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k0ka [10]
3 years ago
11

Please help me please will give brainliest ​

Mathematics
1 answer:
alisha [4.7K]3 years ago
3 0

Answer:

16\sqrt{17}

Step-by-step explanation:

The perimeter is two times width plus two times length:

2\sqrt{17} + 2\cdot 7\sqrt{17} = 2\sqrt{17} + 14\sqrt{17} = 16\sqrt{17}

There is no way to simplify further.

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The temperature in a room begins at 70°F and decreases at a rate of 8 F per hour. Which of the following tables represents this
8_murik_8 [283]

Answer:

there's not options :/ nobody can answer this

6 0
3 years ago
At Alan’s Produce, all produce is 20% off on Tuesdays. If a head of lettuce is normally $2.50, how much would it cost on Tuesday
Rufina [12.5K]

Answer:

$2.00

Step-by-step explanation:

2.50*.20=.5

2.50-.5= $2

7 0
3 years ago
How many solutions does the equation −4y + 4y + 2 = 2 have?
timofeeve [1]
−4y + 4y + 2 = 2
<span> −4y + 4y    = 2 -2
               0 = 0 .  

Anytime you have this format of 0 = 0, just know there are infinitely many.

Any number as a value of y can satisfy the equation.
</span>
Infinitely many
4 0
4 years ago
Read 2 more answers
HELP ME PLZZZZZZZZ ASAP
Alex_Xolod [135]

Answer:

find the common factor

Step-by-step explanation:

1)13

2) 13(1+5)

8 0
2 years ago
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Find the area of the helicoid (or spiral ramp) with vector equation r(u, v) = ucos(v) i + usin(v) j + v k, 0 ≤ u ≤ 1, 0 ≤ v ≤ 9π
Natasha2012 [34]
Let H denote the helicoid parameterized by

\mathbb r(u,v)=u\cos v\,\mathbf i+u\sin v\,\mathbf j+v\,\mathbf k

for 0\le u\le1 and 0\le v\le9\pi. The surface area is given by the surface integral,

\displaystyle\iint_H\mathrm dS=\iint_H\|\mathbf r_u\times\mathbf r_v\|\,\mathrm du\,\mathrm dv

We have

\mathbf r_u=\dfrac{\partial\mathbf r(u,v)}{\partial u}=\cos v\,\mathbf i+\sin v\,\mathbf j
\mathbf r_v=\dfrac{\partial\mathbf r(u,v)}{\partial v}=-u\sin v\,\mathbf i+u\cos v\,\mathbf j+\mathbf k
\implies\mathbf r_u\times\mathbf r_v=\sin v\,\mathbf i-\cos v\,\mathbf j+u\,\mathbf k
\implies\|\mathbf r_u\times\mathbf r_v\|=\sqrt{1+u^2}

So the area of H is

\displaystyle\iint_H\mathrm dS=\int_{v=0}^{v=9\pi}\int_{u=0}^{u=1}\sqrt{1+u^2}\,\mathrm du\,\mathrm dv
=\dfrac{9(\sqrt2+\sinh^{-1}(1))\pi}2
5 0
3 years ago
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