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goblinko [34]
1 year ago
14

Really struggling on this

Mathematics
1 answer:
navik [9.2K]1 year ago
5 0
Angle 4 would be 77degrees because angle two is vertical to angle four.

Angle 3 and Angle 1 are equal because they are vertical to each other.

You would subtract 180 (degrees) minus 77 (degrees) and get 103 (degrees).

So Angle 3 and Angle 1 would both be 103 degrees.

You would get 180 degrees from the line.

You might be interested in
What is the solution set?<br><br> −4.9x+1.3&gt;11.1<br><br> Enter your answer in the box.
Stells [14]
Begin by adding 4.9x to both sides.
1.3> 11.1 + 4.9x This way you never have to remember to turn the > into < when dividing by a minus.

Subtract 11.1 from both sides
-9.8 < 4.9x Divide by 4.9
-9.8 / 4.9 < x
-2 < x

8 0
3 years ago
(d). Use an appropriate technique to find the derivative of the following functions:
natima [27]

(i) I would first suggest writing this function as a product of the functions,

\displaystyle y = fgh = (4+3x^2)^{1/2} (x^2+1)^{-1/3} \pi^x

then apply the product rule. Hopefully it's clear which function each of f, g, and h refer to.

We then have, using the power and chain rules,

\displaystyle \frac{df}{dx} = \frac12 (4+3x^2)^{-1/2} \cdot 6x = \frac{3x}{(4+3x^2)^{1/2}}

\displaystyle \frac{dg}{dx} = -\frac13 (x^2+1)^{-4/3} \cdot 2x = -\frac{2x}{3(x^2+1)^{4/3}}

For the third function, we first rewrite in terms of the logarithmic and the exponential functions,

h = \pi^x = e^{\ln(\pi^x)} = e^{\ln(\pi)x}

Then by the chain rule,

\displaystyle \frac{dh}{dx} = e^{\ln(\pi)x} \cdot \ln(\pi) = \ln(\pi) \pi^x

By the product rule, we have

\displaystyle \frac{dy}{dx} = \frac{df}{dx}gh + f\frac{dg}{dx}h + fg\frac{dh}{dx}

\displaystyle \frac{dy}{dx} = \frac{3x}{(4+3x^2)^{1/2}} (x^2+1)^{-1/3} \pi^x - (4+3x^2)^{1/2} \frac{2x}{3(x^2+1)^{4/3}} \pi^x + (4+3x^2)^{1/2} (x^2+1)^{-1/3} \ln(\pi) \pi^x

\displaystyle \frac{dy}{dx} = \frac{3x}{(4+3x^2)^{1/2}} \frac{1}{(x^2+1)^{1/3}} \pi^x - (4+3x^2)^{1/2} \frac{2x}{3(x^2+1)^{4/3}} \pi^x + (4+3x^2)^{1/2} \frac{1}{ (x^2+1)^{1/3}} \ln(\pi) \pi^x

\displaystyle \frac{dy}{dx} = \boxed{\frac{\pi^x}{(4+3x^2)^{1/2} (x^2+1)^{1/3}} \left( 3x - \frac{2x(4+3x^2)}{3(x^2+1)} + (4+3x^2)\ln(\pi)\right)}

You could simplify this further if you like.

In Mathematica, you can confirm this by running

D[(4+3x^2)^(1/2) (x^2+1)^(-1/3) Pi^x, x]

The immediate result likely won't match up with what we found earlier, so you could try getting a result that more closely resembles it by following up with Simplify or FullSimplify, as in

FullSimplify[%]

(% refers to the last output)

If it still doesn't match, you can try running

Reduce[<our result> == %, {}]

and if our answer is indeed correct, this will return True. (I don't have access to M at the moment, so I can't check for myself.)

(ii) Given

\displaystyle \frac{xy^3}{1+\sec(y)} = e^{xy}

differentiating both sides with respect to x by the quotient and chain rules, taking y = y(x), gives

\displaystyle \frac{(1+\sec(y))\left(y^3+3xy^2 \frac{dy}{dx}\right) - xy^3\sec(y)\tan(y) \frac{dy}{dx}}{(1+\sec(y))^2} = e^{xy} \left(y + x\frac{dy}{dx}\right)

\displaystyle \frac{y^3(1+\sec(y)) + 3xy^2(1+\sec(y)) \frac{dy}{dx} - xy^3\sec(y)\tan(y) \frac{dy}{dx}}{(1+\sec(y))^2} = ye^{xy} + xe^{xy}\frac{dy}{dx}

\displaystyle \frac{y^3}{1+\sec(y)} + \frac{3xy^2}{1+\sec(y)} \frac{dy}{dx} - \frac{xy^3\sec(y)\tan(y)}{(1+\sec(y))^2} \frac{dy}{dx} = ye^{xy} + xe^{xy}\frac{dy}{dx}

\displaystyle \left(\frac{3xy^2}{1+\sec(y)} - \frac{xy^3\sec(y)\tan(y)}{(1+\sec(y))^2} - xe^{xy}\right) \frac{dy}{dx}= ye^{xy} - \frac{y^3}{1+\sec(y)}

\displaystyle \frac{dy}{dx}= \frac{ye^{xy} - \frac{y^3}{1+\sec(y)}}{\frac{3xy^2}{1+\sec(y)} - \frac{xy^3\sec(y)\tan(y)}{(1+\sec(y))^2} - xe^{xy}}

which could be simplified further if you wish.

In M, off the top of my head I would suggest verifying this solution by

Solve[D[x*y[x]^3/(1 + Sec[y[x]]) == E^(x*y[x]), x], y'[x]]

but I'm not entirely sure that will work. If you're using version 12 or older (you can check by running $Version), you can use a ResourceFunction,

ResourceFunction["ImplicitD"][<our equation>, x]

but I'm not completely confident that I have the right syntax, so you might want to consult the documentation.

3 0
2 years ago
Please help apex ansers
goblinko [34]
The factorization of A is y = (x - 8)(x + 7).
The factorization of B is y = (x + 1)(x - 4)(x - 5)

In order to find these, you must first find where each graph crosses the x-axis. In the first problem it does so at 8 and -7. In order to find the correct parenthesis for those, you need to write it out as a statement and then solve for 0. 

x = 8 ---> subtract 8 from both sides
x - 8 = 0
This means we use (x -  8) in our factorization. 

You then need to repeat the process until you have all the pieces. In the second problem, there will be 3 instead of 2 since it crosses the axis 3 times. 
6 0
2 years ago
-3 (x + 10) <br><br> HELP<br> PLEASE
galben [10]

Answer:

−3X−30

Step-by-step explanation:

5 0
2 years ago
Jimmy applied the distributive property to write the equation below.<br><br> 24 + 6 = 6 (4 + 2)
joja [24]

Answer: if you simplify this it becomes 30=28 but this is a false statement so you would say 30≠28

Step-by-step explanation:

≠ means not equal

7 0
2 years ago
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