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3 years ago

Which transformation is represented by the rule (x,y) to (x-9, y+2)

Mathematics

1 answer:

Nitella [24]3 years ago

6 0

Answer:

Translation 9 units to the left, and 2 units up

Step-by-step explanation:

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Sally gave her brother $3.95 in quarters and dimes. if the number of quarters is 6 more than the number of dimes, how many quart

WARRIOR [948]

13 quarters
q=6+d
25q+10d=395
solve for q and get 13

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3 years ago

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Evaluate log38, given log32 ≈ 0.631.

ArbitrLikvidat [17]

$\log_38-\log_32^3=3\log_32\approx3\cdot0.631=1.893\\\\Used:\log_ab^n=n\cdot\log_ab$

4 0

3 years ago

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Which statement describes the inverse of m(x) = x^2 – 17x?

DochEvi [55]

Given:

The function is

$m(x)=x^2-17x$

To find:

The inverse of the given function.

Solution:

We have,

$m(x)=x^2-17x$

Substitute m(x)=y.

$y=x^2-17x$

Interchange x and y.

$x=y^2-17y$

Add square of half of coefficient of y , i.e., $\left(\dfrac{-17}{2}\right)^2$ on both sides,

$x+\left(\dfrac{-17}{2}\right)^2=y^2-17y+\left(\dfrac{-17}{2}\right)^2$

$x+\left(\dfrac{17}{2}\right)^2=y^2-17y+\left(\dfrac{17}{2}\right)^2$

$x+\left(\dfrac{17}{2}\right)^2=\left(y-\dfrac{17}{2}\right)^2$ $[\because (a-b)^2=a^2-2ab+b^2]$

Taking square root on both sides.

$\sqrt{x+\left(\dfrac{17}{2}\right)^2}=y-\dfrac{17}{2}$

Add $\dfrac{17}{2}$ on both sides.

$\sqrt{x+\left(\dfrac{17}{2}\right)^2}+\dfrac{17}{2}=y$

Substitute $y=m^{-1}(x)$ .

$m^{-1}(x)=\sqrt{x+(\dfrac{189}{4}})+\dfrac{17}{2}$

We know that, negative term inside the root is not real number. So,

$x+\left(\dfrac{17}{2}\right)^2\geq 0$

$x\geq -\left(\dfrac{17}{2}\right)^2$

Therefore, the restricted domain is $x\geq -\left(\dfrac{17}{2}\right)^2$ and the inverse function is $m^{-1}(x)=\sqrt{x+(\dfrac{189}{4}})+\dfrac{17}{2}$ .

Hence, option D is correct.

Note: In all the options square of $\dfrac{17}{2}$ is missing in restricted domain.

7 0

3 years ago

Suppose that the position of one particle at time is given by x1=3sin t, y1 = 2 cos t, 0 ≤ t ≤ 2π and the position of a second p

Mashcka [7]

Answer:

there is no collision between the particles

Step-by-step explanation:

for the first particle

x1=3sin t, y1 = 2 cos t, 0 ≤ t ≤ 2π

for the second particle

x2 = -3 + cos t, y2 = 1 + sin t, 0 ≤ t ≤ 2π

then for the collision

x1=x2 → 3*sin t = -3 + cos t → sin t= -1 + (cos t)/3→ 1+ sin t = (1/3)cos t

y1=y2 → 1 + sin t = 2 cos t → (1/3)cos t = 2 cos t →(1/3) = 2

since 1/3 ≠ 2 there is no collision between the particles

6 0

3 years ago

What are two different ways you can show 103?

valina [46]

Answer:

100+3, a hundred three.

Step-by-step explanation:

4 0

3 years ago