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VLD [36.1K]
3 years ago
9

M + 7 ≥ 20, if m = 11

Mathematics
1 answer:
umka21 [38]3 years ago
3 0
Ok so what you have to do is substitute m to 11 so do this 11+7 greater or equal to 20 and that’s not right because 11+7=18 therefore this is incorrect
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What's the error? Jerry says that a cube with edges that measure 10 cm has a volume that is twice as much as a cube with sides t
disa [49]
The quick way to dispute something like this is to simply do the calculation and then create a ratio. 

Cube One (Large Cube)
The formula for a cube is V = e^3 
e = the measurement of an edge. In this case.
e = 10 cm
V = e^3
V = 10^3 = 10*10*10
V = 1000 cm^3

Cube 2 (Small Cube)
V = e^3
e = 5 cm
V = 5*5*5
V = 125 cm^3

Ratio
Large Cube / Small Cube = 1000 / 125 = 8/1.

The difference in size is 8 to 1 not 2 to 1.

Explanation
He's right if he sticks to one side. The ratio of one side of the large cube to the small one is 2 to 1. But once you put that into the formula for volume, three sides are multiplied together and that 2 shows up everytime you multiply the sides together. 

4 0
4 years ago
A Bike shop charges by the hour to rent a bike. Related items are rented for flat fees. Write an expression that represents how
IgorLugansk [536]

Answer:

y= b+ 3a

Step-by-step explanation:

Given data

We are required to rent a bike and a helmet

le the flat fee for bike rental be a

and the flat fee helmet be b

let the number of hours be h

and let the total cost be y

So

y= b+ah

for h = 3

tThen the expression for the total is given as

y= b+ 3a

3 0
3 years ago
Read 2 more answers
according to the graph, what is the approximate average rate of change in the radius of the circle as the area increases from 3
slamgirl [31]

Answer: the answer is 0.125 foot per square foot

Step-by-step explanation:

7 0
3 years ago
Susan earned $8.25 per hour babysitting last summer. She worked 15 hours a week. How much did she earn in 6 weeks?
irina1246 [14]
15*6=90 hours
90 *8.25=742.50
Your answer would be $742.50. Hope it helps!
4 0
3 years ago
Let R be the region in the first quadrant of the​ xy-plane bounded by the hyperbolas xyequals​1, xyequals9​, and the lines yequa
Tema [17]

Answer:

The area can be written as

\int\limits_1^2 \int\limits_1^3 u(\frac{1}{v} - v \, ln(v)) \, du \, dv = 0.2274

And the value of it is approximately 1.8117

Step-by-step explanation:

x = u/v

y = uv

Lets analyze the lines bordering R replacing x and y by their respective expressions with u and v.

  • x*y = u/v * uv = u², therefore, x*y = 1 when u² = 1. Also x*y = 9 if and only if u² = 9
  • x=y only if u/v = uv, And that only holds if u = 0 or 1/v = v, and 1/v = v if and only if v² = 1. Similarly y = 4x if and only if 4u/v = uv if and only if v² = 4

Therefore, u² should range between 1 and 9 and v² ranges between 1 and 4. This means that u is between 1 and 3 and v is between 1 and 2 (we are not taking negative values).

Lets compute the partial derivates of x and y over u and v

x_u = 1/v

x_v = u*ln(v)

y_u = v

y_v = u

Therefore, the Jacobian matrix is

\left[\begin{array}{ccc}\frac{1}{v}&u \, ln(v)\\v&u\end{array}\right]

and its determinant is u/v - uv * ln(v) = u * (1/v - v ln(v))

In order to compute the integral, we can find primitives for u and (1/v-v ln(v)) (which can be separated in 1/v and -vln(v) ). For u it is u²/2. For 1/v it is ln(v), and for -vln(v) , we can solve it by using integration by parts:

\int -v \, ln(v) \, dv = - (\frac{v^2 \, ln(v)}{2} - \int \frac{v^2}{2v} \, dv) = \frac{v^2}{4} - \frac{v^2 \, ln(v)}{2}

Therefore,

\int\limits_1^2 \int\limits_1^3 u(\frac{1}{v} - v \, ln(v)) \, du \, dv = \int\limits_1^2 (\frac{1}{v} - v \, ln(v) ) (\frac{u^2}{2}\, |_{u=1}^{u=3}) \, dv= \\4* \int\limits_1^2 (\frac{1}{v} - v\,ln(v)) \, dv = 4*(ln(v) + \frac{v^2}{4} - \frac{v^2\,ln(v)}{2} \, |_{v=1}^{v=2}) = 0.2274

4 0
3 years ago
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