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2 years ago

In a popular online role playing game, players can create detailed designs for their

Mathematics

1 answer:

Ber [7]2 years ago

5 0

Answer:

2/1

Step-by-step explanation:

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Write an equation of the line that is perpendicular to 5x+20y=10 and passes through the point (8,3)

natali 33 [55]

Y = 32x - 29

You can get this by solving for slope intercept form and then taking the opposite and recipricol of the slope. Solve the rest using point-slope form.

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4 years ago

Parallel and perpendicular Lines faqing help me

liberstina [14]

Perpendicular. y=3x+4

Parallel. y=-1/3 x-8/3

Perpendicular. y=-x-1

Parallel. y=x+11

Perpendicular. y=7/4 x+8

Parallel. y=-7/4 x-9/7

Perpendicular. y=4/5 x-3

Parallel. y=-5/4 x+35/2

yeesh, that took a while :) hope I helped!

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3 years ago

This is the last questionn help

makkiz [27]

Answer:

Step-by-step explanation:

That's the answer....

7 0

3 years ago

NO LINKS!!! Part 2: Find the Lateral Area, Total Surface Area, and Volume. Round your answer to two decimal places.

slava [35]

Answer:

<h3><u>Question 7</u></h3>

<u>Lateral Surface Area</u>

The bases of a triangular prism are the triangles.

Therefore, the Lateral Surface Area (L.A.) is the total surface area excluding the areas of the triangles (bases).

$\implies \sf L.A.=2(10 \times 6)+(3 \times 6)=138\:\:m^2$

<u>Total Surface Area</u>

Area of the isosceles triangle:

$\implies \sf A=\dfrac{1}{2}\times base \times height=\dfrac{1}{2}\cdot3 \cdot \sqrt{10^2-1.5^2}=\dfrac{3\sqrt{391}}{4}\:m^2$

Total surface area:

$\implies \sf T.A.=2\:bases+L.A.=2\left(\dfrac{3\sqrt{391}}{4}\right)+138=167.66\:\:m^2\:(2\:d.p.)$

<u>Volume</u>

$\sf \implies Vol.=area\:of\:base \times height=\left(\dfrac{3\sqrt{391}}{4}\right) \times 6=88.98\:\:m^3\:(2\:d.p.)$

<h3><u>Question 8</u></h3>

<u>Lateral Surface Area</u>

The bases of a hexagonal prism are the pentagons.

Therefore, the Lateral Surface Area (L.A.) is the total surface area excluding the areas of the pentagons (bases).

$\implies \sf L.A.=5(5 \times 6)=150\:\:cm^2$

<u>Total Surface Area</u>

Area of a pentagon:

$\sf A=\dfrac{1}{4}\sqrt{5(5+2\sqrt{5})}a^2$

where a is the side length.

Therefore:

$\implies \sf A=\dfrac{1}{4}\sqrt{5(5+2\sqrt{5})}\cdot 5^2=43.01\:\:cm^2\:(2\:d.p.)$

Total surface area:

$\sf \implies T.A.=2\:bases+L.A.=2(43.01)+150=236.02\:\:cm^2\:(2\:d.p.)$

<u>Volume</u>

$\sf \implies Vol.=area\:of\:base \times height=43.011193... \times 6=258.07\:\:cm^3\:(2\:d.p.)$

8 0

2 years ago

A pound of clementines costs $2.49 and ms. wild buys 1.8 pounds. how much did she spend?

Ann [662]

Answer:

i believe she spent $4.482 on clementines

6 0

2 years ago