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3 years ago

What is the perimeter of the base, p, of this square-based pyramid? (attached image

Mathematics

1 answer:

Katen [24]3 years ago

8 0

Answer:

32in

Step-by-step explanation:

8in + 8in + 8in + 8in = 32in

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Please help me photo include

nikklg [1K]

Answer:

$\frac{-117}{10}$

Step-by-step explanation:

1) Change the mixed fractions into improper fractions:

$-3\frac{1}{5} + (-8\frac{1}{2} )\\= \frac{-16}{5} - \frac{17}{2}$

2) Multiply the first fraction by 2/2 and the second fraction by 5/5 in order to create common factors on the denominator (You can do this because you are essentially multiplying by one, for example 2/2 = 1):

$\frac{-16}{5} *\frac{2}{2} - \frac{-17}{2} * \frac{5}{5} \\ =\frac{-32}{10} - \frac{85}{10}$

3) Simplify the numerator(top part) while keeping the denominator(bottom part) the same:

$\frac{-32}{10} -\frac{85}{10} \\\\=\frac{-32 - 85}{10}\\ \\= \frac{-117}{10}$

5 0

3 years ago

HELP! DUE TODAY!! Will give brainiest

inna [77]

Answer:

the equation is y=0.5x-2

6 0

3 years ago

What is the probability of rolling a 3 on a 6-sided 8)

adelina 88 [10]

Answer:

the chances of rolling a 3 are 1/6

the changes of not doing it is 5/6

7 0

3 years ago

Let z=3+i, then find a. Z² b. |Z| c.<img src="https://tex.z-dn.net/?f=%5Csqrt%7BZ%7D" id="TexFormula1" title="\sq

zysi [14]

Given z = 3 + i, right away we can find

(a) square

z ² = (3 + i )² = 3² + 6i + i ² = 9 + 6i - 1 = 8 + 6i

(b) modulus

|z| = √(3² + 1²) = √(9 + 1) = √10

(d) polar form

First find the argument:

arg(z) = arctan(1/3)

Then

z = |z| exp(i arg(z))

z = √10 exp(i arctan(1/3))

z = √10 (cos(arctan(1/3)) + i sin(arctan(1/3))

Any complex number has 2 square roots. Using the polar form from part (d), we have

√z = √(√10) exp(i arctan(1/3) / 2)

and

√z = √(√10) exp(i (arctan(1/3) + 2π) / 2)

Then in standard rectangular form, we have

$\sqrt z = \sqrt[4]{10} \left(\cos\left(\dfrac12 \arctan\left(\dfrac13\right)\right) + i \sin\left(\dfrac12 \arctan\left(\dfrac13\right)\right)\right)$

and

$\sqrt z = \sqrt[4]{10} \left(\cos\left(\dfrac12 \arctan\left(\dfrac13\right) + \pi\right) + i \sin\left(\dfrac12 \arctan\left(\dfrac13\right) + \pi\right)\right)$

We can simplify this further. We know that z lies in the first quadrant, so

0 < arg(z) = arctan(1/3) < π/2

which means

0 < 1/2 arctan(1/3) < π/4

Then both cos(1/2 arctan(1/3)) and sin(1/2 arctan(1/3)) are positive. Using the half-angle identity, we then have

$\cos\left(\dfrac12 \arctan\left(\dfrac13\right)\right) = \sqrt{\dfrac{1+\cos\left(\arctan\left(\dfrac13\right)\right)}2}$

$\sin\left(\dfrac12 \arctan\left(\dfrac13\right)\right) = \sqrt{\dfrac{1-\cos\left(\arctan\left(\dfrac13\right)\right)}2}$

and since cos(x + π) = -cos(x) and sin(x + π) = -sin(x),

$\cos\left(\dfrac12 \arctan\left(\dfrac13\right)+\pi\right) = -\sqrt{\dfrac{1+\cos\left(\arctan\left(\dfrac13\right)\right)}2}$

$\sin\left(\dfrac12 \arctan\left(\dfrac13\right)+\pi\right) = -\sqrt{\dfrac{1-\cos\left(\arctan\left(\dfrac13\right)\right)}2}$

Now, arctan(1/3) is an angle y such that tan(y) = 1/3. In a right triangle satisfying this relation, we would see that cos(y) = 3/√10 and sin(y) = 1/√10. Then

$\cos\left(\dfrac12 \arctan\left(\dfrac13\right)\right) = \sqrt{\dfrac{1+\dfrac3{\sqrt{10}}}2} = \sqrt{\dfrac{10+3\sqrt{10}}{20}}$

$\sin\left(\dfrac12 \arctan\left(\dfrac13\right)\right) = \sqrt{\dfrac{1-\dfrac3{\sqrt{10}}}2} = \sqrt{\dfrac{10-3\sqrt{10}}{20}}$

$\cos\left(\dfrac12 \arctan\left(\dfrac13\right)+\pi\right) = -\sqrt{\dfrac{10-3\sqrt{10}}{20}}$

$\sin\left(\dfrac12 \arctan\left(\dfrac13\right)+\pi\right) = -\sqrt{\dfrac{10-3\sqrt{10}}{20}}$

So the two square roots of z are

$\boxed{\sqrt z = \sqrt[4]{10} \left(\sqrt{\dfrac{10+3\sqrt{10}}{20}} + i \sqrt{\dfrac{10-3\sqrt{10}}{20}}\right)}$

and

$\boxed{\sqrt z = -\sqrt[4]{10} \left(\sqrt{\dfrac{10+3\sqrt{10}}{20}} + i \sqrt{\dfrac{10-3\sqrt{10}}{20}}\right)}$

3 0

3 years ago

Read 2 more answers

The board of an internet start up company has sixteen members. if one person is in charge of research and one is in charge of ma

Brilliant_brown [7]

The number of ways in which the position of one person in charge of research and one person in charge of marketing can be chosen from the board of 16 members is 240 ways. Computed using permutations.

The permutation is the way of choosing a set of objects from a larger set of objects in a specific sequence.

If we are choosing r number of objects, from n number of objects in a specific sequence, then we follow permutation, and it is given as:

nPr = n!{(n - r)!}.

In the question, we are asked to find the number of ways in which the position of one person in charge of research and one person in charge of marketing can be chosen from the board of 16 members.

Since the position is to be chosen in a specific order, we will follow permutation.

The number of positions to be chosen (r) = 2.

The number of members (n) = 16.

Thus, the number of ways is given as:

nPr = n!{(n - r)!},

or, 16P2 = 16!{(16 - 2)!},

or, 16P2 = 16!/14! = 15*16 = 240.

Thus, the number of ways in which the position of one person in charge of research and one person in charge of marketing can be chosen from the board of 16 members is 240 ways. Computed using permutations.

Learn more about permutations at

brainly.com/question/4658834

#SPJ4

6 0

2 years ago