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Rus_ich [418]
3 years ago
8

Help me Select the correct answer from each drop-down menu.

Mathematics
1 answer:
notsponge [240]3 years ago
6 0

Answer:

Ruiz, slower

Step-by-step explanation:

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3x+4y=24 <br> 2x+5y=2<br> Please help me with this problem
Serga [27]

Answer:

x=16

y=-6

Step-by-step explanation:

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3 years ago
Help, i need you to find the area
stealth61 [152]

Answer:

98

Step-by-step explanation:

Formula Base * Height

7*14=98

5 0
3 years ago
Read 2 more answers
The figure, polygon ABCD is dilated by a factor of 2 to produce A B C D with the origin as the center of dilation
cestrela7 [59]

In the polygon ABCD, the production of point A' and point D' is (-2,-2)  and (4,-2) respectively.

<h3>What is the transformation rule?</h3>

In coordinate planes, the rule of transformation involves interchanging the x and y values on the plane such that:

\mathbf{(x,y) \to  (k_x, k_y)}

where;

  • k = scale factor

From the given information:

  • The scale factor = 2
  • the origin as a center of dilation for image vertices = (0, 0)

Now, using the coordinate vertices:

  • A = (-1, -1)
  • D = (2, -1)

The corresponding coordinate of the image vertices is as follows:

A(-1,-1) → A'(2 × (-1), 2 × (-1))

A' = (-2,-2)

D(2,-1) → A'(2 × (2), 2 × (-1))

D' = (4,-2)

Learn more about the transformation rule in coordinate planes here:

brainly.com/question/4251601

#SPJ1

7 0
2 years ago
Write the equation of the line in fully simplified slope-intercept form.
mixer [17]

Answer:

y=1/2x+6

Step-by-step explanation:

3 0
3 years ago
2. Calculate an expression for dy/dx and d2y/dx2 in terms of t if the parametric pair is given as tan(x) = e^at and e^y = 1 + e^
Ber [7]

I assume a is a constant. If tan(x) = exp(at) (where exp(x) means eˣ), then differentiating both sides with respect to t gives

sec²(x) dx/dt = a exp(at)

Recall that

sec²(x) = 1 + tan²(x)

Then we have

(1 + tan²(x)) dx/dt = a exp(at)

(1 + exp(2at)) dx/dt = a exp(at)

dx/dt = a exp(at) / (1 + exp(2at))

If exp(y) = 1 + exp(2at), then differentiating with respect to t yields

exp(y) dy/dt = 2a exp(2at)

(1 + exp(2at)) dy/dt = 2a exp(2at)

dy/dt = 2a exp(2at) / (1 + exp(2at))

By the chain rule,

dy/dx = dy/dt • dt/dx = (dy/dt) / (dx/dt)

Then the first derivative is

dy/dx = (2a exp(2at) / (1 + exp(2at))) / (a exp(at) / (1 + exp(2at))

dy/dx = (2a exp(2at)) / (a exp(at))

dy/dx = 2 exp(at)

Since dy/dx is a function of t, if we differentiate dy/dx with respect to x, we have to use the chain rule again. Suppose we write

dy/dx = f(t)

By the chain rule, the derivative is

d²y/dx² = df/dx

d²y/dx² = df/dt • dt/dx

d²y/dx² = (df/dt) / (dx/dt)

d²y/dx² = 2a exp(at) / (a exp(at) / (1 + exp(2at)))

d²y/dx² = 2 (1 + exp(2at))

4 0
2 years ago
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