Can an equation have equal expressions

Whats the length of 5 rulers in feet

sergejj [24]

One ruler is 12 inches and 12 inchs= 1 foot so 5 rulers equal 5 feet

6 0

3 years ago

I beg someone help.

erastovalidia [21]

We will begin by working through Part (a): Simplify √18

To simplify a square root such as the one above, we must factor out perfect square numbers (this means we must find a number that multiplies to 18 and gives a whole number when we square root it). Using our knowledge that √9 is a perfect square, we should factor out a 9 from 18, as modeled below:

√18 = √(9*2)

We can separate this square root into two square roots multiplied together, as shown below:

√(9*2) = √9 * √2

Now, we should simplify √9, which equals 3, because 3 * 3 = 9.

√9 * √2 = 3 * √2 = 3√2

Therefore, √18 = 3√2.

Now, we can move on to the next problem: √6 * √15.

To begin this problem, we can multiply the square roots together, which means multiplying the numbers under the radical.

√6 * √15 = √(6*15) = √90

To simplify this, we use the same process as above:

√90 = √(10 * 9) = √9 * √10 = 3√10

Note: We know that this fully simplified because we cannot factor out another perfect square number from the number under the radical (10).

Therefore, your two answers are 3√2 and 3√10.

Hope this helps!

8 0

3 years ago

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Which equation does not have the same solution as the others?

dem82 [27]

<span>The answer is B. X-9=17
Hope this helped! :)</span>

6 0

3 years ago

Read 2 more answers

Problem Solving<br>If seven coins add to $1, what<br>could the coins be?

Alchen [17]

Answer:

2 quarters and 5 dimes

Step-by-step explanation:

(2) 25 cent coins = 50 cents

(5) 10 cent coins = 50 cents

equals $1

4 0

3 years ago

A cylindrical can without a top is made to contain 25 3 cm of liquid. What are the dimensions of the can that will minimize the

Basile [38]

Answer:

Therefore the radius of the can is 1.71 cm and height of the can is 2.72 cm.

Step-by-step explanation:

Given that, the volume of cylindrical can with out top is 25 cm³.

Consider the height of the can be h and radius be r.

The volume of the can is V= $\pi r^2h$

According to the problem,

$\pi r^2 h=25$

$\Rightarrow h=\frac{25}{\pi r^2}$

The surface area of the base of the can is = $\pi r^2$

The metal for the bottom will cost $2.00 per cm²

The metal cost for the base is =$(2.00× $\pi r^2$ )

The lateral surface area of the can is = $2\pi rh$

The metal for the side will cost $1.25 per cm²

The metal cost for the base is =$(1.25× $2\pi rh$ )

$=\$2.5 \pi r h$

Total cost of metal is C= 2.00 $\pi r^2$ + $2.5 \pi r h$

Putting $h=\frac{25}{\pi r^2}$

$\therefore C=2\pi r^2+2.5 \pi r \times \frac{25}{\pi r^2}$

$\Rightarrow C=2\pi r^2+ \frac{62.5}{ r}$

Differentiating with respect to r

$C'=4\pi r- \frac{62.5}{ r^2}$

Again differentiating with respect to r

$C''=4\pi + \frac{125}{ r^3}$

To find the minimize cost, we set C'=0

$4\pi r- \frac{62.5}{ r^2}=0$

$\Rightarrow 4\pi r=\frac{62.5}{ r^2}$

$\Rightarrow r^3=\frac{62.5}{ 4\pi}$

⇒r=1.71

Now,

$\left C''\right|_{x=1.71}=4\pi +\frac{125}{1.71^3}>0$

When r=1.71 cm, the metal cost will be minimum.

Therefore,

$h=\frac{25}{\pi\times 1.71^2}$

⇒h=2.72 cm

Therefore the radius of the can is 1.71 cm and height of the can is 2.72 cm.

6 0

3 years ago