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2 years ago

What is the total surface area of the pyramid??

Mathematics

1 answer:

Dominik [7]2 years ago

4 0

rectangular base= 6*10= 60cm^2

big triangles = 12*5=60, 60*2=120cm^2

small triangles= 12.6*3=37.8, 37.8*2=75.6cm^2

add = 255.6 cm^2

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I need help which triangle theorem do I use for these

irina [24]

Answer:

Step-by-step explanation: B. SSS because it shows us three sides ,C.RHS because it has a right angle, d. AAS because it shows us two sides and a non included side, e. RHS because it has a right angle, f. AAS because it has two angles and non included side.

4 0

2 years ago

Allan keeps tropical fish his aquarium is 4 feet long ,1 foot wideand 2 feet tall each fish needs atleast 0.5ft 3 of water what

Alik [6]

Answer: There are 16 fish that he can keep in the aquarium.

Step-by-step explanation:

Since we have given that

Length of aquarium = 4 feet

Width of aquarium = 1 foot

So, it forms a shape of cuboid.

As we know the formula for "Volume of Cuboid":

$Volume=Length\times Breadth\times Height\\\\Volume=4\times 1\times 2\\\\Volume=8\ ft^3$

According to question, Each fish needs atleast 0.5 cubic feet of water.

So, Number of fish that he can keep in the aquarium is given by

$\frac{8}{0.5}=16$

Hence, there are 16 fish that he can keep in the aquarium.

6 0

2 years ago

50pt! The function g(x)=x^2 is transformed to obtain function h: h(x)=g(x-3).

Vlad1618 [11]

Answer:

It's the Option A.

Step-by-step explanation:

h(x) = g(x - 3)

The -3 moves the whole graph of g(x) 3 units horizontally to the right.

5 0

2 years ago

What is the dollar value of 7 nickels and 8 quarters

victus00 [196]

Answer:

$2.35

Step-by-step explanation:

Nickels are worth 5 cents each and quarters are worth 25 cents.

7 nickels = 7 * 5 cents = 35 cents

8 quarters = 8 * 25 cents = 200 cents

Total = 200 + 35 = 235 cents

There are 100 cents per dollar, so we can find the dollar value by:

235/100 = $2.35

7 0

2 years ago

Can somebody explain how these would be done? The selected answer is incorrect, and I was told "Nice try...express the product b

trapecia [35]

Answer:

Solution ( Second Attachment ) : - 2.017 + 0.656i

Solution ( First Attachment ) : 16.140 - 5.244i

Step-by-step explanation:

Second Attachment : The quotient of the two expressions would be the following,

$6\left[\cos \left(\frac{2\pi }{5}\right)+i\sin \left(\frac{2\pi \:}{5}\right)\right]$ ÷ $2\sqrt{2}\left[\cos \left(\frac{-\pi }{2}\right)+i\sin \left(\frac{-\pi \:}{2}\right)\right]$

So if we want to determine this expression in standard complex form, we can first convert it into trigonometric form, then apply trivial identities. Either that, or we can straight away apply the following identities and substitute,

( 1 ) cos(x) = sin(π / 2 - x)

( 2 ) sin(x) = cos(π / 2 - x)

If cos(x) = sin(π / 2 - x), then cos(2π / 5) = sin(π / 2 - 2π / 5) = sin(π / 10). Respectively sin(2π / 5) = cos(π / 2 - 2π / 5) = cos(π / 10). Let's simplify sin(π / 10) and cos(π / 10) with two more identities,

( 1 ) $\cos \left(\frac{x}{2}\right)=\sqrt{\frac{1+\cos \left(x\right)}{2}}$

( 2 ) $\sin \left(\frac{x}{2}\right)=\sqrt{\frac{1-\cos \left(x\right)}{2}}$

These two identities makes sin(π / 10) = $\frac{\sqrt{2}\sqrt{3-\sqrt{5}}}{4}$ , and cos(π / 10) = $\frac{\sqrt{2}\sqrt{5+\sqrt{5}}}{4}$ .

Therefore cos(2π / 5) = $\frac{\sqrt{2}\sqrt{3-\sqrt{5}}}{4}$ , and sin(2π / 5) = $\frac{\sqrt{2}\sqrt{5+\sqrt{5}}}{4}$ . Substitute,

$6\left[ \left\frac{\sqrt{2}\sqrt{3-\sqrt{5}}}{4}+i\left\frac{\sqrt{2}\sqrt{5+\sqrt{5}}}{4}\right]$ ÷ $2\sqrt{2}\left[\cos \left(\frac{-\pi }{2}\right)+i\sin \left(\frac{-\pi \:}{2}\right)\right]$

Remember that cos(- π / 2) = 0, and sin(- π / 2) = - 1. Substituting those values,

$6\left[ \left\frac{\sqrt{2}\sqrt{3-\sqrt{5}}}{4}+i\left\frac{\sqrt{2}\sqrt{5+\sqrt{5}}}{4}\right]$ ÷ $2\sqrt{2}\left[0-i\right]$

And now simplify this expression to receive our answer,

$6\left[ \left\frac{\sqrt{2}\sqrt{3-\sqrt{5}}}{4}+i\left\frac{\sqrt{2}\sqrt{5+\sqrt{5}}}{4}\right]$ ÷ $2\sqrt{2}\left[0-i\right]$ = $-\frac{3\sqrt{5+\sqrt{5}}}{4}+\frac{3\sqrt{3-\sqrt{5}}}{4}i$ ,

$-\frac{3\sqrt{5+\sqrt{5}}}{4}$ = $-2.01749\dots$ and $\:\frac{3\sqrt{3-\sqrt{5}}}{4}$ = $0.65552\dots$

= $-2.01749+0.65552i$

As you can see our solution is option c. - 2.01749 was rounded to - 2.017, and 0.65552 was rounded to 0.656.

________________________________________

First Attachment : We know from the previous problem that cos(2π / 5) = $\frac{\sqrt{2}\sqrt{3-\sqrt{5}}}{4}$ , sin(2π / 5) = $\frac{\sqrt{2}\sqrt{5+\sqrt{5}}}{4}$ , cos(- π / 2) = 0, and sin(- π / 2) = - 1. Substituting we receive a simplified expression,

$6\sqrt{5+\sqrt{5}}-6i\sqrt{3-\sqrt{5}}$

We know that $6\sqrt{5+\sqrt{5}} = 16.13996\dots$ and $-\:6\sqrt{3-\sqrt{5}} = -5.24419\dots$ . Therefore,

Solution : $16.13996 - 5.24419i$

Which rounds to about option b.

7 0

2 years ago