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irina [24]
2 years ago
13

A solid has twelve more edges than faces. How many vertices does it have?

Mathematics
1 answer:
Nataly_w [17]2 years ago
8 0

Answer:

V=14

Step-by-step explanation:

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1,15,29,43 formula sequence
irga5000 [103]
All you’re doing is adding 14 to each
1+14=15
15+14=29
29+14=42
can i get brainliest answer ?
6 0
3 years ago
Read 2 more answers
PLZZ HELP URGENT
Mashcka [7]

Answer:

5x + 2x.....combine like terms..... = 7x

5x + 2x....subbing in 1                               7x - 1....subbing in 1

5(1) + 2(1) = 5 + 2 = 7                               7(1) - 1 = 7 - 1 = 6

5x + 2x...subbing in 2                                 7x - 1...subbing in 2

5(2) + 2(2) = 10 + 4 = 14                            7(2) - 1 = 14 - 1 = 13

5x + 2x...subbing in 3                                 7x - 1...subbing in 3

5(3) + 2(3) = 15 + 6 = 21                            7(3) - 1 = 21 - 1 = 20

5x + 2x...subbing in 4                                 7x - 1....subbing in 4

5(4) + 2(4) = 20 + 8 = 28                            7(4) - 1 = 28 - 1 = 27

5x + 2x...subbing in 5                                 7x - 1...subbing in 5

5(5) + 2(5) = 25 + 10 = 35                          7(5) - 1 = 35 - 1 = 34

5x + 2x result values are 1 more then 7x - 1 result values

there are no values that will make the 2 expressions equal....

because 5x + 2x = 7x......and the other one is 7x - 1......so the 7x - 1 values will always be 1 number less...because ur subtracting one

Step-by-step explanation:

5 0
3 years ago
Read 2 more answers
Find the length of the curve. R(t) = cos(8t) i + sin(8t) j + 8 ln cos t k, 0 ≤ t ≤ π/4
arsen [322]

we are given

R(t)=cos(8t)i+sin(8t)j+8ln(cos(t))k

now, we can find x , y and z components

x=cos(8t),y=sin(8t),z=8ln(cos(t))

Arc length calculation:

we can use formula

L=\int\limits^a_b {\sqrt{(x')^2+(y')^2+(z')^2} } \, dt

x'=-8sin(8t),y=8cos(8t),z=-8tan(t)

now, we can plug these values

L=\int _0^{\frac{\pi }{4}}\sqrt{(-8sin(8t))^2+(8cos(8t))^2+(-8tan(t))^2} dt

now, we can simplify it

L=\int _0^{\frac{\pi }{4}}\sqrt{64+64tan^2(t)} dt

L=\int _0^{\frac{\pi }{4}}8\sqrt{1+tan^2(t)} dt

L=\int _0^{\frac{\pi }{4}}8\sqrt{sec^2(t)} dt

L=\int _0^{\frac{\pi }{4}}8sec(t) dt

now, we can solve integral

\int \:8\sec \left(t\right)dt

=8\ln \left|\tan \left(t\right)+\sec \left(t\right)\right|

now, we can plug bounds

and we get

=8\ln \left(\sqrt{2}+1\right)-0

so,

L=8\ln \left(1+\sqrt{2}\right)..............Answer

5 0
3 years ago
Nick’s parents purchased their first home in the 1980’s with a 30-year mortgage at 19.5%.
siniylev [52]

<span>$ 2037.40 </span>

<span>Interesting that there was no down payment..........The staggering monthly payment above would be enough to pay off the total mortgage in just over 5 years. Nick's parents would be much better off to save their money for a few years.</span>
7 0
3 years ago
Janet made 75 more cookies to sell at the
UNO [17]

Answer:

200

Step-by-step explanation:

Janet = x + 75

Kathy = x

2x + 75 = 325

2x = 325 - 75

2x = 250

x/Kathy = 125

x + 75/Janet = 125 + 75

= 200

4 0
3 years ago
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