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3 years ago

3. The point where two rays meet to form an angle is called the

Mathematics

1 answer:

Katyanochek1 [597]3 years ago

6 0

The answer is vertex

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Help plz i really appreciate it

yuradex [85]

Answer:

Variant A

Step-by-step explanation:

1)y= - 3x+1

2)y+5= - 3(x-2)

y+5= - 3x+6

y= -3x+1

They are equal

6 0

3 years ago

Find thd <img src="https://tex.z-dn.net/?f=%5Cfrac%7Bdy%7D%7Bdx%7D" id="TexFormula1" title="\frac{dy}{dx}" alt="\frac{dy}{dx}" a

NARA [144]

$x^3y^2+\sin(x\ln y)+e^{xy}=0$

Differentiate both sides, treating $y$ as a function of $x$ . Let's take it one term at a time.

Power, product and chain rules:

$\dfrac{\mathrm d(x^3y^2)}{\mathrm dx}=\dfrac{\mathrm d(x^3)}{\mathrm dx}y^2+x^3\dfrac{\mathrm d(y^2)}{\mathrm dx}$

$=3x^2y^2+x^3(2y)\dfrac{\mathrm dy}{\mathrm dx}$

$=3x^2y^2+6x^3y\dfrac{\mathrm dy}{\mathrm dx}$

Product and chain rules:

$\dfrac{\mathrm d(\sin(x\ln y)}{\mathrm dx}=\cos(x\ln y)\dfrac{\mathrm d(x\ln y)}{\mathrm dx}$

$=\cos(x\ln y)\left(\dfrac{\mathrm d(x)}{\mathrm dx}\ln y+x\dfrac{\mathrm d(\ln y)}{\mathrm dx}\right)$

$=\cos(x\ln y)\left(\ln y+\dfrac1y\dfrac{\mathrm dy}{\mathrm dx}\right)$

$=\cos(x\ln y)\ln y+\dfrac{\cos(x\ln y)}y\dfrac{\mathrm dy}{\mathrm dx}$

Product and chain rules:

$\dfrac{\mathrm d(e^{xy})}{\mathrm dx}=e^{xy}\dfrac{\mathrm d(xy)}{\mathrm dx}$

$=e^{xy}\left(\dfrac{\mathrm d(x)}{\mathrm dx}y+x\dfrac{\mathrm d(y)}{\mathrm dx}\right)$

$=e^{xy}\left(y+x\dfrac{\mathrm dy}{\mathrm dx}\right)$

$=ye^{xy}+xe^{xy}\dfrac{\mathrm dy}{\mathrm dx}$

The derivative of 0 is, of course, 0. So we have, upon differentiating everything,

$3x^2y^2+6x^3y\dfrac{\mathrm dy}{\mathrm dx}+\cos(x\ln y)\ln y+\dfrac{\cos(x\ln y)}y\dfrac{\mathrm dy}{\mathrm dx}+ye^{xy}+xe^{xy}\dfrac{\mathrm dy}{\mathrm dx}=0$

Isolate the derivative, and solve for it:

$\left(6x^3y+\dfrac{\cos(x\ln y)}y+xe^{xy}\right)\dfrac{\mathrm dy}{\mathrm dx}=-\left(3x^2y^2+\cos(x\ln y)\ln y-ye^{xy}\right)$

$\dfrac{\mathrm dy}{\mathrm dx}=-\dfrac{3x^2y^2+\cos(x\ln y)\ln y-ye^{xy}}{6x^3y+\frac{\cos(x\ln y)}y+xe^{xy}}$

(See comment below; all the 6s should be 2s)

We can simplify this a bit by multiplying the numerator and denominator by $y$ to get rid of that fraction in the denominator.

$\dfrac{\mathrm dy}{\mathrm dx}=-\dfrac{3x^2y^3+y\cos(x\ln y)\ln y-y^2e^{xy}}{6x^3y^2+\cos(x\ln y)+xye^{xy}}$

3 0

3 years ago

Compound Inequalities <br> (Help)

sladkih [1.3K]

- 8x + 44 ≥ 60
- 8x ≥ 60 - 44
- 8x ≥ 16
x ≤ - 2

- 4x + 50 < 58
- 4x < 58 - 50
- 4x < 8
x > - 2

Now put the two together. Logically, these two answers since a number cannot be both less than AND greater than - 2.

Your answer should be no solution.

5 0

3 years ago

Megan has $15,000 to invest. She is considering two investment options. Option A pays 3.2% simple interest. Option B pays 4.1% i

True [87]

wheee

Compute each option

option A: simple interest

simple interest is easy

A=I+P

A=Final amount

I=interest

P=principal (amount initially put in)

and I=PRT

P=principal

R=rate in decimal

T=time in years

so given

P=15000

R=3.2% or 0.032 in deecimal form

T=10

A=I+P

A=PRT+P

A=(15000)(0.032)(10)+15000

A=4800+15000

A=19800

Simple interst pays $19,800 in 10 years

Option B: compound interest

for interest compounded yearly, the formula is

$A=P(1+r)^t$

where A=final amount

P=principal

r=rate in decimal form

t=time in years

given

P=15000

r=4.1% or 0.041

t=10

$A=15000(1+0.041)^{10}$

$A=15000(1.041)^{10}$

use your calculator

A=22418.0872024

so after 10 years, she will have $22,418.09 in the compounded interest account

in 10 years, the investment in the simple interest account will be worth $19,800 and the investment in the compounded interest account will be worth$22,418.09

4 0

3 years ago

Read 2 more answers

Rectangle ABCD has vertex coordinates A(1, -2), B(4, -2), C(4,-4), and D(1,

hram777 [196]

The coordinates of C is, (3,-1)
A is the. Answer

4 0

3 years ago