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3 years ago

Find the surface area of the regular pyramid. 10 yd 9 yd 7.8 yd lyd?

Mathematics

2 answers:

Readme [11.4K]3 years ago

8 0

Answer:

The lateral surface area of a regular pyramid is the sum of the areas of its lateral faces. The total surface area of a regular pyramid is the sum of the areas of its lateral faces and its base. The general formula for the lateral surface area of a regular pyramid is where p represents the perimeter of the base and l the slant height. Example 1:Find the lateral surface area of a regular pyramid with a triangular base if each edge of the base measures 8 inches and the slant height is 5 inches.The perimeter of the base is the sum of the sides.p = 3(8) = 24 inchesThe general formula for the total surface area of a regular pyramid is where p represents the perimeter of the base, l the slant height and B the area of the base. Example 2:Find the total surface area of a regular pyramid with a square base if each edge of the base measures 16 inches, the slant height of a side is 17 inches and the altitude is 15 inches.The perimeter of the base is 4s since it is a square.p = 4(16) = 64 inches The area of the base is s2.B = 162 = 256 inches2T. S. A. = There is no formula for a surface area of a non-regular pyramid since slant height is not defined. To find the area, find the area of each face and the area of the base and add them.

Step-by-step explanation:

hope you read it to see if yu can found yours

kobusy [5.1K]3 years ago

4 0

Answer:

234

Step-by-step explanation:

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Doss [256]

Answer:

20 degree

Step-by-step explanation:

x + x + 70 = 110 degree (sum of two opposite interior angle equal to the exterior angle formed)

2x = 110 - 70

x = 40/2

x = 20 degree

3 0

3 years ago

what is the mistake in this problem and why is it wrong? 15+(√9•7)÷13 = 15+(3•7)÷3 = 15+(21)÷3 = 36+3 =12

Serggg [28]

[Edit:}

Okay! So after you have 15+(21)÷3, you have to remember PEMDAS.

PEMDAS is the order in which you solve equations.

1. Parentheses: you solve everything in the parentheses first, all while following the rules of PEMDAS

2. Exponents: after you solve the things in the parentheses, you do the exponents.

3. Then you do Multiplication or Division, solving in the order from left to right.

4. After, you do Addition or Subtraction, solving in the order from left to right.

So using PEMDAS, we'll solve 15+(21)÷3.

We do division before addition, so 21/3 is 7.

Then you add 15 to 7 and get 22 as your final answer.

Hope this helps!

7 0

3 years ago

Suppose that f: R --> R is a continuous function such that f(x +y) = f(x)+ f(y) for all x, yER Prove that there exists KeR su

Pachacha [2.7K]

<h2>Answer with explanation:</h2>

It is given that:

f: R → R is a continuous function such that:

$f(x+y)=f(x)+f(y)------(1)$ ∀ x,y ∈ R

Now, let us assume f(1)=k

Also,

f(0)=0

( Since,

f(0)=f(0+0)

i.e.

f(0)=f(0)+f(0)

By using property (1)

Also,

f(0)=2f(0)

i.e.

2f(0)-f(0)=0

i.e.

f(0)=0 )

Also,

f(2)=f(1+1)

i.e.

f(2)=f(1)+f(1) ( By using property (1) )

i.e.

f(2)=2f(1)

i.e.

f(2)=2k

Similarly for any m ∈ N

f(m)=f(1+1+1+...+1)

i.e.

f(m)=f(1)+f(1)+f(1)+.......+f(1) (m times)

i.e.

f(m)=mf(1)

i.e.

f(m)=mk

Now,

$f(1)=f(\dfrac{1}{n}+\dfrac{1}{n}+.......+\dfrac{1}{n})=f(\dfrac{1}{n})+f(\dfrac{1}{n})+....+f(\dfrac{1}{n})\\\\\\i.e.\\\\\\f(\dfrac{1}{n}+\dfrac{1}{n}+.......+\dfrac{1}{n})=nf(\dfrac{1}{n})=f(1)=k\\\\\\i.e.\\\\\\f(\dfrac{1}{n})=k\cdot \dfrac{1}{n}$

Also,

when x∈ Q

i.e. $x=\dfrac{p}{q}$

Then,

$f(\dfrac{p}{q})=f(\dfrac{1}{q})+f(\dfrac{1}{q})+.....+f(\dfrac{1}{q})=pf(\dfrac{1}{q})\\\\i.e.\\\\f(\dfrac{p}{q})=p\dfrac{k}{q}\\\\i.e.\\\\f(\dfrac{p}{q})=k\dfrac{p}{q}\\\\i.e.\\\\f(x)=kx\ for\ all\ x\ belongs\ to\ Q$