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Marta_Voda [28]
3 years ago
13

This is the equation of a quadratic function. y−2x2=−3 Which statement describes the graph of the function? Question 2 options:

The graph opens upward and has a maximum value of –3. The graph opens upward and has a minimum value of –3. The graph opens downward and has a minimum value of –3. The graph opens downward and has a maximum value of –3.
Mathematics
1 answer:
Salsk061 [2.6K]3 years ago
3 0

Did you ever get the answer?

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If you place a 26-foot ladder against the top of a 10-foot building, how many feet will the bottom of the ladder be from the bot
True [87]

Answer:

24 foot

Step-by-step explanation:

We are given that

Length of ladder=26 foot

Height of building=10 foot

We have to find the distance between the bottom of ladder and the bottom of building.

Pythagorus theorem:

(Hypotenuse)^2=(Base)^2+(Perpendicular\;side)^2

We have hypotenuse =AC=26 foot

Perpendicular side =AB=10 foot

Base=BC

Substitute the values in the given formula

(26)^2=(10)^2+(BC)^2

676=100+(BC)^2

(BC)^2=676-100=576

BC=\sqrt{576}=24 (Take positive because length is always positive)

BC=24 foot

Hence, the bottom of the ladder will be 24 foot  from the bottom of the building.

8 0
3 years ago
What is the fraction for What is the fraction for 0.21
gavmur [86]
The fraction for 0.21 would be 21/100
3 0
3 years ago
How do you solve the system of equations for y=2x-4 y=4x-10?
SVEN [57.7K]
Set the equal to each other

2x-4=4x-10
6=2x
x=3
y=2(3)-4
  =6-4
y =2
8 0
3 years ago
What's the flux of the vector field F(x,y,z) = (e^-y) i - (y) j + (x sinz) k across σ with outward orientation where σ is the po
emmasim [6.3K]
\displaystyle\iint_\sigma\mathbf F\cdot\mathrm dS
\displaystyle\iint_\sigma\mathbf F\cdot\mathbf n\,\mathrm dS
\displaystyle\iint_\sigma\mathbf F\cdot\left(\frac{\mathbf r_u\times\mathbf r_v}{\|\mathbf r_u\times\mathbf r_v\|}\right)\|\mathbf r_u\times\mathbf r_v\|\,\mathrm dA
\displaystyle\iint_\sigma\mathbf F\cdot(\mathbf r_u\times\mathbf r_v)\,\mathrm dA

Since you want to find flux in the outward direction, you need to make sure that the normal vector points that way. You have

\mathbf r_u=\dfrac\partial{\partial u}[2\cos v\,\mathbf i+\sin v\,\mathbf j+u\,\mathbf k]=\mathbf k
\mathbf r_v=\dfrac\partial{\partial v}[2\cos v\,\mathbf i+\sin v\,\mathbf j+u\,\mathbf k]=-2\sin v\,\mathbf i+\cos v\,\mathbf j

The cross product is

\mathbf r_u\times\mathbf r_v=\begin{vmatrix}\mathbf i&\mathbf j&\mathbf k\\0&0&1\\-2\sin v&\cos v&0\end{vmatrix}=-\cos v\,\mathbf i-2\sin v\,\mathbf j

So, the flux is given by

\displaystyle\iint_\sigma(e^{-\sin v}\,\mathbf i-\sin v\,\mathbf j+2\cos v\sin u\,\mathbf k)\cdot(\cos v\,\mathbf i+2\sin v\,\mathbf j)\,\mathrm dA
\displaystyle\int_0^5\int_0^{2\pi}(-e^{-\sin v}\cos v+2\sin^2v)\,\mathrm dv\,\mathrm du
\displaystyle-5\int_0^{2\pi}e^{-\sin v}\cos v\,\mathrm dv+10\int_0^{2\pi}\sin^2v\,\mathrm dv
\displaystyle5\int_0^0e^t\,\mathrm dt+5\int_0^{2\pi}(1-\cos2v)\,\mathrm dv

where t=-\sin v in the first integral, and the half-angle identity is used in the second. The first integral vanishes, leaving you with

\displaystyle5\int_0^{2\pi}(1-\cos2v)\,\mathrm dv=5\left(v-\dfrac12\sin2v\right)\bigg|_{v=0}^{v=2\pi}=10\pi
5 0
3 years ago
Ms. Parker baked 6 dozen cookies. She gave 2 dozen cookies to her sister, 1 À dozen
Lyrx [107]

Answer:

Step-by-step explanation:

She gave out three dozen and she had six dozen so she has three dozen left three dozen 3 * 12 she has 36 cookies

6 0
3 years ago
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