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3 years ago

How do I solve the inequality for x in this problem: 7 ≤ -9 - 4x

Mathematics

2 answers:

katovenus [111]3 years ago

7 0

Answer:

-4 less than or equal to x

Step-by-step explanation:

Sedaia [141]3 years ago

5 0

Answer:

same thing as the above answer

Step-by-step explanation:

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What is the correct symbol?

Nadusha1986 [10]

Answer:

Since 10/9 is greater than 1, multiplying by 10/9 makes the value larger

Step-by-step explanation:

Step 1: Solve the fraction

10/9 = 1.1112

Therefore 10/9 > 1

Step 2: Multiple the fraction by itself

10/9 x 10/9 = 100/81

Convert fraction to decimals

100/81 = 1.2345678901.....

1.234567901 > 1.1112

Therefore 10/9 x 10/9 is bigger than 10/9

3 0

3 years ago

How do you write the algebraic notation when the scale factor is 3

Alja [10]

Answer:

(x,y) = (3x,3y)

Step-by-step explanation:

3 0

2 years ago

please i have been stuck on thos math problem for 123 minutes ago please help please I'm stressing out please I urgently need he

alisha [4.7K]

Answer:

it could be 1. c 2. a 3. d 4. b

or it could be the answers i did for last time in the pic: (that is my old account that got banned)

but i think the answers that are right are the ones i did rn of 1. c 2. a 3. d 4. b

8 0

3 years ago

Select all of the potential solution(s) of the equation 2log5x = log54.

tankabanditka [31]

Answer:

-2 and 2

Step-by-step explanation: just did the problem

4 0

3 years ago

Show that ( 2xy4 + 1/ (x + y2) ) dx + ( 4x2 y3 + 2y/ (x + y2) ) dy = 0 is exact, and find the solution. Find c if y(1) = 2.

fredd [130]

$\dfrac{\partial\left(2xy^4+\frac1{x+y^2}\right)}{\partial y}=8xy^3-\dfrac{2y}{(x+y^2)^2}$

$\dfrac{\partial\left(4x^2y^3+\frac{2y}{x+y^2}\right)}{\partial x}=8xy^3-\dfrac{2y}{(x+y^2)^2}$

so the ODE is indeed exact and there is a solution of the form $F(x,y)=C$ . We have

$\dfrac{\partial F}{\partial x}=2xy^4+\dfrac1{x+y^2}\implies F(x,y)=x^2y^4+\ln(x+y^2)+f(y)$

$\dfrac{\partial F}{\partial y}=4x^2y^3+\dfrac{2y}{x+y^2}=4x^2y^3+\dfrac{2y}{x+y^2}+f'(y)$

$f'(y)=0\implies f(y)=C$

$\implies F(x,y)=x^2y^3+\ln(x+y^2)=C$

With $y(1)=2$ , we have

$8+\ln9=C$

$\boxed{x^2y^3+\ln(x+y^2)=8+\ln9}$

8 0

2 years ago