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3 years ago

14

Hope u help me pls I love u

1 answer:

QveST [7]3 years ago

8 0

Answer:c= a +b

Step-by-step explanation:

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Find the value of X<br> PLEASE HELP ASAP!!!!!

MAVERICK [17]

Answer:

-10

Step-by-step explanation:

the upper right is the same (105°), therefore the left 75°

now you have 180 = 75 + x + 115

therefore: -10

6 0

3 years ago

PLS HELPP WHAT IS THE ANSWER???

Elena-2011 [213]

Answer:

-220

Step-by-step explanation:

6 0

2 years ago

Read 2 more answers

100 POINTS!! OMG YOU NEED TO SOLVE THIS!

natita [175]

Answer:

The correct answer is: $\Delta BDE \cong \Delta BFK$ by <em>rule</em> ASA rule of congruence.

Step-by-step explanation:

First let us prove $\Delta BDE \cong \Delta BFK$ by rule ASA (rule of congruence).

<u>Congruent side:</u>

$\overline{BD} \cong \overline{BF}$ (Given)

<u>Congruent angles:</u>

1. By definition of perpendicular,

$\angle{BFK} = 90 \textdegree \ (Since \ \overline{FK} \ is \ perpendicular \ to \ \overline{AB} \ (\overline{FK} \perp \overline{AB} =Given))$

Also,

$\angle{BDE} = 90 \textdegree \ (Given)$

Therefore,

$\angle{BDE} \cong \angle{BFK}$

or you can say,

$\angle{D} \cong \angle{F}$

2. Common angle between $\Delta BDE \ and \ \Delta BFK$ is $\angle{B}$

In a nutshell, in $\Delta BDE$ ,

$\angle{B}$ (Angle)

$\overline{BD}$ (Side)

$\angle{BDE}$ (Angle)

are congruent to the following angle, side and angle of $\Delta BFK$ :

$\angle{B}$ (Angle)

$\overline{BF}$ (Side)

$\angle{BFK}$ (Angle)

Therefore, by ASA rule of congruence, we can say $\Delta BDE \cong \Delta BFK$ .

<em>Since both triangles are congruent</em>, the sides $\overline{ED}$ and $\overline{FK}$ are also congruent.

5 0

3 years ago

Read 2 more answers

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

Why do banks pay you interest?

Anastasy [175]

Step-by-step explanation:

option C . they are borrowing your money and rewarding you for keeping money with them

6 0

2 years ago