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3 years ago

Please help will give brainliest

Mathematics

1 answer:

olchik [2.2K]3 years ago

5 0

Answer:

7. f₍ₓ₋₂₎ - 7

9. f₍₋ₓ₎ + 3

11. Vertical stretch by the factor of 12 and translated 2 units up

Step-by-step explanation:

This is my best guess and wish it is good. Otherwise report it.

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Find the sine pls help

Nat2105 [25]

Answer:

7/25 is correct answer dude

7 0

3 years ago

−2x (x + 5 ) for x = −3

ValentinkaMS [17]

Answer:

Step-by-step explanation:

(-2)(-3)(-3+5)

6(2)

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4 0

3 years ago

The distance between Taylor’s house and her school is 300,000 centimeters.

Sergio [31]

Answer:

30 km

Step-by-step explanation:

1 kilometer = 100 meters.

1 meter = 100 cm

therefore, 1 km = 10,000

and then, 300000 cm is 30 km

7 0

4 years ago

NASA launches a rocket at t = 0 seconds. Its height, in meters above sea-level, as a function of time is given by h ( t ) = − 4.

Oliga [24]

Answer:

$\displaystyle 1)48.2 \: \: \text{sec}$

$\rm \displaystyle 2)3021.6 \: m$

Step-by-step explanation:

<h3>Question-1:</h3>

so when flash down occurs the rocket will be in the ground in other words the elevation(height) from ground level will be 0 therefore,

to figure out the time of flash down we can set h(t) to 0 by doing so we obtain:

$\displaystyle - 4.9 {t}^{2} + 229t + 346 = 0$

to solve the equation can consider the quadratic formula given by

$\displaystyle x = \frac{ - b \pm \sqrt{ {b}^{2} - 4 ac} }{2a}$

so let our a,b and c be -4.9,229 and 346 Thus substitute:

$\rm\displaystyle t = \frac{ - (229) \pm \sqrt{ {229}^{2} - 4.( - 4.9)(346)} }{2.( - 4.9)}$

remove parentheses:

$\rm\displaystyle t = \frac{ - 229 \pm \sqrt{ {229}^{2} - 4.( - 4.9)(346)} }{2.( - 4.9)}$

simplify square:

$\rm\displaystyle t = \frac{ - 229 \pm \sqrt{ 52441- 4( - 4.9)(346)} }{2.( - 4.9)}$

simplify multiplication:

$\rm\displaystyle t = \frac{ - 229 \pm \sqrt{ 52441- 6781.6} }{ - 9.8}$

simplify Substraction:

$\rm\displaystyle t = \frac{ - 229 \pm \sqrt{ 45659.4} }{ - 9.8}$

by simplifying we acquire:

$\displaystyle t = 48.2 \: \: \: \text{and} \quad - 1.5$

since time can't be negative

$\displaystyle t = 48.2$

hence,

at 48.2 seconds splashdown occurs

<h3>Question-2:</h3>

to figure out the maximum height we have to figure out the maximum Time first in that case the following formula can be considered

$\displaystyle x _{ \text{max}} = \frac{ - b}{2a}$

let a and b be -4.9 and 229 respectively thus substitute:

$\displaystyle t _{ \text{max}} = \frac{ - 229}{2( - 4.9)}$

simplify which yields:

$\displaystyle t _{ \text{max}} = 23.4$

now plug in the maximum t to the function:

$\rm \displaystyle h(23.4)- 4.9 {(23.4)}^{2} + 229(23.4)+ 346$

simplify:

$\rm \displaystyle h(23.4) = 3021.6$

hence,

about 3021.6 meters high above sea-level the rocket gets at its peak?

5 0

3 years ago

Find the curl of ~V ~V = sin(x) cos(y) tan(z) i + x^2y^2z^2 j + x^4y^4z^4 k

ch4aika [34]

Given

$\vec v = f(x,y,z)\,\vec\imath+g(x,y,z)\,\vec\jmath+h(x,y,z)\,\vec k \\\\ \vec v = \sin(x)\cos(y)\tan(z)\,\vec\imath + x^2y^2z^2\,\vec\jmath+x^4y^4z^4\,\vec k$

the curl of $\vec v$ is

$\displaystyle \nabla\times\vec v = \left(\frac{\partial h}{\partial y}-\frac{\partial g}{\partial z}\right)\,\vec\imath - \left(\frac{\partial h}{\partial x}-\frac{\partial f}{\partial z}\right)\,\vec\jmath + \left(\frac{\partial g}{\partial x}-\frac{\partial f}{\partial y}\right)\,\vec k$

$\nabla\times\vec v = \left(4x^4y^3z^4-2x^2y^2z\right)\,\vec\imath \\\\ - \left(4x^3y^4z^4-\sin(x)\cos(y)\sec^2(z)\right)\,\vec\jmath \\\\ + \left(2xy^2z^2+\sin(x)\sin(y)\tan(z)\right)\,\vec k$

$\nabla\times\vec v = \left(4x^4y^3z^4-2x^2y^2z\right)\,\vec\imath \\\\ + \left(\sin(x)\cos(y)\sec^2(z)-4x^3y^4z^4\right)\,\vec\jmath \\\\ + \left(2xy^2z^2+\sin(x)\sin(y)\tan(z)\right)\,\vec k$

7 0

3 years ago

Please help will give brainliest ​

Please help will give brainliest