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3 years ago

Find the lateral area for the regular pyramid.

Mathematics

2 answers:

lord [1]3 years ago

6 0

Answer:

Lateral area of a regular pyramid with square is 12.64 square unit

Step-by-step explanation:

Given : A regular pyramid with base is a square.

We have to find the lateral area for the regular pyramid.

Lateral area of a regular pyramid with square = $a\sqrt{a^2+4h^2}$

Where, a is side of square

and h is height of pyramid.

Substitute, a = 2

and h = 3

we have,

Lateral area of a regular pyramid with square = $2\sqrt{(2)^2+4(3)^2}$

Simplify, we have,

Lateral area of a regular pyramid with square = 12.64

Thus, The Lateral area of a regular pyramid with square is 12.64 square unit

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Question

Find the lateral area for the regular pyramid.

fiasKO [112]3 years ago

3 0

2x2+2x $\sqrt{(2/2) x^{2} +3 x^{2} } +2 x \sqrt{(2/2) x^{2} +3 x^{2} }$
The answer is about 16.649

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(8.11×107)–(4.3×107)

Ivanshal [37]

Answer: 407.67

Explanation:

(8.11×107)–(4.3×107) =

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407.67

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3 years ago

Use the relationship between the

motikmotik

The equation x° = 180° - (37°+ 53°) can be used to find the value of x.

Step-by-step explanation:

Step 1; There are three lines in the given diagram. There are a baseline and two other lines. Out of the other two lines, one extends above and below the baseline whereas the other extends only above. Since the baseline is horizontal and the others are at angles we have the sum of all the three angles as 180° i.e. 37°, x°, and 53°.

37° + x° + 53° = 180°.

Step 2; To solve the value of x, we keep the unknown value at the left-hand side whereas all the known values are taken to the right side of the equation.

x° = 180° - (37°+ 53°).

So the fifth option can be used to determine the value of x.

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3 years ago

Which coordinate pairs make the equation x-9y=12 true?

asambeis [7]

Answer:

(12,0) and (3, -1) and (0,-4/3)

Step-by-step explanation:

To find if a point makes an equation true, just substitute its x and y values in in the equation and simplify.

1) Let's test the point (12,0). Substitute 12 for x and 0 for y in the equation:

$x-9y=12\\(12)-9(0)=12\\12-0=12\\12=12$

12 does equal 12, so (12,0) makes the equation true.

2) Let's test the point (0, 12). Substitute 0 for x and 12 for y in the equation:

$x-9y=12\\(0)-9(12) = 12\\0-108 = 12\\-108 = 12$

However, -108 does not equal 12, so (0,12) does not make the equation true.

3) Let's test the point (3, -1). Substitute 3 for x and -1 for y in the equation:

$x-9y=12\\(3)-9(-1) = 12\\3 + 9 = 12\\12 = 12$

12 does equal 12, so (3, -1) makes the equation true.

4) Let's test the point (0, -4/3). Substitute 0 for x and -4/3 for y in the equation:

$(0) -9(-\frac{4}{3} ) = 12\\0 + \frac{36}{3} = 12\\0 + 12 = 12 \\12 = 12$

12 does equal 12, so (0, -4/3) makes the equation true.

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3 years ago

Can you write the equation for a circle if the given point does not lie on the circle? Explain .

Pachacha [2.7K]

Answer:

Step-by-step explanation:

The equation of a circle with center (a,b) and radius r is given as:

${(x - a)}^{2} + {(y - b)}^{2} = {r}^{2}$

If a given point (x,y) does not lie on this circle, it will not satisfy its equation.

This means the distance from the point to the center is not equal to the radius.

It is either less or greater than the radius.

Hence you cannot write the equation of the circle.

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