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3 years ago

Anyone know the answer?? ( I got confused half way)

Mathematics

2 answers:

Liula [17]3 years ago

8 0

Yes the answer is eight and three forths

weqwewe [10]3 years ago

7 0

The answer is 8 and 3 fourths which would be written as 8 3/4

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Which pair of numbers has a GCF of 3. 3and 18, 12 and 18, 8 and 24, 1 and 3

Alinara [238K]

3 and 18 | 12&18 GCF = 6; 8&24 GCF = 4; 1&3 GCF = 3

4 0

3 years ago

Bài4 mình không biết làm mọi người help mee

aivan3 [116]

Sorry I can’t read that hand writing if I could just write it on brainly

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3 years ago

To rent a certain meeting room, a college charges a reservation fee of $31 and an additional fee of $7.40 per hour. The chemistr

Mamont248 [21]

T= # of rental hours
Reservation Fee= $31
Per Hour Fee= $7.40

Multiply number of hours rented by the cost per hour, add that total to the reservation fee. The total needs to be less than $97.60.

Res Fee + ($ per hr * # hrs) < $97.60

$31 + $7.40t < $97.60
subtract 31 from both sides

7.40t < 66.60
divide both sides by 7.40

t < 9

ANSWER: To spend less than $97.60, they need to rent the room less than 9 hours.

Hope this helps! :)

7 0

3 years ago

Without dividing,estimates the quotient for 3,328 divide by 6

Katarina [22]

Answer:

For this equation let's round the 3328 to a 3300.

Step-by-step explanation:

If we remove the 2 zeros, we remain with a 33. 33/6 is 5.5 Now if we add back 2 places we get 550. I know this is written badly, but I hope it helped.

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3 years ago

Read 2 more answers

In the midpoint rule for triple integrals we use a triple riemann sum to approximate a triple integral over a box b, where f(x,

Lana71 [14]

The sub-boxes will have dimensions $\frac{2-0}{2} \times \frac{2-0}{2} \times \frac{2-0}{2} =1\times1\times1=1 \ cubic \ units$

x sub-intervals are 0 to 1 and 1 to 2. Midpoints are at $x= \frac{1}{2}$ and $x= \frac{3}{4}$
y sub-intervals are 0 to 1 and 1 to 2. Midpoints are at $y= \frac{1}{2}$ and $y= \frac{3}{4}$
z sub-intervals are 0 to 1 and 1 to 2. Midpoints are at $z= \frac{1}{2}$ and $z= \frac{3}{4}$ 

Let $f(x,y,z)=\cos{(xyz)}$

$\int\limits \int\limits \int\limits {f(x,y,z)} \, dV \approx f\left( \frac{1}{2} , \frac{1}{2} , \frac{1}{2} \right)+f\left( \frac{1}{2} , \frac{1}{2} , \frac{3}{4} \right)+f\left( \frac{1}{2} , \frac{3}{4} , \frac{1}{2} \right)+f\left( \frac{1}{2} , \frac{3}{4} , \frac{3}{4} \right)$
$+f\left( \frac{3}{4} , \frac{1}{2} , \frac{1}{2} \right)+f\left( \frac{3}{4} , \frac{1}{2} , \frac{3}{4} \right)+f\left( \frac{3}{4} , \frac{3}{4} , \frac{1}{2} \right)+f\left( \frac{3}{4} , \frac{3}{4} , \frac{3}{4} \right) \\ \\ \approx\cos{ \frac{1}{8} }+\cos{ \frac{3}{16} }+\cos{ \frac{3}{16} }+\cos{ \frac{9}{32} }+\cos{ \frac{3}{16} }+\cos{ \frac{9}{32} }+\cos{ \frac{9}{32} }+\cos{ \frac{27}{64} } \\ \\ \approx0.9922+0.9825+0.9825+0.9607+0.9825+0.9607+0.9607 \\ +0.9123 \\ \\ \approx\bold{7.734}$

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3 years ago