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2 years ago

Find the slope of the line - type you answer AS A WHOLE NUMBER (NO fraction)

Mathematics

2 answers:

n200080 [17]2 years ago

5 0

Answer:

2/1

Step-by-step explanation:

Mkey [24]2 years ago

4 0

2/1 is the answer of the problem

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Y=x^2+5x-8 to factored form

almond37 [142]

Answer:

Factor and set each factor equal to zero. Y=x2+5x−8

Step-by-step explanation:

5 0

1 year ago

Square root of 2tanxcosx-tanx=0

kobusy [5.1K]

If you're using the app, try seeing this answer through your browser: brainly.com/question/3242555

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Solve the trigonometric equation:

$\mathsf{\sqrt{2\,tan\,x\,cos\,x}-tan\,x=0}\\\\ \mathsf{\sqrt{2\cdot \dfrac{sin\,x}{cos\,x}\cdot cos\,x}-tan\,x=0}\\\\\\ \mathsf{\sqrt{2\cdot sin\,x}=tan\,x\qquad\quad(i)}$

Restriction for the solution:

$\left\{ \begin{array}{l} \mathsf{sin\,x\ge 0}\\\\ \mathsf{tan\,x\ge 0} \end{array} \right.$

Square both sides of (i):

$\mathsf{(\sqrt{2\cdot sin\,x})^2=(tan\,x)^2}\\\\ \mathsf{2\cdot sin\,x=tan^2\,x}\\\\ \mathsf{2\cdot sin\,x-tan^2\,x=0}\\\\ \mathsf{\dfrac{2\cdot sin\,x\cdot cos^2\,x}{cos^2\,x}-\dfrac{sin^2\,x}{cos^2\,x}=0}\\\\\\ \mathsf{\dfrac{sin\,x}{cos^2\,x}\cdot \left(2\,cos^2\,x-sin\,x \right )=0\qquad\quad but~~cos^2 x=1-sin^2 x}$

$\mathsf{\dfrac{sin\,x}{cos^2\,x}\cdot \left[2\cdot (1-sin^2\,x)-sin\,x \right]=0}\\\\\\ \mathsf{\dfrac{sin\,x}{cos^2\,x}\cdot \left[2-2\,sin^2\,x-sin\,x \right]=0}\\\\\\ \mathsf{-\,\dfrac{sin\,x}{cos^2\,x}\cdot \left[2\,sin^2\,x+sin\,x-2 \right]=0}\\\\\\ \mathsf{sin\,x\cdot \left[2\,sin^2\,x+sin\,x-2 \right]=0}$

Let

$\mathsf{sin\,x=t\qquad (0\le t$

So the equation becomes

$\mathsf{t\cdot (2t^2+t-2)=0\qquad\quad (ii)}\\\\ \begin{array}{rcl} \mathsf{t=0}&\textsf{ or }&\mathsf{2t^2+t-2=0} \end{array}$

Solving the quadratic equation:

$\mathsf{2t^2+t-2=0}\quad\longrightarrow\quad\left\{ \begin{array}{l} \mathsf{a=2}\\ \mathsf{b=1}\\ \mathsf{c=-2} \end{array} \right.$

$\mathsf{\Delta=b^2-4ac}\\\\ \mathsf{\Delta=1^2-4\cdot 2\cdot (-2)}\\\\ \mathsf{\Delta=1+16}\\\\ \mathsf{\Delta=17}$

$\mathsf{t=\dfrac{-b\pm\sqrt{\Delta}}{2a}}\\\\\\ \mathsf{t=\dfrac{-1\pm\sqrt{17}}{2\cdot 2}}\\\\\\ \mathsf{t=\dfrac{-1\pm\sqrt{17}}{4}}\\\\\\ \begin{array}{rcl} \mathsf{t=\dfrac{-1+\sqrt{17}}{4}}&\textsf{ or }&\mathsf{t=\dfrac{-1-\sqrt{17}}{4}} \end{array}$

You can discard the negative value for t. So the solution for (ii) is

$\begin{array}{rcl} \mathsf{t=0}&\textsf{ or }&\mathsf{t=\dfrac{\sqrt{17}-1}{4}} \end{array}$

Substitute back for t = sin x. Remember the restriction for x:

$\begin{array}{rcl} \mathsf{sin\,x=0}&\textsf{ or }&\mathsf{sin\,x=\dfrac{\sqrt{17}-1}{4}}\\\\ \mathsf{x=0+k\cdot 180^\circ}&\textsf{ or }&\mathsf{x=arcsin\bigg(\dfrac{\sqrt{17}-1}{4}\bigg)+k\cdot 360^\circ}\\\\\\ \mathsf{x=k\cdot 180^\circ}&\textsf{ or }&\mathsf{x=51.33^\circ +k\cdot 360^\circ}\quad\longleftarrow\quad\textsf{solution.} \end{array}$

where k is an integer.

I hope this helps. =)

3 0

2 years ago

Find the derivative of the following. please show the steps when you answer :1) f(x) = 8xe^x2) y= 5xe^x^43) f(x)= x^8+5/x4) f(t)

borishaifa [10]

Answer:

Since,

$\frac{d}{dx}x^n = nx^{n-1}$

$\frac{d}{dx}(f(x).g(x)) = f(x).\frac{d}{dx}(g(x)) + g(x).\frac{d}{dx}(f(x))$

$\frac{d}{dx}(\frac{f(x)}{g(x)})=\frac{g(x).f'(x) - f(x) g'(x)}{(g(x))^2}$

1) $y = 8x e^x$

Differentiating with respect to x,

$\frac{dy}{dx}=8( x \times e^x + e^x) = 8(xe^x + e^x) = 8e^x(x+1)$

2) $y = 5x e^{x^4}$

Differentiating w. r. t x,

$\frac{dy}{dx}=5(x\times 4x^3 e^{x^4}+e^{x^4})=5e^{x^4}(4x^4+1)$

3) $y = x^8 + \frac{5}{x^4}$

Differentiating w. r. t. x,

$\frac{dy}{dx}=8x^7 - \frac{5}{x^5}\times 4 = 8x^7 - \frac{20}{x^5}=\frac{8x^{12}-20}{x^5}$

4) $f(t) = te^{11}-6t^5$

Differentiating w. r. t. t,

$f'(t) = e^{11} - 30t^4$

5) $g(p) = p\ln(2p+3)$

Differentiating w. r. t. p,

$g'(p) = p\frac{1}{2p+3}(2) + \ln(2p+3) = \frac{2p}{2p+3}+\ln(2p+3)$

6) $z = (te^{6t}+e^{5t})^7$

Differentiating w. r. t. t,

$\frac{dz}{dt}=7(te^{6t}+e^{5t})^6 ( 6te^{6t}+e^{6t} + 5e^{5t})$

7) $w =\frac{2y + y^2}{7+y}$

Differentiating w. r. t. y,

$\frac{dw}{dy} = \frac{(7+y)(2+2y)-(2y+y^2)}{(7+y)^2} = \frac{14 + 2y + 14y +2y^2 - 2y - y^2}{(7+y)^2}=\frac{14+14y+y^2}{(7+y)^2}$

7 0

2 years ago

The quality-control department of Starr Communications, the manufacturer of video-game cartridges, has determined from records t

notsponge [240]

Answer: (A) The probability that a cartridge purchased will have a video or audio defect is 1.9%

(B) The probability that a cartridge purchased will not have a video or audio defect is zero.

Step-by-step explanation: The data given shows that 1.2% (or 120) cartridges have video defects, 0.9% have audio defects (or 90) and 0.2% (or 20) have both audio and video defects.

The possible outcomes for all events (audio defects and video defects) is derived as 120 plus 90 which is equals 210 possibilities (or possible outcomes).

Therefore the probability of having an audio defect is calculated as follows;

P(Audio) = Number of required outcomes/Number of all possible outcomes

P(Audio) = 90/210

P(Audio) = 3/7

Also the probability of having a video defect is derived as follows;

P(Video) = Number of required outcomes/Number of all possible outcomes

P(Video) = 120/210

P(Video) = 4/7

However we should take note of the fact that 0.2% or 20 of the cartridges in the sample size has both audio and video defects. Hence the probability that a cartridge has both audio and video defects is calculated as;

P(Audio and Video) = Number of required outcomes/Number of all possible outcomes

P(Audio and Video) = 20/210

P(Audio and Video) = 2/21

To calculate the probability that a cartridge bought would have either an audio or a video defect would mean to add both probabilities together, but we MUST SUBTRACT the probability of having both an audio defect and video defect (that is P{Audio and Video}). The reason is that this is already included in both probabilities and we need to avoid double counting. Hence we have;

(A); P(Video OR Audio defect) = P(Audio) + P(Video) - P(Audio and Video)

P(Video OR Audio defect) = (3/7 + 4/7) - 2/21

P(Video OR Audio defect) = 1 - 2/21

P((Video OR Audio defect) = 19/21

Therefore the probability that a cartridge purchased will have a video or audio defect is 190, or better still 1.9%.

(B): From all possibilities shown, which is 210 possibilities of either events, we have determined that 120 will be the probability of having an audio defect and 90 will be the probability of having a video defect. Therefore the probability that a cartridge purchased will not fall into any of either possibilities is zero.

6 0

2 years ago

1/2 cubed? Please help...

Fudgin [204]

1/2 cubed= 1/2 times 1/2 times 1/2
1/2 time 1/2 = 1/4
1/4 times 1/2 = 1/8

A: 1/2 cubed equals 1/8

6 0

3 years ago