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mel-nik [20]
2 years ago
9

How do i find these in surface area with work??

Mathematics
1 answer:
Monica [59]2 years ago
5 0

Answer:

Just multiply. the numbers that they r showing u is for the surface.

Step-by-step explanation:

I do it all the time and get it right.

PLS GIVE ME BRAINILIEST IT WOULD REALLY HELP ME!

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Which quadratic function has a wider graph than y=2x^2?
Trava [24]

Answer:

y = 1/2 x²

Step-by-step explanation:

The coefficient of the first term in a quadratic, in our case here, x², will tell us how the graph stretches. This is akin to the slope within the linear graph. Similar to the slope, the smaller the coefficient value, or value of slope m, the shallower the angle.

When discussing quadratics, the larger the coefficient of our x² term, the steeper, and skinnier the graph. If we want to look for a graph that is wider than y = 2x², then we need to find a graph with a coefficient that is less than 2.

Our only option then is

y = 1/2 x²

8 0
3 years ago
How do you solve an equation when yhe variable is being divided by a whole number
valentinak56 [21]
For example, if x/2=7, x would be equal to 7x2=14, bringing the whole number the variable is divided by to the other side of the equation.
4 0
3 years ago
Find the extreme values of the function f(x, y) = 4x2 + 6y2 on the circle x2 + y2 = 1.
julia-pushkina [17]

Answer with Step-by-step explanation:

We are given that

f(x,y)=4x^2+6y^2

Let g(x,y)=x^2+y^2=1

We have to find the extreme values  of the given function

\nabla f(x,y)

\nabla g(x,y)=

Using Lagrange multipliers

\nabla f(x,y)=\lambda \nabla g(x,y)

f_x=\lambda g_x

8x=\lambda 2x

Possible value x=0 or \lambda=4

If x=0 then substitute the value in g(x,y)

Then, we get y=\pm 1

f_y=\lambda g_y

12y=\lambda 2y

If \lambda=4 and substitute in the equation

Then , we  get possible value of y=0

When y=0 substitute in g(x,y) then we get

x=\pm 1

Hence, function has possible  extreme values at points (0,1),(0,-1), (1,0) and (-1,0).

f(0,1)=6

f(0,-1)=6

f(1,0)=4

f(-1,0)=4

Therefore, the maximum value of f  on the circlex^2+y^2=1  is f(0,\pm1)=6 and minimum value of f(\pm1,0)=4

7 0
3 years ago
Quadrilateral H’ is the image of quadrilateral H after a sequence of transformations. If quadrilateral H’ is congruent to quadri
Nataly_w [17]
The last one is correct



A translation of 6 units down a rotation and then a reflection






good luck
8 0
3 years ago
Read 2 more answers
Can someone help please
marishachu [46]

Answer:

The answer is 2 -option b

5 0
2 years ago
Read 2 more answers
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