A customer buys five cans of baked beans, four pocket of brown sugar and two pockets from the supermarket. He gives the cashier 150 and gets 26 in change, a can of baked beans costs half as much as pocket of biscuits whilst a packet of sugar is three rand cheaper than a packet of biscuits.1. how much did all items cost
2. calculate the cost of a can of baked beans, sugar, and biscuits
3. How much will eight cans of baked beans, one packet of sugar and six pockets of biscuits cost?

Answers

Answer 1

Answer: The question has 3 unknowns thus we need three equations to fully solve the problem.
We let x = the price of a can of baked beans, y = the price of a packet of sugar and z = the price of a packet of biscuit
From the given, we generate the following equations;
<1> 5x+4y+2z = Amount spent = 150-26 = 124
<2> z = 2x
<3> y = z -3
We use equation 2 and 3 to substitute to equation 1 in order to obtain the value of x.
5x + 4 (2x - 3) + 2(2x) = 124
5x + 8x - 12 + 4x = 124
17x = 136
x = 8 (price of a can of baked beans)
y= (2 x 8) - 3 = 13 (price of a packet of sugar)
z= 2 x 8 = 16 (price of a pocket of biscuit)

For the new set of conditions,
8x + y + 6z = (8 x 8) + 13 +(6 x 16) = 173 (Cost for the new set of conditions)

Answer 2

Answer: let's make b the can of baked beans.

Then biscuits are twice its prize, 2b.

s, the packet of sugar is 2b-3 - thre rand cheaper than a packet of biscuits.

1) the custommer bought:
buys five cans of baked beans, four pocket of brown sugar and two pockets

5b+4* (2b-3)+2*2b

now we calculate:

5b+4* (2b-3)+2*2b=5b+8b-12+4b=13b-12+4b=17b-12

this is equal to 150-26=124

so 17b-12=124
17b=136
b=8
So the beans cost 8, biscuits cost 8*2=16, and sugar 16-3=13
2) this would be:
8+13+16=21+16=37
3)This would be:

8*8+13+6*16=64+13+96=173

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Square root of 64 over 25

Answers

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Knowing that $64 = 8^2\quad\text{and}\quad25 = 5^2$ , we have

$\sqrt{(64)/(25)} = (√(64))/(√(25)) = \boxed{(8)/(5)}$

so $\sqrt{(64)/(25) }$

that is equal to $( √(64))/( √(25)) = (8)/(5)$

Which expression is equivalent to 7a2b + 10a2b2 + 14a2b3?

Answers

The given expression can be simplified in many ways by grouping like terms. The simplest form is obtained by factoring out a²b which gives us the following expression.

a²b(7 + 10b +14b²)

Answer:

a²b(7 + 10b +14b²)

Step-by-step explanation:

$7a^2b + 10a^2b^2 + 14a^2b^3$

To find out equivalent expression we need to factor the given expression

To factor it first we find out GCF(greatest common factor)

7 a^2b -------> 7*a*a*b

10a^2b^2 ----->5*2 * a*a*b*b

14a^2b^3 -------> 7*2 *a*a*b*b*b

To get GCF we look at the factors that are in common

GCF = a*a*b that is a^2b

We factor out GCF a^2b, write a^2b outside and divide all the terms by a^2b

$a^2b(7 +10b +14b^2)$

7+10b+14b^2 cannot be factor . its prime

The sum of 3 consecutive odd integers is 87.Find the equation used to solve this problem and the three integers.

Answers

the first odd integer is x. The next odd integer will be x+2. and the next is x+4. Then we can say that (x)+(x+2)+(x+4)=87
working that out yields 3x+6=87
and thus 3x=81
and by consequence x=27

and thus our 3 odd integers are 27,29 and 31

Let be y - 2, y and y + 2 the 3 consecutives odd integers;
There sum is 3y, and, in the same time is 87 => y = 87/3 => y = 29 =>
y - 2 = 27; y + 2 = 31;
The 3 integers are : 27, 29, 31.

Please help Evaluate the function when x=4

Blue- f(4) = 2(4) - 5 = 3

Teal- f(4) = 4

Yellow- f(4) = 1/2(4) + 1 = 3

Red- f(4) = 2(4) - 5 = -1