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3 years ago

Determine if the card below represents a reflection over the x-axis or a

Mathematics

1 answer:

salantis [7]3 years ago

6 0

Answer:

reflected over x axis

Step-by-step explanation:

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A rock is dropped from a height of 25 feet. The function h=-16t^2 +25 gives the height h of the rock after t seconds. When does

scZoUnD [109]

Answer:

5 seconds

Step-by-step explanation:

The ground is 25 eet, so 25=-16t^2+400

-16t^2=-400, 16t^2=400

t^2=25, dividing both sides by 16

t=5 seconds.

3 0

3 years ago

Rigid transformations how to know when a figure is dilation transformation, rotation and dilation?

Naddika [18.5K]

Answer:

the shape could be congruent or similar to its preimage. There are basically four types of transformations: Rotation; Translation; Dilation; Reflection; Definition of Transformations. Transformations could be rigid (where the shape or size of preimage is not changed) and non-rigid (where the size is changed but the shape remains the same).

Step-by-step explanation:

7 0

2 years ago

Help me out with this

ArbitrLikvidat [17]

Answer:

What is 2x + 5 (y-1) when x=8 and y=5

Step-by-step explanation:

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5 0

2 years ago

Read 2 more answers

Find the max and min values of f(x,y,z)=x+y-z on the sphere x^2+y^2+z^2=81

Anton [14]

Using Lagrange multipliers, we have the Lagrangian

$L(x,y,z,\lambda)=x+y-z+\lambda(x^2+y^2+z^2-81)$

with partial derivatives (set equal to 0)

$L_x=1+2\lambda x=0\implies x=-\dfrac1{2\lambda}$
$L_y=1+2\lambda y=0\implies y=-\dfrac1{2\lambda}$
$L_z=-1+2\lambda z=0\implies z=\dfrac1{2\lambda}$
$L_\lambda=x^2+y^2+z^2-81=0\implies x^2+y^2+z^2=81$

Substituting the first three equations into the fourth allows us to solve for $\lambda$ :

$x^2+y^2+z^2=\dfrac1{4\lambda^2}+\dfrac1{4\lambda^2}+\dfrac1{4\lambda^2}=81\implies\lambda=\pm\dfrac1{6\sqrt3}$

For each possible value of $\lambda$ , we get two corresponding critical points at $(\mp3\sqrt3,\mp3\sqrt3,\pm3\sqrt3)$ .

At these points, respectively, we get a maximum value of $f(3\sqrt3,3\sqrt3,-3\sqrt3)=9\sqrt3$ and a minimum value of $f(-3\sqrt3,-3\sqrt3,3\sqrt3)=-9\sqrt3$ .

5 0

2 years ago

What expression is equivalent to -3+2/3y-4-1/3y

Neko [114]

Answer:

The simplified form of the given expression is $-7 + (\frac{1}{3})y$

Step-by-step explanation:

Here, the given expression is:

$-3 + (\frac{2}{3}) y - 4 - (\frac{1}{3})y$

Now to simplify the given expression, perform operations on LIKE TERMS:

We get:

$-3 + (\frac{2}{3}) y - 4 - (\frac{1}{3})y =( -3 - 4) + [(\frac{2}{3}) y- (\frac{1}{3})y]\\= - 7 + [(\frac{2}{3}) -(\frac{1}{3})]y = -7 + [\frac{2-1}{3}]y\\ = -7 + (\frac{1}{3})y\\ \implies -3 + (\frac{2}{3}) y - 4 - (\frac{1}{3})y = -7 + (\frac{1}{3})y$

Hence the simplified form of the given expression is $-7 + (\frac{1}{3})y$

6 0

3 years ago