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3 years ago

2 and one half cakes are divided up into 18 pieces for guests how much cake will each person get

Mathematics

1 answer:

natali 33 [55]3 years ago

4 0

Well if you were to do 18 ÷ 2 it would be 8. But since this is 8 and a half let's use 2.5. So, let's do this.

18 ÷ 2.5 = 7.2.

But since cake can't be exactly that we can estimate to 7 since it is the closest so, the answer is 7 pieces of cake.

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Help me solve this problem

Elenna [48]

Answer:

=, there are equal

Step-by-step explanation: < means less than and > means greater than and = is just equal.

Use your calculator to the square root of 125.

The answer is 11.180...... and so on. But we just leave it as 11. Therefore, 11 and 11 are equal. That's the answer to your question.

4 0

2 years ago

jada has already taken 4 pages of notes on her own, and she will take 2 pages out during each class session. In all, how many cl

stich3 [128]

Answer:

Step-by-step explanation:

I don't totally understand you question but if its 4 pages originally then she takes 2 pages every class, then

56-4= 52 (56 pages minus the 4 she already took).

52/2= 26 (52 pages divided by 2 pages per class)

4 0

3 years ago

The first term of a geometric sequence is 32, and the 5th term of the sequence is 818 .

sammy [17]

Answer:

$32,24,18,\frac{27}{2} ,\frac{81}{8}$

Step-by-step explanation:

Let x, y , and z be the numbers.

Then the geometric sequence is $32,x,y,z,\frac{81}{8}$

Recall that term of a geometric sequence are generally in the form:

$a,ar,ar^2,ar^3,ar^4$

This implies that:

a=32 and $ar^4=\frac{81}{8}$

Substitute a=32 and solve for r.

$32r^3=\frac{81}{8}$

$r^4=\frac{81}{256}$

Take the fourth root to get:

$r=\sqrt[4]{\frac{81}{256} }$

$r=\frac{3}{4}$

Therefore $x=32*\frac{3}{4} =24$

$y=24*\frac{3}{4} =18$

$z=18*\frac{3}{4} =\frac{27}{2}$

8 0

3 years ago

Which two fractions have a sum of −7/30 ? Drag and drop the appropriate fraction into each box.

Brrunno [24]

Well you will need two number that are divisible by 30 and add up to -7 so they will have to be a negative.

I would try -2 and -5 which are both divisible by 30 and ad up to -7 so...

-2/30 + -5/30 = -7/30

5 0

3 years ago

Not sure what to do here can someone please help

kumpel [21]

Answer:

x = 2

Step-by-step explanation:

These equations are solved easily using a graphing calculator. The attachment shows the one solution is x=2.

<h3>Squaring</h3>

The usual way to solve these algebraically is to isolate radicals and square the equation until the radicals go away. Then solve the resulting polynomial. Here, that results in a quadratic with two solutions. One of those is extraneous, as is often the case when this solution method is used.

$\sqrt{x+2}+1=\sqrt{3x+3}\qquad\text{given}\\\\(x+2)+2\sqrt{x+2}+1=3x+3\qquad\text{square both sides}\\\\2\sqrt{x+2}=(3x+3)-(x+3)=2x\qquad\text{isolate the root term}\\\\x+2=x^2\qquad\text{divide by 2, square both sides}\\\\x^2-x-2=0\qquad\text{write in standard form}\\\\(x-2)(x+1)=0\qquad\text{factor}$

The solutions to this equation are the values of x that make the factors zero: x=2 and x=-1. When we check these in the original equation, we find that x=-1 does not work. It is an extraneous solution.

x = -1: √(-1+2) +1 = √(3(-1)+3) ⇒ 1+1 = 0 . . . . not true

x = 2: √(2+2) +1 = √(3(2) +3) ⇒ 2 +1 = 3 . . . . true . . . x = 2 is the solution

<h3>Substitution</h3>

Another way to solve this is using substitution for one of the radicals. We choose ...

$u=\sqrt{x+2}\qquad\text{requires $u\ge0$}\\\\u^2-2=x\qquad\text{solve for x}\\\\u+1=\sqrt{3(u^2-2)+3}\qquad\text{substitute for x in the original equation}\\\\(u+1)^2=3u^2-3\qquad\text{square both sides, simplify a little}\\\\2u^2-2u-4=0\qquad\text{subtract $(u+1)^2$}\\\\2(u-2)(u+1)=0\qquad\text{factor}$

Solutions to this equation are ...

u = 2, u = -1 . . . . . . the above restriction on u mean u=-1 is not a solution

The value of x is ...

x = u² -2 = 2² -2

x = 2 . . . . the solution to the equation

_____

<em>Additional comment</em>

Using substitution may be a little more work, as you have to solve for x in terms of the substituted variable. It still requires two squarings: one to find the value of x in terms of u, and another to eliminate the remaining radical. The advantage seems to be that the extraneous solution is made more obvious by the restriction on the value of u.

6 0

2 years ago