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3 years ago

How to do this question I need to go to school on 1/9

Mathematics

2 answers:

ELEN [110]3 years ago

5 0

$7.9 cost of one pack
10 packs cost is $79
She pays $29 + Voucher

From the total $79 we substract the $29 she paid with cash or card.

79-29=50

So the value of the voucher is $50

PS: celia should cut on chocolate. Not healthy for her :)

Shkiper50 [21]3 years ago

3 0

1 pack = $7.90

10 packs = $7.90 x 10 = $79

Celia pays $29 plus a voucher to equal $79.

29 + (voucher) = 79

(voucher) = 50 subtracted 29 from both sides

Answer: $50

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3 years ago

32) Find the value of 5x° + 2x-1 when x = 4.

SashulF [63]

Answer:

5x4°x7

Step-by-step explanation:

5x°+2x-1 when x =4

5x4°+2x4-1

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3 years ago

Points A and B have coordinates (8, 3) and (p, q) respectively. The equation of the perpendicular

DaniilM [7]

Answer:

p = -4, q = -3

Step-by-step explanation:

y = -2x +4 ... (1) perpendicular bisector of AB, slope = -2

slope of AB = 1/2

Line AB pass (8,3): (y-3) / (x-8) = 1/2

AB equation: y-3 = 1/2(x-8) y = 1/2x - 1 ... (2)

(2)-(1): 5/2 x = 5 x = 2

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(8+p)/2 = 2

p = -4

(3+q)/2 = 0

q = -3

4 0

3 years ago

(5x10,000) +(2x100) +(9x1) +(3x1/1000)

zubka84 [21]

Scientific notation:
5.0209003 x 10^4

7 0

3 years ago

From a piece of tin in the shape of a square 6 inches on a side, the largest possible circle is cut out. What is the ratio of th

wel

Answer:

$\sf \dfrac{1}{4} \pi \quad or \quad \dfrac{7}{9}$

Step-by-step explanation:

The width of a square is its side length.

The width of a circle is its diameter.

Therefore, the largest possible circle that can be cut out from a square is a circle whose diameter is equal in length to the side length of the square.

Formulas

$\sf \textsf{Area of a square}=s^2 \quad \textsf{(where s is the side length)}$

$\sf \textsf{Area of a circle}=\pi r^2 \quad \textsf{(where r is the radius)}$

$\sf \textsf{Radius of a circle}=\dfrac{1}{2}d \quad \textsf{(where d is the diameter)}$

If the diameter is equal to the side length of the square, then:
$\implies \sf r=\dfrac{1}{2}s$

Therefore:

$\begin{aligned}\implies \sf Area\:of\:circle & = \sf \pi \left(\dfrac{s}{2}\right)^2\\& = \sf \pi \left(\dfrac{s^2}{4}\right)\\& = \sf \dfrac{1}{4}\pi s^2 \end{aligned}$