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2 years ago

Solve the inequality.

Mathematics

2 answers:

zysi [14]2 years ago

8 0

Answer:

?= 2

X=1

Step-by-step explanation:

3(1+1)<9

3×2<9

6<9

Pls mark brainliest.

sp2606 [1]2 years ago

6 0

Answer:

3x+3 <9

x<3

because

the question said

3x+3<9

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2x>4+10 (greater than or equal to)

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4 years ago

Help Please I’ll Give Brainliest :)

Mnenie [13.5K]

Answer:

5 tables

Step-by-step explanation:

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3 years ago

Read 2 more answers

All whole numbers are also rational numbers?

KATRIN_1 [288]

Yes all whole numbers are also rational number because they can be represented as a ratio of itself and 1, or as ratio of two other whole numbers.

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4 = 8 / 2

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3 years ago

X + y = 4<br> y = 2x + 1<br> solve using the substitution method

quester [9]

X+(2x+1)=4
3x+1=4
x=1
1+y=4
y=3
(1,3)

3 0

3 years ago

Find the solution of the initial value problem<br><br> dy/dx=(-2x+y)^2-7 ,y(0)=0

Leokris [45]

Substitute $v(x)=-2x+y(x)$ , so that $\dfrac{\mathrm dv}{\mathrm dx}=-2+\dfrac{\mathrm dy}{\mathrm dx}$ . Then the ODE is equivalent to

$\dfrac{\mathrm dv}{\mathrm dx}+2=v^2-7$

which is separable as

$\dfrac{\mathrm dv}{v^2-9}=\mathrm dx$

Split the left side into partial fractions,

$\dfrac1{v^2-9}=\dfrac16\left(\dfrac1{v-3}-\dfrac1{v+3}\right)$

so that integrating both sides is trivial and we get

$\dfrac{\ln|v-3|-\ln|v+3|}6=x+C$

$\ln\left|\dfrac{v-3}{v+3}\right|=6x+C$

$\dfrac{v-3}{v+3}=Ce^{6x}$

$\dfrac{v+3-6}{v+3}=1-\dfrac6{v+3}=Ce^{6x}$

$\dfrac6{v+3}=1-Ce^{6x}$

$v=\dfrac6{1-Ce^{6x}}-3$

$-2x+y=\dfrac6{1-Ce^{6x}}-3$

$y=2x+\dfrac6{1-Ce^{6x}}-3$

Given the initial condition $y(0)=0$ , we find

$0=\dfrac6{1-C}-3\implies C=-1$

so that the ODE has the particular solution,

$\boxed{y=2x+\dfrac6{1+e^{6x}}-3}$

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3 years ago