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3 years ago

list all the integer x that satisfy both the simultaneous linear inequalities 5-x<3 and x÷2 +3<5

Mathematics

1 answer:

Evgen [1.6K]3 years ago

6 0

For the first one any number less than 2 works
for the second on any number less than 4 works

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What is y + 20 < 180? please help me

taurus [48]

Answer:

y < 160

Step-by-step explanation:

y + 20 < 180

-20 < -20

y < 160

y is less than 160

5 0

3 years ago

Read 2 more answers

(5x to the power of 2 -6x+2)+9x to the power of 2 +10x-7)

xxTIMURxx [149]

Answer:

$45x^{4} -4x^{3} -77x^{2} +62x-14$

Step-by-step explanation:

Im assuming the question was $(5x^{2} -6x+2)(9x^{2} +10x-7)$ if thats not the case then im wrong

6 0

3 years ago

0.4 x 10 =....<br>1.25 x 10 = <br>97.465 x10 =

chubhunter [2.5K]

Answer:

0.4 × 10 = 4

1.25 × 10 = 12.5

97.465 × 10 = 974.65

3 0

3 years ago

A large tank is filled to capacity with 300 gallons of pure water. brine containing 5 pounds of salt per gallon is pumped into t

Licemer1 [7]

If $A(t)$ is the amount of salt in the tank at time $t$ , then the rate at which this amount changes over time is given by the ODE

$A'(t)=\dfrac{3\text{ gal}}{1\text{ min}}\cdot\dfrac{5\text{ lb}}{1\text{ gal}}-\dfrac{3\text{ gal}}{1\text{ min}}\cdot\dfrac{A(t)\text{ lb}}{300+(3-3)t\text{ gal}}$

$A'(t)+\dfrac1{100}A(t)=15$

We're told that the tank initially starts with no salt in the water, so $A(0)=0$ .

Multiply both sides by an integrating factor, $e^{t/100}$ :

$e^{t/100}A'(t)+\dfrac1{100}e^{t/100}A(t)=15e^{t/100}$
$\left(e^{t/100}A(t)\right)'=15e^{t/100}$
$e^{t/100}A(t)=1500e^{t/100}+C$
$A(t)=1500+Ce^{-t/100}$

Since $A(0)=0$ , we have

$0=1500+C\implies C=-1500$

so that the amount of salt in the tank over time is given by

$A(t)=1500(1-e^{-t/100})$

After 10 minutes, the amount of salt in the tank is

$A(10)=1500(1-e^{-1/10})\approx142.74\text{ lb}$

8 0

4 years ago

6j-k=23<br>3k+6j=11<br>Solve j and k using the elimination method on the simultaneous equations

frez [133]

Answer:

Step-by-step explanation:

6j - k = 23

6j + 3k = 11.....multiply by -1

---------------

6j - k = 23

-6j - 3k = -11 (result of multiplying by -1)

---------------

-4k = 12

k = -12/4

k = - 3 <=====

6j - k = 23

6j - (-3) = 23

6j + 3 = 23

6j = 23 - 3

6j = 20

j = 20/6

j = 10/3 <===

solution is : j = 10/3 and k = -3

5 0

3 years ago