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3 years ago

A number c is less than 5 and greater than or equal to -2. Write and graph the inequality.

Mathematics

1 answer:

UkoKoshka [18]3 years ago

5 0

Answer:

5>c<_ -2

c can be 4 ,3 ,2 ,1 ,0 ,-1 ,-2

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two buses leave a station at the same time and travel in opposite directions. One bus travels 15 km/h faster than the other. If

Marina CMI [18]

The speed of slower bus is 118 km/hr and speed of faster bus is 133 km/hr

<h3>What is speed?</h3>

Speed is a scalar quantity that refers to "how fast an object is moving".

Given that, two buses leave a station at the same time and travel in opposite directions, one bus travels 15 km/h faster than the other, the two buses are 1004 kilometres apart after 4 hours,

Speed = distance/ time

Let the speed of the slower bus be v and the faster bus be (v+15)

According to the question,

vt + (v+15)t = 1004

vt + vt + 15t = 1004

t = 4

4v + 4v + 60 = 1004

8v = 944

v = 118

Hence, The speed of slower bus is 118 km/hr and speed of faster bus is 133 km/hr

For more references on speed, click;

brainly.com/question/7359669

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8 0

1 year ago

Need helps plsssssssssssssss

slavikrds [6]

Answer:

Step-by-step explanation:

4^3 is just 4*4*4

so 4*4=16

and 16*4=64

TADAAAAAAAAAA

6 0

3 years ago

Read 2 more answers

Wich is the missing number in the prime factorization will rate branleist

gogolik [260]

The answer is a. 5. Good luck

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3 years ago

At Central High School, there are 48 football players and 20 cheerleaders. What is the ratio of cheerleaders to football players

Ahat [919]

Answer:

5 : 12

Step-by-step explanation:

Cheerleaders : football players

20 : 48

5 : 12

6 0

4 years ago

Read 2 more answers

Solve the equation <img src="https://tex.z-dn.net/?f=%2835%20x%5E%7B4%7D%20y%2B14%20x%5E%7B5%7D%20y-2%20y%5E%7B3%7D-4x%20y%5E%7B

jeka94

$\underbrace{35x^4y+14x^5y-2y^3-4xy^3}_M\,\mathrm dx+\underbrace{7x^5-6xy^2}_N\,\mathrm dy=0$

$M_y=35x^4+14x^5-6y^2-12xy^2$
$N_x=35x^4-6y^2$

$\dfrac{N_x-M_y}N=\dfrac{-14x^5+12xy^2}{7x^5-6xy^2}=-2$

This suggests an integrating factor depending on $x$ only is possible, and given by

$\mu(x)=\exp\left(-\displaystyle\int\frac{N_x-M_y}N\,\mathrm dx\right)=e^{2x}$

Distributing across the ODE, we end up with

$\underbrace{(35x^4y+14x^5y-2y^3-4xy^3)e^{2x}}_{M^*}\,\mathrm dx+\underbrace{(7x^5-6xy^2)e^{2x}}_{N^*}\,\mathrm dy=0$

The equation is now exact, with

${M^*}_y={N^*}_x=(35x^4+14x^5-6y^2-12xy^2)e^{2x}$

Now we find the solution:

$F_x=M^*$
$F=\displaystyle\int(35x^4+14x^5-2y^3-4xy^3)e^{2x}\,\mathrm dx$
$F=(7x^5y-2xy^3)e^{2x}+f(y)$

$F_y=N^*$
$(7x^5-6xy^2)e^{2x}+f'(y)=(7x^5-6xy^2)e^{2x}$
$f'(y)=0$
$\implies f(y)=C$

The general solution is then

$F(x,y)=(7x^5y-2xy^3)e^{2x}=C$

7 0

4 years ago