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3 years ago

Find the surface area of the regular pyramid.

Mathematics

1 answer:

Paraphin [41]3 years ago

3 0

Area of all triangles : 10 x 12 = 120 120 divided by 2 = 60 then, 60 x 4 = 240

Area of the Square: 12 x 12 = 144

Surface Area of the Pyramid: 240 + 144 = 384 square millimeters!!

Hope This helps :)

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Twelve education students, in groups of four, are taking part in a student-teacher program.

eduard

Answer:

B. 330

Step-by-step explanation:

got it right on edge :)

4 0

3 years ago

A circle has a radius that is 4 centimeters long. If a central angle has a measure of 3 radians, what is the length of the arc t

kirill115 [55]

The total length of the circle is calculated by the following formula:

Lenght = 2.π.r

As r = 4c

Then,

Lenght = 2.π.4

Lenght = 8π centimetrs

Now use the rule of 3

But we have to know:

...<..> ...

Then,

8π ________ 2π <= 360°

x ________ 3π

2π . x = 8π . 3π

x = (8π . 3π) / 2π

x = (8 . 3π)/ 2

x = 12π centimenters

x ~ 37,69 c

4 0

3 years ago

Read 2 more answers

Given the two points what is the slope of the line?<br> (4,2) (10.4)

zlopas [31]

$m = \frac{y2-y1}{x2-x1}\\ m=\frac{4-2}{10-4} \\m = \frac{2}{6} \\\\m=\frac{1}{3}$

Therefore, the slope of the line is 1/3. In case if you want the equation too.

$y=mx+b\\y=\frac{1}{3}x+b\\ 2=\frac{1}{3}(4)+b\\ 2=\frac{4}{3}+b\\ b=-\frac{4}{3}+2\\ b=-\frac{4}{3} +\frac{6}{3}\\ b=\frac{2}{3}$

Therefore, the equation is y=1/3x+2/3

6 0

3 years ago

Select the correct answer from the drop-down menu.

lidiya [134]

Answer:

6y² - 20y + 6

Step-by-step explanation:

Use FOIL method

(3y - 1)(y-3) = 3y*y + 3y*(-3) + (-1)*y + (-1)*(-3)

= 3y²- 9y - y + 3 {add the like terms}

= 3y² - 10y + 3

2(3y - 1)(y - 3) = 2 ( 3y² - 10y + 3)

= 2*3y² - 2*10y + 2*3

= 6y² - 20y + 6

6 0

3 years ago

Help please! Will pick the Brainliest.<br> (sorry if it’s sideways)

KatRina [158]

$(x^2y^8)^\frac{2}{3}=x^{2\cdot\frac{2}{3}}y^{8\cdot\frac{2}{3}}=x^{\frac{4}{3}}y^{\frac{16}{3}}=x^{1\frac{1}{3}}y^{5\frac{1}{3}}=x^1x^\frac{1}{3}y^5y^\frac{1}{3}=xy^5\sqrt[3]{x}\sqrt[3]{y}=\boxed{xy^5\sqrt[3]{xy}}$

Answer B)

$\frac{1}{3}\left(\frac{2}{15}x-\frac{2}{3}\right)>x+\frac{1}{5}\quad|\cdot3\\\\\\\frac{2}{15}x-\frac{2}{3}>3x+\frac{3}{5}\quad|\cdot15\\\\\\2x-\frac{30}{3}>45x+\frac{45}{5}\\\\2x-10>45x+9\\\\-10-9>45x-2x\\\\-19>43x\quad|:43\\\\\boxed{x$

Answer B)

$\sqrt{14xy}\cdot\sqrt{12}\cdot\sqrt{30y}=\sqrt{14\cdot12\cdot30}\cdot\sqrt{xy^2}=\sqrt{2\cdot7\cdot3\cdot4\cdot2\cdot3\cdot5}\cdot\sqrt{xy^2}=\\\\=\sqrt{2^2\cdot3^2\cdot2^2\cdot5\cdot7}\cdot\sqrt{xy^2}=2\cdot3\cdot2\cdot\sqrt{35}\cdot y\cdot\sqrt{x}=\boxed{12y\sqrt{35x}}$

3 0

3 years ago