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Marysya12 [62]
2 years ago
11

HELP PLEASEEEEEEEEEEEEEEEEEEEEEE ITS A GRADE

Mathematics
2 answers:
tigry1 [53]2 years ago
7 0

Answer:

B for the first graph (as x increases, y increases), D for the second graph (as x increases, y decreases), D for the third graph (minimum at y = -9), and A for the forth graph (the graph opens up because the a value in the equation is positive)

laiz [17]2 years ago
3 0

Answer:

as x increases, y increses

the parabola grows wider and taller, meaning that the y and x values will both increase:)

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Translate the word phrase into a math expression.<br><br> three times the sum of a number and four.
larisa86 [58]
3(x+4) is what i came up with
6 0
3 years ago
3x^2+11x+6 in factor form
Novosadov [1.4K]

Answer: (3x+2)(x+3)

Step-by-step explanation: Factor by grouping.

I hope this helps you out.

8 0
3 years ago
Read 2 more answers
PLEASE HELP:
Kryger [21]
Alright so lets start with an arbitrary amount of students. Just to help us visualize the problem.

Say, 100 students for the first year when it was founded?

So far,
1996 - 100 students

From 96' to 97', it doubles. So:
1996 - 100 students
1997 - 200 students

From 97' to 98', it doubles AGAIN.
1996 - 100 students
1997 - 200 students
1998 - 400 students

So, what the percentage increase from 100, to 400?

Well, 100 x 4 gives us 400, so it's a 400 percent increase.
3 0
3 years ago
Read 2 more answers
Please help ASAP thank you
koban [17]
R - 18 = 50

r = 50 + 18 = 68
Hence Rick is 68 yrs old
7 0
2 years ago
Consider f and c below. f(x, y, z) = yzexzi + exzj + xyexzk,
Jet001 [13]

Answer:

f(x,y,z)=ye^{xz}+C  

Step-by-step explanation:

We can write the given expression as :

\vec f(x,y,z)=yze^{xz}\,\vec\imath+e^{xz}\,\vec\jmath+xye^{xz}\,\vec k

As given,   f = ∇f.

∇f = \dfrac{\partial f}{\partial x}i  + \dfrac{\partial f}{\partial y}j  +\dfrac{\partial f}{\partial z}k 

We can write the partial derivative with respect to x, y and z.

\dfrac{\partial f}{\partial x}=yze^{xz}       ___(Equation 1)

\dfrac{\partial f}{\partial y}=e^{xz}               ______(Equation 2)

\dfrac{\partial f}{\partial z}=xye^{xz}             ______(Equation 3)

Take equation 2 and integrate with respect to y,

\dfrac{\partial f}{\partial y}=e^{xz}

f(x,y,z)=ye^{xz}+a(x,z)           ----------Equation 4

Derivate both sides w.r.t x , we get :

\frac{d}{dx}(yze^{xz})=yze^{xz}+\dfrac{\partial a}{\partial x}

or

\dfrac{\partial a}{\partial x}=0

integrate

a(x,z)=b(z)

put in equation 4 ,

we get :

f(x,y,z)=ye^{xz}+b(z)

take derivative wrt z

\frac{d}{dz} (ye^{xz}+b(z))\impliesxye^{xz}=xye^{xz}+\frac{db}{dz}

we can take here:

\frac{db}{dz} = 0

integrate:

\int\ {\frac{db}{dz} } \, =\int0

b(z) = C

The function can be written as :

from equation 4 :

f(x,y,z)=ye^{xz}+C  

Where C is a constant.

4 0
3 years ago
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