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garri49 [273]
3 years ago
6

What is the image of (0,-4) after a dilation by a scale factor of 1/2centered at the origin?

Mathematics
1 answer:
Alika [10]3 years ago
5 0

Answer:

The image after the dilation is (0,-2)

Step-by-step explanation:

To get the image of g he dilation, what we have to do is to divide the given coordinate by 2 or simply multiply each of the given coordinates by 1/2

Mathematically, we have this as;

(0 * 1/2 , -4 * 1/2)

= (0, -2)

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In the question, the angles proves that this triangle is an isoceles triangle.

So we can set the two sides equal: x + 4 = 3x - 8

Add -x+8 to both sides of the equation: 12 = 2x

Divide both sides by 2: x = 6

We find that x=6. AC is also x, so, it is 6 too.
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Find the area of the shaded figure. The area is square units (AREA OF COMPOSITE FIGURES)​
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Answer:

28 square units

Step-by-step explanation:

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3 years ago
Help pls 2x - 2 = -18
nexus9112 [7]

Answer:

x=-8

Step-by-step explanation:

We are solving for the variable, x, so let us <u>isolate</u> all terms of x. Let us start by moving the numerical value 2 to the right side, thereby <u>isolating</u> the term 2x on the left:

2x= -18+2

2x=-16

So close! Let us divide both sides by 2 now:

x= -8

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8 0
3 years ago
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Which of the following ordered triples is the solution to the system of equations?
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3 years ago
Solve: (3x^2-y)dx + (4y^3-x)dy =0 and find the solution passing through (1,1).
nordsb [41]

Step-by-step explanation:

The given equation is

(3x^{2}-y)dx+(4y^{3}-x)dy=0\\M(x,y)dx+N(x,y)dy=0

As a check for exactness we have

\frac{\partial N}{\partial x}=\frac{\partial M}{\partial y}\\\\\therefore \frac{\partial N}{\partial x}=\frac{\partial (4y^{3}-x)}{\partial x} =-1\\\\\frac{\partial M}{\partial y}=\frac{\partial (3x^{3}-y)}{\partial y} =-1\\\\\therefore \frac{\partial N}{\partial x}=\frac{\partial M}{\partial y}=-1

Hence the given equation is an exact differential equation and thus the solution is given by

thus the solution is given by

u(x,y)=\int M(x,y)\partial x+\phi (y)\\\\u(x,y)=\int (3x^{2}-y)\partial x+\phi (y,c)\\\\u(x,y)=x^{3}-xy+\phi (y,c)\\\\

Similarly we have

u(x,y)=\int N(x,y)\partial y+\phi (x,c)\\\\u(x,y)=\int (4y^{3}-x)\partial y+\phi (x,c)\\\\u(x,y)=y^{4}-xy+\phi (x,c)\\\\

Comparing both the solutions we infer

\phi (x,c)=x^{3}+c

Hence the solution becomes

u(x,y)=x^{3}+y^{4}-xy=c

given boundary condition is that it passes through (1,1) hence

1^{3}+1^{4}-1=c\\\\\therefore c=1

thus solution is

u(x,y)=x^{3}+y^{4}-xy=1

4 0
3 years ago
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