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larisa [96]
3 years ago
9

A turtle pack that normally sells for $39 is on sale for 33% off. Find the

Mathematics
1 answer:
Paul [167]3 years ago
3 0

Answer:

67% discounted

26

555.2

Step-by-step explanation:

39 - 13 = 26

48 x 1.15 = 55.2

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Jerome is writing a coordinate proof to show that the midsegment of a trapezoid is parallel to its bases. He starts by assigning
Bingel [31]

Answer:

coordinates of R(b,c)

slope of NM is 0.

Step-by-step explanation:

The midpoint of the line joining the points (x₁,y₁) and (x₂,y₂) is given by

(\frac{x_{1}+x_{2}}{2},\frac{y_{1}+y_{2}}{2})

The midpoint of the line segment NK is given by

(\frac{0+2b}{2},\frac{0+2c}{2})

∴R(b,c)

The slope of the line joining the points (x₁,y₁) and (x₂,y₂) is given by

m =\frac{y_{2}-y_{1}}{x_{2}-x_{1}}

Slope of RS is

m_{RS} =\frac{c-c}{a+d-b} = 0

Hence

coordinates of R(b,c)

and slope of NM is 0.

8 0
3 years ago
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I need help on this on this onr too anyone mind helping?
V125BC [204]
The correct answer should be 23 3/4 miles
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4 0
3 years ago
Zadanie optymalizacyjne z matematyki. Proszę o rozwiązanie i wyjaśnienie
Yakvenalex [24]

Volume of the pyramid:

V=\dfrac{s^2h}3

Perimeter of the cross-section:

40=\sqrt2\,s+2\sqrt{\dfrac{s^2}2+h^2}=\sqrt2\left(s+\sqrt{s^2+2h^2}\right)

\implies h=\sqrt{\dfrac{(20\sqrt2-s)^2-s^2}2}=\sqrt{400-20\sqrt2\,s}

Area of the cross-section:

P=\dfrac12(\sqrt2\,s)h=\dfrac{sh}{\sqrt2}

\implies P=\dfrac{s\sqrt{400-20\sqrt2\,s}}{\sqrt2}=s\sqrt{200-10\sqrt2\,s}

First derivative test:

\dfrac{\mathrm dP}{\mathrm ds}=\dfrac{20\sqrt2-3s}{\sqrt{4-\dfrac{\sqrt2}5s}}=0\implies s=\dfrac{20\sqrt2}3

Then the height of the cross-section/pyramid is

h=\sqrt{400-20\sqrt2\,s}=\dfrac{20}{\sqrt3}

The volume of the pyramid that maximizes the cross-sectional area P is

V=\dfrac{\left(\frac{20\sqrt2}3\right)^2\frac{20}{\sqrt3}}3=\dfrac{16000}{27\sqrt3}

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3 years ago
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Harrizon [31]
The equation that models this relationship is y = 3x + 2.
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Four yardequal Blank feet
jonny [76]
1 yard = 3 feet

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So 4 yards is equal to 12 feet.
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