3d + 8 = 2d - 7 equals d = -15.

First, subtract 2d from both sides. Your problem should look like: 3d + 8 - 2d = -7.
Second, simplify 3d + 8 - 2d to get d + 8. Your problem should look like: d + 8 = -7.
Third, subtract 8 from both sides. Your problem should look like: d = -7 - 8.
Fourth, simplify -7 - 8 to get -15. Your problem should look like: d = -15, which is your answer.

Hopefully this helps, I'm not exactly sure what you meant by "is equations," but this is how you solve the problem.

6 0

2 years ago

Y=arccos(1/x)<br><br> Please help me do them all! I don’t know derivatives :(

Stells [14]

Answer:

$f(x) = {sec}^{ - 1} x \\ let \: y = {sec}^{ - 1} x \rightarrow \: x = sec \: y\\ \frac{dx}{dx} = \frac{d(sec \: y)}{dx} \\ 1 = \frac{d(sec \: y)}{dx} \times \frac{dy}{dy} \\ 1 = \frac{d(sec \: y)}{dy} \times \frac{dy}{dx} \\1 = tan \: y.sec \: y. \frac{dy}{dx} \\ \frac{dy}{dx} = \frac{1}{tan \: y.sec \: y} \\ \frac{dy}{dx} = \frac{1}{ \sqrt{( {sec}^{2} \: y - 1}) .sec \: y} \\ \frac{dy}{dx} = \frac{1}{ |x | \sqrt{ {x }^{2} - 1} } \\ \therefore \frac{d( {sec}^{ - 1}x) }{dx} = \frac{1}{ |x | \sqrt{ {x }^{2} - 1} } \\ \frac{d( {sec}^{ - 1}5x) }{dx} = \frac{1}{ |5x | \sqrt{25 {x }^{2} - 1} }\\\\y=arccos(\frac{1}{x})\Rightarrow cosy=\frac{1}{x}\\x=secy\Rightarrow y=arcsecx\\\therefore \frac{d( {sec}^{ - 1}x) }{dx} = \frac{1}{ |x | \sqrt{ {x }^{2} - 1} }$

4 0

2 years ago

the question is in the picture, i need the weight of train a and b i’ll give the right answer brainlist :))

frez [133]

Answer:

a 106

b 92

Step-by-step explanation:

Hope this helps!

6 0

2 years ago

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