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2 years ago

Please answer, serious answers only !

Mathematics

2 answers:

nadya68 [22]2 years ago

6 0

Answer:

Step-by-step explanation:

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djyliett [7]2 years ago

3 0

Answer:

Step-by-step explanation:

Those 2 angles (the one you are given and the one you want to find) are vertically opposite.

Vertically opposite angles are equal.

Therefore the answer is 64 degrees.

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What are the first five multiples of seven

Mkey [24]

7,14,21,28,35
Multiply 7 by 1-5

7 0

3 years ago

Read 2 more answers

Roselyn is driving to visit her family, which live 150 150150 kilometers away. Her average speed is 60 6060 kilometers per hour.

Sauron [17]

Answer:

She can go 120 km before she runs out of fuel

It will take 2 hours.

Step-by-step explanation:

150 km is the distance

60 km/ h is the speed

The gas tank is 20 liters

We can go 6 km per liter

Fuel costs .60 dollars per liter

We need to determine how far she can go on a tank of gas

20 liters * 6 km / liter = 120 km

She can go 120 km before she runs out of fuel

120 km = 60 km/ h * x hours

Divide each side by 60

120/60 = x

2 hours

8 0

2 years ago

8. Solve x/10 = –7. A. x = –0.7 B. x= –17 C. x = –70 D. x = 3

ANEK [815]

For this problem, I believe you would just have to plug in for x.

For example, -17 divided by 10 and so on.

The answer should be C, -70. -70 divided by 10 equals -7.

4 0

3 years ago

How do you solve this

olchik [2.2K]

The answer would have to be 4x^2y

4 0

3 years ago

Differentiating a Logarithmic Function in Exercise, find the derivative of the function. See Examples 1, 2, 3, and 4.

Leni [432]

Answer: $\dfrac{2x^2-1}{x(x^2-1)}$

Step-by-step explanation:

The given function : $y=\ln(x(x^2 - 1)^{\frac{1}{2}})$

$\Rightarrow\ y=\ln x+\ln (x^2-1)^{\frac{1}{2}}$ [ $\because \ln(ab)=\ln a +\ln b$ ]

$\Rightarrow y=\ln x+\dfrac{1}{2}\ln (x^2-1)}$ [ $\because \ln(a)^n=n\ln a$ ]

Now , Differentiate both sides with respect to x , we will get

$\dfrac{dy}{dx}=\dfrac{1}{x}+\dfrac{1}{2}(\dfrac{1}{x^2-1})\dfrac{d}{dx}(x^2-1)$ (By Chain rule)

[Note : $\dfrac{d}{dx}(\ln x)=\dfrac{1}{x}$ ]

$\dfrac{1}{x}+\dfrac{1}{2}(\dfrac{1}{x^2-1})(2x-0)$

[ $\because \dfrac{d}{dx}(x^n)=nx^{n-1}$ ]

$=\dfrac{1}{x}+\dfrac{1}{2}(\dfrac{1}{x^2-1})(2x) = \dfrac{1}{x}+\dfrac{x}{x^2-1}\\\\\\=\dfrac{(x^2-1)+(x^2)}{x(x^2-1)}\\\\\\=\dfrac{2x^2-1}{x(x^2-1)}$

Hence, the derivative of the given function is $\dfrac{2x^2-1}{x(x^2-1)}$ .

8 0

3 years ago