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Ad libitum [116K]
3 years ago
13

A square box is filled with two layers of identical square pieces of chocolate. Kirill had eaten all 20 pieces in the upper laye

r along the walls of the box, how many pieces of chocolate are left in the box?​
Mathematics
2 answers:
Natali [406]3 years ago
8 0

Answer: 50? Hope this helped.

Vikentia [17]3 years ago
4 0

Answer:

52.

Step-by-step explanation:

Since it is a square box, along the walls has

20 chocolates, only a 6x6 has 20 along the walls.

So 36+36 would be 72 and 72 - 20 is 52.

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Can someone please help me !!
motikmotik

Answer:

a x = 14

Step-by-step explanation:

because she gets every 14 dollars one hour.

hope this helps you

4 0
2 years ago
Sam is hired for a 20-day period. On days that he works, he earns $60. For each day that he does not work, $30 is subtracted fro
Alenkinab [10]

Answer:

He did not work for 6 days

Step-by-step explanation:

Sam is hired for a 20-day period.

Let x be the no. of days he did not work

So, No. of days he worked = 20-x

he earns per working day = $60

So, he earns for (20-x) days = 60(20-x)

For each day that he does not work, $30 is subtracted from his earnings.

So, Amount deducted for x days = 30x

So, His earning for 20 days =60(20-x)-30x

We are given that At the end of the 20-day period, he received $660

So, 60(20-x)-30x=660

1200-60x-30x=660

1200-90x=660

1200-660=90x

540=90x

\frac{540}{90}=x

6=x

So, he did not work for 6 days

So, Option C is true

5 0
3 years ago
Solve the given differential equation by separation of variables. dy dx = xy + 5x − y − 5 xy − 2x + 6y − 12
Solnce55 [7]

solution:

Consider the differential equation DE

Dy/dx = xy + 5x –y -5 /xy -2x + 6y -12

Write the DE as the follows.

Dy/dx = x(y+5) -1(y+5)/x(y-2) +6(y-2)

Dy/dx = (x-1) (y+5)/(x+6)(y-2)

Separate the variables.

y-2/y+5 dy = x-1/x+6 dx

integrate on the both sides,

∫y-2/y+5 dy = ∫x-1/x+6 dx

7in(x+6) -7in(y+5) = x-y+c

In(x+6)∧7 –in(y+5)∧7 = x-y+c , using bIna =Inab

In [(x+6)∧7/(y+5)∧7] = x-y+c   ,using Ina – Inb = In(b/a)

eIn[(x+6)∧7/(y+5)∧7] = ex-y+c    , taking exponents on both sides

(x+6)∧7/(y+5)∧7 = ec.ex-y       ,use eInx = x

(x+6)∧7/(y+5)∧7 = c1ex-y , take ec =c1

Hence, the solution of the DE is (x+6)∧7/(y+5)∧7 = c1ex-y


3 0
3 years ago
A group of adults and students went on a class trip to Washington, DC. The number of male students was 1 more than 7 times the n
liberstina [14]
Given:
adults = x
male = 7x + 1
female = (7x + 1)/2
total number of people = 82

x + (7x + 1) + [(7x +1)/2] = 82
2x/2 + (14x + 2)/2 + (7x + 1)/2 = 82
2x + 14x + 2 + 7x + 1 = 82 * 2
23x + 3 = 164
23x = 164 - 3
23x = 161
x = 161/23
x = 7

adults = x = 7
males = 7x + 1 = 7(7) + 1 = 49 + 1 = 50
females = (7x+1)/2 = 50/2 = 25

7 + 50 + 25 = 82

6 0
3 years ago
Rewrite the quadratic function in vertex form.<br> Y=2x^2+4x-1
Fantom [35]

Answer:

\large\boxed{y=2(x+1)^2-3}

Step-by-step explanation:

The vertex form of an equation of a parabola:

y=a(x-h)^2+k

(h, k) - vertex

We have

y=2x^2+4x-1=2\left(x^2+2x-\dfrac{1}{2}\right)

We must use the formula: (a+b)^2=a^2+2ab+b^2\qquad(*)

2\left(x^2+2(x)(1)-\dfrac{1}{2}\right)=2\bigg(\underbrace{x^2+2(x)(1)+1^2}_{(*)}-1^2-\dfrac{1}{2}\bigg)\\\\=2\left((x+1)^2-1-\dfrac{1}{2}\right)=2\left((x+1)^2-\dfrac{3}{2}\right)

Use the distributive formula a(b + c) = ab + ac

2(x+1)^2+2\left(-\dfrac{3}{2}\right)=2(x+1)^2-3

5 0
2 years ago
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