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4 years ago

Find the quotient and remainder 255÷6

Mathematics

2 answers:

Olegator [25]4 years ago

7 0

..255 : 6 = 42
-24
====
...15
..-12
====
R 3

$255:6=42 \ R. \ 3$
or
$255:6=42 \frac{3}{6}=42\frac{1}{2}$

ludmilkaskok [199]4 years ago

3 0

42.5 or 42 with a remainder of 3

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Alicia wants to bake a cake with ⁹/10 grams of flour.However,she only has ⅓ grams of flour she needs.How much does she have?

natali 33 [55]

Answer:

We know that:

She needs 9/10 grams of flour for the cake (this quantity does not make a lot of sense, maybe is written incorrectly, but is the only info we have, so let's solve the problem with this)

She only has 1/3 of the amount she needs.

How much flour does she have?

Well, she needs 9/10 grams, and she has one-third of that.

Then she has (1/3) times (9/10) grams, this is:

F = (1/3)*(9/10) grams

F = (1*9)/(3*10) grams

F = 9/30 grams

That is the amount of flour that she has.

6 0

3 years ago

Solve the following quadratics. State the FACTORS AND SOLUTIONS. 1. 2x^2 - 7x + 3 2. 3x^2 + 7x +2

tekilochka [14]

Answer:

1. x = 3, 1/2 (solutions); (x - 3)(2x - 1) (factors)

2. x = -1/3, -2 (solutions); (3x + 1)(x + 2) (factors)

Step-by-step explanation:

1. 2x^2 - 7x + 3

To solve problem 1, you will need to identify your a, b, and c values in this quadratic function.

Since this problem is in standard form, it will be easy to identify these values. The standard form of a quadratic function is ax^2 + bx + c.

The a value is 2, the b value is -7, and the c value is 3 if we use our standard form and see which numbers are plugged into it.

Since we know that

a = 2
b = -7
c = 3

we can use the quadratic formula: $x = \frac{-b~\pm~\sqrt{b^2~-~4ac} }{2a}$

Substitute the a, b, and c values into the quadratic formula: $x=\frac{-(-7)\pm\sqrt{(-7)^2-4(2)(3)} }{2(2)}$

Now simplify using the laws of pemdas: $x=\frac{7\pm\sqrt{(49)-(24)} }{4}$

Simplify even further: $x=\frac{7\pm\sqrt{(25)} }{4} \rightarrow x=\frac{7\pm (5) }{4}$

Now split this equation into two equations to solve for x: $x=\frac{12 }{4} ~~and~~ x=\frac{2 }{4}$

12/4 can be simplified to 3, and 2/4 can be simplified to 1/2.

This means your solutions to problem 1 is 3, 1/2.

$\boxed {x=3,\frac{1}{2} }$

There is also another way to solve for the quadratic functions, and this was by factoring.

If you factor 2x^2 - 7x + 3 using the bottoms-up method, you will get (x - 3)(2x - 1).

After factoring, solving for the solutions is simple because all you have to do is set each factor to 0.

x - 3 = 0
2x - 1 = 0

After solving for x by adding 3 to both sides, or by adding 1 to both sides then dividing by 2, you will end up with the same solutions: x = 3 and x = 1/2.

2. 3x^2 + 7x + 2

To save time I'll be using the bottoms-up factoring method, but remember to refer back to problem 1 (quadratic formula) if you prefer that method.

Factor this quadratic function using the bottoms-up method. After factoring you will have (3x + 1)(x + 2). These are your factors.

Now to solve for x and find the solutions of the quadratic function, you will set both factors equal to 0.