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2 years ago

Please help me somebody! <3

Mathematics

2 answers:

Misha Larkins [42]2 years ago

8 0

I believe it would be -17m-4/8

Tems11 [23]2 years ago

7 0

Answer: 85 m + 20

Step-by-step explanation:

Make fractions have same, common denominator. LCD = 40

7/8 = 35/40

9/10 = 36/40

2/5 = 16/40

3/1 = 120 / 40

$\frac{35}{40}m+\frac{36}{40}-\frac{120}{40}m - \frac{16}{40}$

35m + 36 - 120m - 16

85 m + 20

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Pls answer for my hw thanks

allsm [11]

Answer:

27.86 cm

Step-by-step explanation:

use a^2+b^2=c^2

plug in 10 to a and 26 to b

10^2+26^2=c^2

then simplify

100+676=c^2

776=c^2

then find the square root of 776 which is 27.856... cm

this can be simplified to 27.86 cm

change it to 28 cm if your teacher wants it rounded to the whole number

7 0

2 years ago

Find any stationary points of the graph of <img src="https://tex.z-dn.net/?f=y%20%3D%202x%5E2%20%2B%20e%5E-x%5E4" id="TexFormula

melomori [17]

$y = 2x {}^{2} + e {}^{ - x {}^{4} }$

$\frac{dy}{dx} = 4x +( e {}^{ - x {}^{4} } )( - 4x {}^{3} )$

$\frac{dy}{dx} = 4x - 4x {}^{3} e { }^{ - x {}^{4} }$

$\frac{dy}{dx} = 4x(1 - x {}^{2} e {}^{ - x {}^{4} } )$

$\frac{dy}{dx} = 0 \\ 4x = 0 \: \: \: \: \: 1 - x {}^{2} e {}^{ - x {}^{4} } = 0$

$4x = 0 \\ x = 0 \: \: \: \: \: \: y(0) = 0 + e {}^{0} = 1$

<h2> $\frac{x {}^{2} }{e {}^{x {}^{4} } } = 1 \\ x {}^{2} = e {}^{x {}^{4} } \\ 2ln(x) = x {}^{4} \\ these \: 2 \: functions \: do \: not \: intersect$ </h2><h2>Stationary point ( 0 , 1 )</h2>

<h2 />

6 0

1 year ago

Select the correct choice and fill in the answer box

Anvisha [2.4K]

To begin with, let us look at a few definitions that will help

A relation is a function if each x-value is paired with exactly one y-value. A vertical line test on a graph can be used to determine whether a relation is a function.

If we use a graph to check, we will have

We can see that there is no overlapping of coordinates. The table satisfies the vertical line test.

Hence, it is a function

The domain and range of function is the set of all possible inputs and outputs of a function respectively. The domain and range of a function y = f(x) is given as domain= {x ,x∈R }, range= {f(x), x∈Domain}.

The domain of the function, D is given by

$D=\mleft\lbrace-1,0,1,2,3\mright\rbrace$

The range, R is given by

$R=\mleft\lbrace-6,-1,2,5,8\mright\rbrace$

8 0

8 months ago

(−6x2−x)−(−4x2+10x−1) perform operation

weqwewe [10]

Answer:

Step-by-step explanation:

-6x^2 - x + 4x^2 - 10x + 1 = -2x^2 - 11x + 1

8 0

2 years ago

On an alien planet with no atmosphere, acceleration due to gravity is given by g = 12m/s^2. A cannonball is launched from the or

almond37 [142]

Answer:

a) $\vec r (t) = \left[(90\cdot \cos \theta)\cdot t \right]\cdot i + \left[(90\cdot \sin \theta)\cdot t -6\cdot t^{2} \right]\cdot j$ , b) $\theta = \frac{\pi}{4}$ , c) $y_{max} = 84.375\,m$ , $t = 3.75\,s$ .

Step-by-step explanation:

a) The function in terms of time and the inital angle measured from the horizontal is:

$\vec r (t) = [(v_{o}\cdot \cos \theta)\cdot t]\cdot i + \left[(v_{o}\cdot \sin \theta)\cdot t -\frac{1}{2}\cdot g \cdot t^{2} \right]\cdot j$

The particular expression for the cannonball is:

$\vec r (t) = \left[(90\cdot \cos \theta)\cdot t \right]\cdot i + \left[(90\cdot \sin \theta)\cdot t -6\cdot t^{2} \right]\cdot j$

b) The components of the position of the cannonball before hitting the ground is:

$x = (90\cdot \cos \theta)\cdot t$

$0 = 90\cdot \sin \theta - 6\cdot t$

After a quick substitution and some algebraic and trigonometric handling, the following expression is found:

$0 = 90\cdot \sin \theta - 6\cdot \left(\frac{x}{90\cdot \cos \theta} \right)$

$0 = 8100\cdot \sin \theta \cdot \cos \theta - 6\cdot x$

$0 = 4050\cdot \sin 2\theta - 6\cdot x$

$6\cdot x = 4050\cdot \sin 2\theta$

$x = 675\cdot \sin 2\theta$

The angle for a maximum horizontal distance is determined by deriving the function, equalizing the resulting formula to zero and finding the angle:

$\frac{dx}{d\theta} = 1350\cdot \cos 2\theta$

$1350\cdot \cos 2\theta = 0$

$\cos 2\theta = 0$

$2\theta = \frac{\pi}{2}$

$\theta = \frac{\pi}{4}$

Now, it is required to demonstrate that critical point leads to a maximum. The second derivative is:

$\frac{d^{2}x}{d\theta^{2}} = -2700\cdot \sin 2\theta$

$\frac{d^{2}x}{d\theta^{2}} = -2700$

Which demonstrates the existence of the maximum associated with the critical point found before.

c) The equation for the vertical component of position is:

$y = 45\cdot t - 6\cdot t^{2}$

The maximum height can be found by deriving the previous expression, which is equalized to zero and critical values are found afterwards:

$\frac{dy}{dt} = 45 - 12\cdot t$

$45-12\cdot t = 0$

$t = \frac{45}{12}$

$t = 3.75\,s$

Now, the second derivative is used to check if such solution leads to a maximum:

$\frac{d^{2}y}{dt^{2}} = -12$

Which demonstrates the assumption.

The maximum height reached by the cannonball is:

$y_{max} = 45\cdot (3.75\,s)-6\cdot (3.75\,s)^{2}$

$y_{max} = 84.375\,m$

7 0

2 years ago