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2 years ago

I need helppp!!!!!!!

Mathematics

2 answers:

Marina86 [1]2 years ago

7 0

i dont get that either lol pain. why does school have to be so hard

Otrada [13]2 years ago

7 0

I believe the answer is 1.2 !!

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ACCTUALLY HELP ME??!!!!!!! NEEDED FOR TOMORROW AND 50 POINTS! will award brainliest for first answer!!

drek231 [11]

Answer:

a:c = 35:24

a:c = 20:27

a:c = 35:22

a:c = 28:27

Step-by-step explanation:

a:b = 7:3

Using cross products

3a = 7b

Divide by 7

3a/7 = b

Now we want

8b = 5c

Substitute in 3a/7 for b

8 (3a/7) = 5c

24/7a = 5c

Multiply by 7

24/7a *7 = 5*7c

24a = 35c

Divide by c

24 a/c = 35

Divide by 24

a/c = 35/24

a:c = 35:24

a:b = 4:9

Using cross products

9a = 4b

Divide by 4

9a/4 = b

Now we want

3b = 5c

Substitute in 9a/4 for b

3 (9a/4) = 5c

27/4a = 5c

Multiply by 4

27/4a *4 = 5*4c

27a = 20c

Divide by c

27 a/c = 20

Divide by 27

a/c = 20/27

a:c = 20:27

b:c = 5:11

Using cross products

11b = 5c

Divide by 11

b = 5c/11

Now we want

2a = 7b

Substitute in 5c/11 for b

2a = 7(5c/11)

2a = 35c/11

Multiply by 11

2a*11 = 35c

22a = 35c

Divide by c

22 a/c = 35

Divide by 22

a/c = 35/22

a:c = 35:22

b:c = 14:3

Using cross products

3b = 14c

Divide by 3

b = 14c/3

Now we want

9a = 2b

Substitute in 14c/3 for b

9a = 2(14c/3)

9a = 28c/3

Multiply by 3

9a*3 = 28c

27a = 28c

Divide by c

27 a/c = 28

Divide by 27

a/c = 28/27

a:c = 28:27

7 0

3 years ago

a regular rectangular pyramid has a base and lateral faces that are congruent equilateral triangles. it has a lateral surface ar

Margarita [4]

A pyramid is regular if its base is a regular polygon, that is a polygon with equal sides and angle measures.
(and the lateral edges of the pyramid are also equal to each other)

Thus a regular rectangular pyramid is a regular pyramid with a square base, of side length say x.

The lateral faces are equilateral triangles of side length x.

The lateral surface area is 72 cm^2, thus the area of one face is 72/4=36/2=18 cm^2.

now we need to find x. Consider the picture attached, showing one lateral face of the pyramid.

by the Pythagorean theorem:

$h= \sqrt{ x^{2} - (x/2)^{2}}= \sqrt{ x^{2}- x^{2}/4}= \sqrt{3x^2/4}= \frac{ \sqrt{3} }{2}x$

thus,

$Area_{triangle}= \frac{1}{2}\cdot base \cdot height\\\\18= \frac{1}{2}\cdot x \cdot \frac{ \sqrt{3} }{2}x\\\\ \frac{18 \cdot 4}{ \sqrt{3}}=x^2$

thus:

$x^2 =\frac{18 \cdot 4}{ \sqrt{3}}= \frac{18 \cdot 4 \cdot\ \sqrt{3} }{3}=24 \sqrt{3}$ (cm^2)

but $x^{2}$ is exactly the base area, since the base is a square of sidelength = x cm.

So, the total surface area = base area + lateral area = $24 \sqrt{3}+72$ cm^2

Answer: $24 \sqrt{3}+72$ cm^2