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3 years ago

The two lines below interest as shown. What is the value of x ?

Mathematics

1 answer:

FinnZ [79.3K]3 years ago

3 0

These two angles are vertical angles. Vertical angles are congruent, so you have to make an equation.

2x + 29 = 9x - 13

7x = 42

x = 6 degrees

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Maslowich

Is this a glitch or I don’t see a question? I only see dots “……”

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2 years ago

What is the geometric mean between 7 and 12

Tresset [83]

Answer:

Geometric Mean = 9.17

Step-by-step explanation:

Here,

a = 7

b = 12

Geometric mean = root(ab)

= root( 7 x 12)

= root(84)

= 9.17

6 0

2 years ago

Read 2 more answers

The types of Geometry that mathematicians study are:

Fiesta28 [93]

Option B i.e. Non-Euclidean

8 0

3 years ago

PLEASE HELP ASAP!!!!

sweet-ann [11.9K]

Answer:

1 to 5

211 to 1500

11 to 300

Step-by-step explanation:

1.8 to 9 = 18/90 = 1(18)/5(18) = 1/5 = 1 to 5

2.11 to 15 = 211/1500 = 211 to 1500

3.30 to 90 = 330/9000 = 110/3000 = 11/300 = 11 to 300

4 0

3 years ago

Dy/dx if y = Ln (2x3 + 3x).

NeTakaya

Answer:

$\frac{6x^2+3}{2x^3+3x}$

Step-by-step explanation:

You need to apply the chain rule here.

There are few other requirements:

You will need to know how to differentiate $\ln(u)$ .

You will need to know how to differentiate polynomials as well.

So here are some rules we will be applying:

Assume $u=u(x) \text{ and } v=v(x)$

$\frac{d}{dx}\ln(u)=\frac{1}{u} \cdot \frac{du}{dx}$

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

$\text{ constant multiply rule } \frac{d}{dx}c\cdot u=c \cdot \frac{du}{dx}$

$\text{ sum/difference rule } \frac{d}{dx}(u \pm v)=\frac{du}{dx} \pm \frac{dv}{dx}$

Those appear to be really all we need.

Let's do it:

$\frac{d}{dx}\ln(2x^3+3x)=\frac{1}{2x^3+3x} \cdot \frac{d}{dx}(2x^3+3x)$

$\frac{d}{dx}(\ln(2x^3+3x)=\frac{1}{2x^3+3x} \cdot (\frac{d}{dx}(2x^3)+\frac{d}{dx}(3x))$

$\frac{d}{dx}(\ln(2x^3+3x)=\frac{1}{2x^3+3x} \cdot (2 \cdot \frac{dx^3}{dx}+3 \cdot \frac{dx}{dx})$

$\frac{d}{dx}(\ln(2x^3+3x)=\frac{1}{2x^3+3x} \cdot (2 \cdot 3x^2+3(1))$

$\frac{d}{dx}(\ln(2x^3+3x)=\frac{1}{2x^3+3x} \cdot (6x^2+3)$

$\frac{d}{dx}(\ln(2x^3+3x)=\frac{6x^2+3}{2x^3+3x}$

I tried to be very clear of how I used the rules I mentioned but all you have to do for derivative of natural log is derivative of inside over the inside.

Your answer is $\frac{dy}{dx}=\frac{(2x^3+3x)'}{2x^3+3x}=\frac{6x^2+3}{2x^3+3x}$ .

3 0

3 years ago