All categories

2 years ago

Helpppp me please I forgot how to do this

Mathematics

2 answers:

galben [10]2 years ago

5 0

Answer:

-11

Step-by-step explanation:

plz mark as brainliest.

adell [148]2 years ago

4 0

Answer:

-11

Step-by-step explanation:

the absolute value of 33 is 33, and the absolute value of 3 is 3, but the negative sign is outside of the absolute value bars, so the answer is negative. the simpler equation shouldlook like this

33÷-3=___

and 33÷-3= -11

You might be interested in

-0.75p-2=0.25 Help me please

Art [367]

First you change the number that is not with "p" and it changes with the opposite sign

-0.75p = 0.25 + 2

-0.75p = 2.25

then as -0.75 and p are multiplying, you pass -0.75 as dividing

p = 2.25/-0.75

p = 3

6 0

2 years ago

Read 2 more answers

Josiah invests $360 into an account that accrues 3% interest annually. Assuming no deposits or withdrawals are made, which equat

Natalka [10]

y = 360(1.03)x is the answer

6 0

2 years ago

Read 2 more answers

which of the following could result in higher living costs for the homeowner that the renter doesnt have?

maw [93]

Well, Homeowners have to pay property taxes, maintnece fees, plus all utilities. Renters have to pay utilities and sometimes maintenance fees if the property does not have a property manager. Because a rental agent finds a tenant for a fee, a property manager takes care and collects rent for the absentee owner.

8 0

3 years ago

Read 2 more answers

Describe the graph the inequality on a number line: -7 ≥ g - 21 I need help on this its due today and ill give large sum of poin

Gekata [30.6K]

Answer:

Step-by-step explanation:

5 0

2 years ago

Find the second derivative of 2x^3-3y^2=8

Umnica [9.8K]

Answer:

$\frac{d^2y}{dx^2}=\frac{x(2y-x^3)}{y^3}$

Step-by-step explanation:

Find the first implicit derivative using implicit differentiation

$2x^3-3y^2=8\\\\6x^2-6y\frac{dy}{dx}=0\\ \\-6y\frac{dy}{dx}=-6x^2\\ \\\frac{dy}{dx}=\frac{x^2}{y}$

Use the substitution of dy/dx to find the second derivative (d²y/dx²)

$\frac{d^2y}{dx^2}=\frac{(y)(2x)-(x^2)(\frac{dy}{dx})}{y^2}\\ \\\frac{d^2y}{dx^2}=\frac{2xy-(x^2)(\frac{x^2}{y})}{y^2}\\\\\frac{d^2y}{dx^2}=\frac{2xy-\frac{x^4}{y}}{y^2}\\\\\frac{d^2y}{dx^2}=\frac{x(2y-x^3)}{y^3}$

5 0

2 years ago