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2 years ago

Solve for x. −23(3x−4)+3x=5619/6 A. 19/6 B. 11/6 C. 21/6 D. 29/6

Mathematics

1 answer:

MrRa [10]2 years ago

3 0

Answer:

Step-by-step explanation:

−23(3x−4)+3x=5619/6 Remove brackets on the left. Divide by 6 on the right.

-69x - 92 + 3x = 936.5 Combine like terms on the left.

-66x - 92 = 936.5 Add 92 to both sides

-66x = 936.5 + 92

-66x = 1028.5 Divide by -66

x = -15 7/12

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Solve for x going step by step .

OleMash [197]

Answer:

x=11

Step-by-step explanation:

We know that the angle is equal to 1/2 the difference of the long arc length between where the line segments touch the circle and the small arc length where the line segments touch the circle

They long distance is 360 - (86+74) = 200

The small distance is 86

4x+13 = 1/2 ( 200 - 86)

4x+13 = 1/2 (114)

4x+13 = 57

Subtract 13 from each side

4x+13-13 = 57-13

4x= 44

Divide by 4

4x/4 = 44/4

x=11

6 0

2 years ago

Find the derivative of ln(secx+tanx)

Sliva [168]

If you're using the app, try seeing this answer through your browser: brainly.com/question/3000160

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Find the derivative of

$\mathsf{y=\ell n(sec\,x+tan\,x)}\\\\\\ \mathsf{y=\ell n\!\left(\dfrac{1}{cos\,x}+\dfrac{sin\,x}{cos\,x} \right )}\\\\\\ \mathsf{y=\ell n\!\left(\dfrac{1+sin\,x}{cos\,x} \right )}$

You can treat y as a composite function of x:

$\left\{\! \begin{array}{l} \mathsf{y=\ell n\,u}\\\\ \mathsf{u=\dfrac{1+sin\,x}{cos\,x}} \end{array} \right.$

so use the chain rule to differentiate y:

$\mathsf{\dfrac{dy}{dx}=\dfrac{dy}{du}\cdot \dfrac{du}{dx}}\\\\\\ \mathsf{\dfrac{dy}{dx}=\dfrac{d}{du}(\ell n\,u)\cdot \dfrac{d}{dx}\!\left(\dfrac{1+sin\,x}{cos\,x}\right)}$

The first derivative is 1/u, and the second one can be evaluated by applying the quotient rule:

$\mathsf{\dfrac{dy}{dx}=\dfrac{1}{u}\cdot \dfrac{\frac{d}{dx}(1+sin\,x)\cdot cos\,x-(1+sin\,x)\cdot \frac{d}{dx}(cos\,x)}{(cos\,x)^2}}\\\\\\ \mathsf{\dfrac{dy}{dx}=\dfrac{1}{u}\cdot \dfrac{(0+cos\,x)\cdot cos\,x-(1+sin\,x)\cdot (-\,sin\,x)}{(cos\,x)^2}}$

Multiply out those terms in parentheses:

$\mathsf{\dfrac{dy}{dx}=\dfrac{1}{u}\cdot \dfrac{cos\,x\cdot cos\,x+(sin\,x+sin\,x\cdot sin\,x)}{(cos\,x)^2}}\\\\\\ \mathsf{\dfrac{dy}{dx}=\dfrac{1}{u}\cdot \dfrac{cos^2\,x+sin\,x+sin^2\,x}{(cos\,x)^2}}\\\\\\ \mathsf{\dfrac{dy}{dx}=\dfrac{1}{u}\cdot \dfrac{(cos^2\,x+sin^2\,x)+sin\,x}{(cos\,x)^2}\qquad\quad (but~~cos^2\,x+sin^2\,x=1)}\\\\\\ \mathsf{\dfrac{dy}{dx}=\dfrac{1}{u}\cdot \dfrac{1+sin\,x}{(cos\,x)^2}}$

Substitute back for $\mathsf{u=\dfrac{1+sin\,x}{cos\,x}:}$

$\mathsf{\dfrac{dy}{dx}=\dfrac{1}{~\frac{1+sin\,x}{cos\,x}~}\cdot \dfrac{1+sin\,x}{(cos\,x)^2}}\\\\\\ \mathsf{\dfrac{dy}{dx}=\dfrac{cos\,x}{1+sin\,x}\cdot \dfrac{1+sin\,x}{(cos\,x)^2}}$

Simplifying that product, you get

$\mathsf{\dfrac{dy}{dx}=\dfrac{1}{1+sin\,x}\cdot \dfrac{1+sin\,x}{cos\,x}}\\\\\\ \mathsf{\dfrac{dy}{dx}=\dfrac{1}{cos\,x}}$

∴ $\boxed{\begin{array}{c}\mathsf{\dfrac{dy}{dx}=sec\,x} \end{array}}\quad\longleftarrow\quad\textsf{this is the answer.}$

I hope this helps. =)

Tags: <em>derivative composite function logarithmic logarithm log trigonometric trig secant tangent sec tan chain rule quotient rule differential integral calculus</em>

3 0

3 years ago

Determine the value of x in the diagram.

Igoryamba

The answer to the question is 114

8 0

3 years ago

Write the inequality shown by the graph with the boundary line

Bingel [31]

Answer:

Step-by-step explanation:

since the shaded area is above the line y must be > or >= 1/4x-4, so the symbol must be > or >=. since the line isn't dashed, points on the line also fit the inequality meaning that the answer is >=

3 0

2 years ago

Solve for x: A=bx+cw

ozzi

Answer:

The answer is x=A-cw/b.

I need to add more characters so don't worry

8 0

2 years ago

Read 2 more answers