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Oduvanchick [21]
4 years ago
15

30 children are going on a trip it cost £5 including lunch some children take their own packed lunch they pay only £3 the 30 chi

ldren pay 110 altogether how many children take their own packed lunch?
Mathematics
2 answers:
nikitadnepr [17]4 years ago
7 0
No enough info
you can just say that five brought there own lunch while 19 didnt
ivolga24 [154]4 years ago
4 0
Firstly this question can not be solved As 110 divide by 5 is 22 So I'll make up a new question 30 children are going on a trip it cost £5 including lunch some children take their own packed lunch they pay only £3 the 30 children pay 90 altogether. The first step is to see how many pupils pay the full fair that would be 18.
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Can you help plz 7 + 56 y
ivolga24 [154]

I think it's 56y + 7, is what I'm pretty sure of.

8 0
3 years ago
Read 2 more answers
Let R = [ 0 , 1 ] × [ 0 , 1 ] R=[0,1]×[0,1]. Find the volume of the region above R R and below the plane which passes through th
Bond [772]

The three vectors \langle0,0,1\rangle, \langle1,0,8\rangle, and \langle0,1,9\rangle each terminate on the plane. We can get two vectors that lie on the plane itself (or rather, point in the same direction as vectors that do lie on the plane) by taking the vector difference of any two of these. For instance,

\langle1,0,8\rangle-\langle0,0,1\rangle=\langle1,0,7\rangle

\langle0,1,9\rangle-\langle0,0,1\rangle=\langle0,1,8\rangle

Then the cross product of these two results is normal to the plane:

\langle1,0,7\rangle\times\langle0,1,8\rangle=\langle-7,-8,1\rangle

Let (x,y,z) be a point on the plane. Then the vector connecting (x,y,z) to a known point on the plane, say (0, 0, 1), is orthogonal to the normal vector above, so that

\langle-7,-8,1\rangle\cdot(\langle x,y,z\rangle-\langle0,0,1\rangle)=0

which reduces to the equation of the plane,

-7x-8y+z-1=0\implies z=7x+8y+1

Let z=f(x,y). Then the volume of the region above R and below the plane is

\displaystyle\int_0^1\int_0^1(7x+8y+1)\,\mathrm dx\,\mathrm dy=\boxed{\frac{17}2}

6 0
3 years ago
5(3a+b)−2(3a+b) in simplest form<br> I NEED HELP!
nordsb [41]

Answer:

9a + 3b

Is the answer

Step-by-step explanation:

5(3a + b) - 2(3a + b) \\  = 15a + 5b - 6a - 2b \\  = 9a + 3b

( - ) \times ( + ) = ( - ) \\

That is why

( - 2) \times b =  - 2

And

( - 2) \times 3a = ( - 6)

7 0
3 years ago
Please help me with question number 2
Mice21 [21]

Answer:

2. The area of the side walk is approximately 217 m²

3. The distance away from the sprinkler the water can spread is approximately 11 feet

4. The area of the rug is 49.6

Step-by-step explanation:

2. The dimensions of the flower bed and the sidewalks are;

The diameter of the flower bed = 20 meters

The width of the circular side walk, x = 3 meters

Therefore, the diameter of the outer edge of the side walk, D, is given as follows

D = d + 2·x (The width of the side walk is applied to both side of the circular diameter)

∴ D  = 20 + 2×3 = 26

The area of the side walk = The area of the sidewalk and the side walk = The area of the flower bed

∴ The area of the side walk, A = π·D²/4 - π·d²/4

∴ A = 3.14 × 26²/4 - 3.14 × 20²/4 = 216.66

By rounding to the nearest whole number, the area of the side walk, A ≈ 217 m²

3. Given that the area formed by the circular pattern, A = 379.94 ft.², we have;

Area of a circle = π·r²

∴ Where 'r' represents how far it can spread, we have;

π·r² = 379.94

r = √(379.94 ft.²/π) ≈ 10.997211 ft.

Therefore, the distance away from the sprinkler the water can spread, r ≈ 11 feet

4. The circumference of the rug = 24.8 meters

The circumference of a circle, C = 2·π·r

Where;

r = The radius of the circle

π = 3.1

∴ For the rug of radius 'r', C = 2·π·r = 24.8

r = 24.8/(2·π) = 12.4/π = 12.4/3.1 = 4

The area = π·r²

∴ The area of the rug = 3.1 × 4² = 49.6.

8 0
3 years ago
A person places $37700 in an investment account earning an annual rate of 5%,
lorasvet [3.4K]

Answer:

8,229,437 cents

Step-by-step explanation:

Using the compound interest formula;

A = P(1+r)^n

Given

Principal invested = $37700

rate r = 5% = 0.05

Time t = 16years

Substitute into the formula

A = 37700(1+0.05)^16

A  = 37700(1.05)^16

A = 37700(2.1829)

A = 82,294.37

Hence the amount of money, to the  nearest cent, in the account after 16 years is 8,229,437 cents

4 0
3 years ago
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