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sweet [91]
3 years ago
11

Solve the function operations problem

Mathematics
1 answer:
madreJ [45]3 years ago
6 0
I can solve it for you if you’re still there
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Use the rule (x,y) (3x,2y) to find the image for the preimage defined by the given points.
Elina [12.6K]

Answer: The points of the images are  (9,10), (15,6), (6,4)  and the image is not a rigid motion because the shape changes in side.

Step-by-step explanation:

Since it gives you the scale factor then find they coordinates by multiplying the coordinates by the scare factor.

A(3,5) → (3*3,5*2) → (9,10)

B( 5,3) → (5*3, 3*2) → (15,6)

C ( 2,3)→ (2*3, 2*2)→ ( 6,4)

4 0
3 years ago
In baseball, Shintaro had 4 hits in his last 12 times at bat.
swat32

Answer:

The probability of shintaro is 4/12 simplified which is 1/3

Step-by-step explanation:

It's like saying.. Someone had won 4 races every 12 race

You divide both side of 4/12 till you can't divide anymore

4/12

3/9

2/6

1/3

7 0
3 years ago
Can somebody answer this for me plz its my homework.​
timofeeve [1]

Answer:

The rule for a reflection over the y -axis is (x,y)→(−x,y) .

The rule for a reflection over the x -axis is (x,y)→(x,−y) .

sorry I can't help you with number 4, hopefully this helps!

6 0
3 years ago
Am i correct? calculus
Advocard [28]
Check the picture below.

now, the "x" is a constant, the rocket is going up, so "y" is changing and so is the angle, but "x" is always just 15 feet from the observer.  That matters because the derivative of a constant is zero.

now, those are the values when the rocket is 30 feet up above.

\bf tan(\theta )=\cfrac{y}{x}\implies tan(\theta )=\cfrac{y}{15}\implies tan(\theta )=\cfrac{1}{15}\cdot y
\\\\\\
\stackrel{chain~rule}{sec^2(\theta )\cfrac{d\theta }{dt}}=\cfrac{1}{15}\cdot \cfrac{dy}{dt}\implies \cfrac{1}{cos^2(\theta )}\cdot\cfrac{d\theta }{dt}=\cfrac{1}{15}\cdot \cfrac{dy}{dt}
\\\\\\
\boxed{\cfrac{d\theta }{dt}=\cfrac{cos^2(\theta )\frac{dy}{dt}}{15}}\\\\
-------------------------------\\\\


\bf cos(\theta )=\cfrac{adjacent}{hypotenuse}\implies cos(\theta )=\cfrac{15}{15\sqrt{5}}\implies cos(\theta )=\cfrac{1}{\sqrt{5}}\\\\
-------------------------------\\\\
\cfrac{d\theta }{dt}=\cfrac{\left( \frac{1}{\sqrt{5}} \right)^2\cdot 11}{15}\implies \cfrac{d\theta }{dt}=\cfrac{\frac{1}{5}\cdot 11}{15}\implies \cfrac{d\theta }{dt}=\cfrac{\frac{11}{5}}{15}\implies \cfrac{d\theta }{dt}=\cfrac{11}{5}\cdot \cfrac{1}{15}
\\\\\\
\cfrac{d\theta }{dt}=\cfrac{11}{75}

8 0
3 years ago
Does any one know the answer to this thank you
makvit [3.9K]

Answer:

The First choice

Step-by-step explanation:

8 0
3 years ago
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