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3 years ago

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Six pipes are laid end to form a row.each pipe is 9 1/3 inches long.how long is the row of pipes in feet?

1 answer:

laiz [17]3 years ago

3 0

six pipes are laid end to end to form a row. Each pipe is 9 1/3 inches long. how long is the row of pipes, in feet?

Answer:

4.67 feet or 4 feet 8 inches or $4\frac{2}{3}$ feet

Step-by-step explanation:

Let x be the length of the row pipes.

Given:

Number of pipes = 6

Each pipe has length $9\frac{1}{3}$ inches. and six pipes are laid end to end to form a row.

The total length of the row is equal to length of the 6 pipes.

So, total length of row = Number of pipes $\times$ Length of the pipe

$x=\frac{28}{3}\times 6$

$x=28\times 2$

$x=56$ inches

convert inch into feet.

$x=\frac{56}{12}$

$x=4.67 feet$

Therefore, The length of the row is 4.67 feet or 4 feet 8 inches or $4\frac{2}{3}$ feet .

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Step-by-step explanation:

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2 years ago

Which three lengths could be the lengths of the sides of a triangle?

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Answer:

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Step-by-step explanation:

The three lengths that could be the lengths of the sides of a triangle, must satisfy the following condition;

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2 years ago

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3 years ago

Read 2 more answers

Will mark brainliest!!!plz helppp

muminat

Answer:

(5,-6)

Step-by-step explanation:

ONE WAY:

If $f(x)=x^2-6x+3$ , then $f(x-2)=(x-2)^2-6(x-2)+3$ .

Let's simplify that.

Distribute with $-6(x-2)$ :

$f(x-2)=(x-2)^2-6x+12+3$

Combine the end like terms $12+3$ :

$f(x-2)=(x-2)^2-6x+15$

Use $(x-b)^2=x^2-2bx+b^2$ identity for $(x-2)^2$ :

$f(x-2)=x^2-4x+4-6x+15$

Combine like terms $-4x-6x$ and $4+15$ :

$f(x-2)=x^2-10x+19$

We are given $g(x)=f(x-2)$ .

So we have that $g(x)=x^2-10x+19$ .

The vertex happens at $x=\frac{-b}{2a}$ .

Compare $x^2-10x+19$ to $ax^2+bx+c$ to determine $a,b,\text{ and } c$ .

$a=1$

$b=-10$

$c=19$

Let's plug it in.

$\frac{-b}{2a}$

$\frac{-(-10)}{2(1)}$

$\frac{10}{2}$

$5$

So the $x-$ coordinate is 5.

Let's find the corresponding $y-$ coordinate by evaluating our expression named $g$ at $x=5$ :

$5^2-10(5)+19$

$25-50+19$

$-25+19$

$-6$

So the ordered pair of the vertex is (5,-6).

ANOTHER WAY:

The vertex form of a quadratic is $a(x-h)^2+k$ where the vertex is $(h,k)$ .

Let's put $f$ into this form.

We are given $f(x)=x^2-6x+3$ .

We will need to complete the square.

I like to use the identity $x^2+kx+(\frac{k}{2})^2=(x+\frac{k}{2})^2$ .

So If you add something in, you will have to take it out (and vice versa).

$x^2-6x+3$

$x^2-6x+(\frac{6}{2})^2+3-(\frac{6}{2})^2$

$(x+\frac{-6}{2})^2+3-3^2$

$(x+-3)^2+3-9$

$(x-3)^2+-6$

So we have in vertex form $f$ is:

$f(x)=(x-3)^2+-6$ .

The vertex is (3,-6).

So if we are dealing with the function $g(x)=f(x-2)$ .

This means we are going to move the vertex of $f$ right 2 units to figure out the vertex of $g$ which puts us at (3+2,-6)=(5,-6).

The $y-$ coordinate was not effected here because we were only moving horizontally not up/down.

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