Help me please !!!!!!!

Mathematics

2 answers:

Natalka [10]3 years ago

7 0

Answer: D

Step-by-step explanation:

h(x) is f(x) for x moving 4 units right.

h(x)=f(x-4) =√(x-4)

Answer D

zlopas [31]3 years ago

5 0

Answer:

Step-by-step explanation:

Given f(x) then f(x + a) is a horizontal translation of f(x)

• If a > 0 then a shift left of a units

• If a < 0 then a shift right of a units

Here h(x) is f(x) shifted 4 units , that is a > 0 , then

h(x) = $\sqrt{x-4}$ → D

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Please help me me with this please actually help me please and thank you

Vsevolod [243]

Answer:

A. (-1, 3)

Step-by-step explanation:

To solve, find which point in the options are in the double shaded region (the region that is shaded the darkest which is the region above the point where the lines cross).

A. (-1, 3) is the correct answer because it falls in the correct region.

B. (-4, 1) is not the correct answer because it does not fall ABOVE both lines like the question says.

C. (0, 1) is not the correct answer because it falls in the 2nd darkest region.

6 0

2 years ago

An airplane travels about 1,500 Km in a straight line between Toronto and Winnipeg. However, the plane must climb to 10,000 m du

Anna007 [38]

Answer:416

Step-by-step explanation: Based on the given problem above, it requires a trigonometric solution. So here it goes:

The climb distance is:200 * Sec [ArcTan [10/200]] = 10√ 401 km

and the decent distance is:300 * Sec [ArcTan [10/300]] = 10√901 km

So add the given results above with 500 km and this will be the additional distance that plane moves through the air.

The answer would be in the unit meters.

((10√401 +10√901−500)∗1000)=416 meters

Hope this is the answer that you are looking for.

7 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}}\\\\
-------------------------------\\\\
$

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