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1 year ago

△RST is dilated with the rule DT,1/3 (x, y), where the center of dilation is T(3, –2).

Mathematics

2 answers:

dmitriy555 [2]1 year ago

6 0

△RST is dilated with the rule DT,1/3 (x, y), where the center of dilation is T(3, –2).

The distance between the x-coordinates of R and T is 3 .

The distance between the y-coordinates of R and T is 6.

R' is 1 unit left,2 units up from T, so the coordinates of R' are (2,0)

Gala2k [10]1 year ago

6 0

Answer:

△RST is dilated with the rule DT,1/3 (x, y), where the center of dilation is T(3, –2). The distance between the x-coordinates of R and T is 3. See below.

Step-by-step explanation:

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Answer:

After converting Raj will have $31.88 USD

Step-by-step explanation:

1 unit rupee = 0.01594 USD

It is given that,

Raj is visiting the United States and needs to convert 2000 rupees to US dollars

<u>Convert 2000 rupees to USD</u>

To find USD we have to multiply rupees with 0.01594

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2 years ago

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If mBC = (9x-53) and mCD = (2x + 45) find mBAD

Rom4ik [11]

Answer:

$m\angle BAD=73^o$

Step-by-step explanation:

The picture of the question in the attached figure

step 1

Find the value of x

Let

O ----> the center of the circle

we know that

Triangle BOC≅Triangle COD

$m\angle BOC=arc\ BC$ ----> by central angle

$m\angle COD=arc\ CD$ ----> by central angle

$m\angle BOC=m\angle COD$

therefore

$arc\ BC=arc\ CD$

substitute the given values

$(9x-53)^o=(2x+45)^o$

solve for x

$9x-2x=45+53\\7x=98\\x=14$

step 2

Find the measure of angle BAD

we know that

The inscribed angle is half that of the arc it comprises.

$m\angle BAD=\frac{1}{2} [arc\ BC+arc\ CD]$

$arc\ BC=9(14)-53=73^o$

$arc\ CD=2(14)+45=73^o$

substitute

$m\angle BAD=\frac{1}{2} [73^o+73^o]=73^o$

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2 years ago

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2 years ago

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Find the point (x,y) of x2+14xy+49y2=100 that is closest to the origin and lies in the first quadrant.

Gala2k [10]

Notice that

$x^2+14xy+49y^2=(x+7y)^2$

so the constraint is a set of two lines,

$(x+7y)^2=100\implies\begin{cases}x+7y=10\\x+7y=10\end{cases}$

and only the first line passes through the first quadrant.

The distance between any point $(x,y)$ in the plane is $\sqrt{x^2+y^2}$ , but we know that $\sqrt{f(x,y)}$ and $f(x,y)$ share the same critical points, so we need only worry about minimizing $x^2+y^2$ . The Lagrangian for this problem is then

$L(x,y,\lambda)=x^2+y^2+\lambda(x+7y-10)$

with partial derivatives (set equal to 0)

$L_x=2x+\lambda=0$

$L_y=2y+7\lambda=0$

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We have

$L_y-7L_x=2y-14x=0\implies y=7x$

which tells us that

$x+7y-10=0\iff x+49x=10\implies x=\dfrac15\implies y=\dfrac75$

so that $\left(\dfrac15,\dfrac75\right)$ is a critical point. The Hessian for the target function $x^2+y^2$ is

$H(x,y)=\begin{bmatrix}2&0\\0&2\end{bmatrix}$

which is positive definite for all $x,y$ , so the critical point is the site of a minimum. The minimum distance itself (which we don't seem to care about for this problem, but we might as well state it) is $\sqrt{\left(\dfrac15\right)^2+\left(\dfrac75\right)^2}=2$ .

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