Please help!!!

Which situation best fits the following inequality? 40 ≥ ≥ 6.5x - 14

zepelin [54]

40+40+40

6.80÷2

and if you do this 

you will end up 

getting your answers 

6 0

2 years ago

Lamar used 7/8 of a cup of raisins to make snacks mix. he used 3/4 of a cup of raisins to make cookies what expression could be

Alinara [238K]

The answer is B.

In order to add the two values together, you must find the least common multiple for the denominators. The denominators are 8 and 4, and the least common multiple of 8 and 4 is 8. Because you have to multiply 4 by 2 to get 8, you must also multiply 3 by 2 to get 6. This will give you 6/8.

6 0

2 years ago

A car moving at a constant rate travel 96 miles in 2 hours at the car maintains this rate how many miles will the car travel in

salantis [7]

Answer:

speed = distance / time

At 96miles in 2 hours

the speed = 96miles/ 2hours = 48miles per hour

at 5 hours, Distance = speed x time

distance = 48 x 5 = 240miles

Step-by-step explanation:

step 1 : determine the speed per hour

speed = distance / time

step 2 : after getting the speed of the car per hour, the distance cover in 5hours can now be determine using

distance = speed x time

8 0

2 years ago

Read 2 more answers

I made 2 calls and was on the phone for a total of 46 minutes. One call was 10 mins. longer than the other. How long was each c

wlad13 [49]

The one call that was ten min more is 56 min

7 0

2 years ago

Read 2 more answers

Find the average value of f(x+y)=sin(x+y) over

Troyanec [42]

For each given region $D$ , the average value of $f$ over $D$ is the integral of $f$ over $D$ divided by the area of $D$ . In both cases, $D$ is a rectangle so the area is trivial.

a. The area of $D$ is $\dfrac\pi3\cdot\dfrac{5\pi}3=\dfrac{5\pi^2}9$ . The integral of $f$ over $D$ is

$\displaystyle\int_0^{5\pi/3}\int_0^{\pi/3}\sin(x+y)\,\mathrm dx\,\mathrm dy$

$\displaystyle=\int_0^{5\pi/3}\left(\cos y-\cos\left(y+\dfrac\pi3\right)\right)\,\mathrm dy$

$=\sin\dfrac{5\pi}3-\sin\left(\dfrac{5\pi}3+\dfrac\pi3\right) - \sin0+\sin\dfrac\pi3=0$

So the average value of $f$ over this region is 0.

b. The area of $D$ is $\dfrac{2\pi}3\cdot\dfrac{7\pi}6=\dfrac{7\pi^2}9$ . The integral of $f$ is

$\displaystyle\int_0^{7\pi/6}\int_0^{2\pi/3}\sin(x+y)\,\mathrm dx\,\mathrm dy$

$=\displaystyle\int_0^{7\pi/6}\left(\cos y-\cos\left(y+\frac{2\pi}3\right)\right)\,\mathrm dy$

$=\sin\dfrac{7\pi}6-\sin\left(\dfrac{7\pi}6+\dfrac{2\pi}3\right) - \sin0+\sin\dfrac{2\pi}3=\dfrac{\sqrt3}2$

The average vale of $f$ over this region is then $\dfrac{\frac{\sqrt3}2}{\frac{7\pi^2}9}=\dfrac{9\sqrt3}{14\pi^2}$ .

4 0

2 years ago