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Nookie1986 [14]
4 years ago
12

Ray is boiling water at 99°C. Which of the following best describes 99°C? please help

Mathematics
1 answer:
kolezko [41]4 years ago
7 0
99+99=198 so it is 198°C
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Evaluate<img src="https://tex.z-dn.net/?f=10%5E%7B6%7D" id="TexFormula1" title="10^{6}" alt="10^{6}" align="absmiddle" class="la
katovenus [111]
121,000,000,000
this is because:
10^6= 1,000,000
10^3= 1,000
121^1= 121
3 0
3 years ago
If 3/4 of the library books are fiction and 5/7 are paperback what fraction of the books are fiction and paperback
Bess [88]

Answer:

15/38

Step-by-step explanation:

assuming the fraction of paperback in the fiction category is also 5/7, the fraction of the books which are fiction and paperback would just be the 2 fractions multiplied together.

3 0
3 years ago
Is the solution of this equation extraneous?
UkoKoshka [18]

Answer:

x = - 3 is extraneous

Step-by-step explanation:

Given

\sqrt{x+7} - 1 = x  ( add 1 to both sides )

\sqrt{x+7} = x + 1 ( square both sides )

x + 7 = (x + 1)² ← distribute right side

x + 7 = x² + 2x + 1 ( subtract x + 7 from both sides )

0 = x² + x - 6 ← in standard form

0 = (x + 3)(x - 2) ← in factored form

Equate each factor to zero and solve for x

x + 3 = 0 ⇒ x = - 3

x - 2 = 0 ⇒ x = 2

As a check

Substitute these values into the equation and if both sides are equal then they are the solutions.

x = - 3 : \sqrt{-3+7} - 1 = \sqrt{4} - 1 = 2 - 1 = 1 ≠ - 3

x = 2 : \sqrt{2+7} - 1  = \sqrt{9} - 1 = 3 - 1 = 2

x = 2 is a solution and x = - 3 is extraneous

6 0
3 years ago
Read 2 more answers
Find the arc-length parametrization of the curve that is the intersection of the elliptic cylinder x 2 + y 2/2 = 1 and the plane
storchak [24]

Answer:

f(\theta) = (cos(\frac{\theta}{\sqrt2}), \sqrt2 sin(\frac{\theta}{\sqrt2}), cos(\frac{\theta}{\sqrt2})-2)

0 ≤ Ф ≤ 4π.

Step-by-step explanation:

since x²+y²/2 = 1, then x²+s² = 1, with s = (y/√2)². Hence, (x,s) = (cos(Ф),sin(Ф)) and (x,y,z) = (cos(Ф),√2 sin(Ф), cos(Ф)-2). This expression evaluated in zero gives as result (1,0,-1). The derivate of this function is (-sin(Ф),√2 cos(Ф), -sen(Ф))

the norm of the derivate is √(sin²(Ф) + 2cos²(Ф)+sin²(Ф)) = √2. In order to make the norm equal to 1, i will divide Ф by √2, so that a √2 is dividing each term after derivating.

We take

f(\theta) = (cos(\frac{\theta}{\sqrt2}), \sqrt2 sin(\frac{\theta}{\sqrt2}), cos(\frac{\theta}{\sqrt2})-2)

Note that

  • f(0) = (1,0,-1)
  • f'(\theta) = (\frac{sin(\frac{\theta}{\sqrt2})}{\sqrt2}, cos(\frac{\theta}{\sqrt2})}, \frac{sin(\frac{\theta}{\sqrt2})}{\sqrt2})

Whose square norm is 1/2cos²(Ф/2)+sen²(Ф/2)+1/2cos²(Ф/2) = 1. This is te parametrization that we wanted.

The values from Ф range between 0 an 4π, because the argument of the sin and cos is Ф/2, not Ф, Ф/2 should range between 0 and 2π.

6 0
3 years ago
Pls answer 1 + 1 <br><br> this is too hard
salantis [7]

Answer:

<h2>ANSWER</h2>
  • 2

Step-by-step explanation:

1+1=2

ONE PLUS ONE EQUALS TWO

5 0
2 years ago
Read 2 more answers
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