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3 years ago

Urgent Please help!!!!!!!

Mathematics

2 answers:

MakcuM [25]3 years ago

5 0

Answer:

d) 5 lollipops each and 3 left over

Step-by-step explanation:

To find out how many each of them get, you need to find the total amount of lollipops to be shared equally.

Total No. of lollipops = 3 + 5 + 7 + 4 + 9

= 28

No. of lollipops each of them get = 28 ÷ 5

= 5 R3

(To find the remainder of something, you have to know how much is the quotient (which is 5 in this case), multiply it by the divisor (which is also 5 in this case) then minus that number from the total (which is 28 in this case) )

Additional working:

5 × 5 = 25

28 - 25 = 3 (remainder)

olya-2409 [2.1K]3 years ago

4 0

Answer:

it is D

Step-by-step explanation:

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Use the given transformation to evaluate the integral. (15x + 15y) dA R , where R is the parallelogram with vertices (−1, 4), (1

MA_775_DIABLO [31]

Answer:

$\int_R 15x+15y dA = \frac{8}{16875}$

Step-by-step explanation:

Recall the following: x = 15u+15v, y = -60u+15v. So, x-y = 75u. Then u = (x-y)/75. 4x+y = 75v. Then v = (4x+y)/75.

We will see how this transformation maps the region R to a new region in the u-v domain. To do so, we will see where the transformation maps the vertices of the region.

(-1,4) -> ((-1-4)/75,(4(-1)+4)/75) = (-1/15, 0)

(1,-4)->(1/15,0)

(3,-2)->(1/15,2/15)

(1,6)->(-1/15,2/15)

That is, the new region in the u-v domain is a rectangle where $\frac{-1}{15}\leq u \leq \frac{1}{15}, 0\leq v \leq \frac{2}{15}$ .

We will calculate the jacobian of the change variables. That is

$\left |\begin{matrix} \frac{du}{dx}& \frac{du}{dy}\\ \frac{dv}{dx}& \frac{dv}{dy}\end{matrix}\right|$ (we are calculating the determinant of this matrix). The matrix is

$\left |\begin{matrix} \frac{1}{75}& \frac{-1}{75}\\ \frac{4}{75}& \frac{1}{75}\end{matrix}\right|=(\frac{1}{75^2})(1+4) = \frac{1}{15\cdot 75}$ (the in-between calculations are omitted).

We will, finally, do the calculations.

Recall that

$15x+15y = 15(15u+15v) + 15(-60u+15v) = (15^2-15\cdot 60 )u+2\cdot 15^2v = 15^2(-3)u+2\cdot 15^2 v$

We will use the change of variables theorem. So,

$\int_R 15x+15y dA = \int_{\frac{-1}{15}}^{\frac{1}{15}}\int_{0}^{\frac{2}{15}} 15^2(-3)u+2\cdot 15^2 v \cdot (\frac{1}{15^2\cdot 5}) dv du = \int_{\frac{-1}{15}}^{\frac{1}{15}}\int_{0}^{\frac{2}{15}}\frac{-3}{5}u+\frac{2}{5}v dvdu$

This si because we are expressing the original integral in the new variables. We must multiply by the jacobian to guarantee that the change of variables doesn't affect the value of the integral. Then,

$\int_{\frac{-1}{15}}^{\frac{1}{15}}\int_{0}^{\frac{2}{15}}\frac{-3}{5}u+\frac{2}{5}v dvdu = \int_{\frac{-1}{15}}^{\frac{1}{15}}\frac{-3}{5}u\cdot \frac{2}{15} + \frac{2}{5}\cdot \left.\frac{v^2}{2}\right|_{0}^{\frac{2}{15}}du = \frac{-3}{5}\left.\frac{u^2}{2}\right|_{\frac{-1}{15}}^{\frac{1}{15}}\cdot \frac{2}{15} + \frac{2}{5}\cdot \left.\frac{v^2}{2}\right|_{0}^{\frac{2}{15}} = \frac{8}{16875}$

5 0

4 years ago

What is the volume of the right rectangular prism?

Sladkaya [172]

250 m because you multiple the base by the height

7 0

2 years ago

At a school with 100? students, 33 were taking? Arabic, 32 ?Bulgarian, and 40 Chinese. 9 students take only? Arabic, 12 take onl

Stels [109]

Answer: The answer is 4 and 32.

Step-by-step explanation: Let "A", "B" and "C" represents the set of students who were taking Arabic, Bulgarian and Chinese respectively.

The, according to the given information, we have

$n(A)=33,~~n(B)=32,~~n(C)=40,~~n(A\cap B)=14.$

Let 'p' represents the number of students who take all the three languages, then

$n(A\cap B\cap C)=p.$

Also,

$n(A)-n(A\cap B)-n(A\cap C)+n(A\cap B\cap C)=9\\\\\Rightarrow n(A\cap B)+n(A\cap C)=24+p~~~~~~~~~~~~~~(a),\\\\n(A\cap B)+n(B\cap C)=20+p~~~~~~~~~~~~~~(b),\\\\n(B\cap C)+n(A\cap C)=20+p~~~~~~~~~~~~~~~(c).$

From here, we get after subtracting equation(c) from (b) that

$n(A\cap B)=n(A\cap C)=14.$

Therefore,

$p=14+14-24=4,$ and from equation (a), we find

$n(B\cap C)=24-14=10.$

Thus,

$n(A\cap B\cap C)=4$ and

$n(A\cup B\cup C)=n(A)+n(B)+n(C)-n(A\cap B)-n(B\cap C)-n(A\cap C)+n(A\cap B\cap C)\\\\\Rightarrow n(A\cap B\cap C)=33+32+40-9-12-20+4\\\\\Rightarrow n(A\cap B\cap C)=68.$