evenly. 

Your answer will be "never."

Hope this helped you out! :-)
5 0
4 years ago
Read 2 more answers
A circle has a radius of 6 inches and a central angle of 60°. What is the measure of the arc length associated with this angle?
Readme [11.4K]

Answer:

2\pi

Step-by-step explanation:

first you take 60°/360°= 1/6   then you take 1/6 and multiple it by 2\pi and that equals \frac{\pi }{3}   finally take  \frac{\pi }{3}  times the radius (6) and that should equal 2\pi

PLATO USERS #platolivesmatters

6 0
3 years ago
Bill and Pete shared 1/2 of a cake. Bill got to eat twice as much cake as Pete. What fraction of the whole cake did Bill eat?
KatRina [158]

Answer:

Step-by-step explanation:

Total cake shared=1/2

Pete share=x

Bill= twice as much as Pete

=2x

Pete's share+ Bill's share=total share

x+2x=1/2

3x=1/2

x=1/2÷3

x=1/2×1/3

=1/6

Pete's share=x=1/6

Bill's share=2x

=2(1/6)

=2/6

=1/3

3 0
3 years ago
Solve for k. show work!
dolphi86 [110]

Answer:

71 degrees

Step-by-step explanation:

A right angle is 90 degrees, so the answer is 90-19=71 degrees.

4 0
4 years ago
Compute the surface area of the portion of the sphere with center the origin and radius 4 that lies inside the cylinder x^2+y^2=
Tom [10]

Answer:

16π

Step-by-step explanation:

Given that:

The sphere of the radius = x^2 + y^2 +z^2 = 4^2

z^2 = 16 -x^2 -y^2

z = \sqrt{16-x^2-y^2}

The partial derivatives of Z_x = \dfrac{-2x}{2 \sqrt{16-x^2 -y^2}}

Z_x = \dfrac{-x}{\sqrt{16-x^2 -y^2}}

Similarly;

Z_y = \dfrac{-y}{\sqrt{16-x^2 -y^2}}

∴

dS = \sqrt{1 + Z_x^2 +Z_y^2} \ \ . dA

=\sqrt{1 + \dfrac{x^2}{16-x^2-y^2} + \dfrac{y^2}{16-x^2-y^2} }\ \ .dA

=\sqrt{ \dfrac{16}{16-x^2-y^2}  }\ \ .dA

=\dfrac{4}{\sqrt{ 16-x^2-y^2}  }\ \ .dA

Now; the region R = x² + y² = 12

Let;

x = rcosθ = x; x varies from 0 to 2π

y = rsinθ = y; y varies from 0 to \sqrt{12}

dA = rdrdθ

∴

The surface area S = \int \limits _R \int \ dS

=  \int \limits _R\int  \ \dfrac{4}{\sqrt{ 16-x^2 -y^2} } \ dA

= \int \limits^{2 \pi}_{0} } \int^{\sqrt{12}}_{0} \dfrac{4}{\sqrt{16-r^2}} \  \ rdrd \theta

= 2 \pi \int^{\sqrt{12}}_{0} \ \dfrac{4r}{\sqrt{16-r^2}}\ dr

= 2 \pi \times 4 \Bigg [ \dfrac{\sqrt{16-r^2}}{\dfrac{1}{2}(-2)} \Bigg]^{\sqrt{12}}_{0}

= 8\pi ( - \sqrt{4} + \sqrt{16})

= 8π ( -2 + 4)

= 8π(2)

= 16π

4 0
3 years ago
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