All categories

3 years ago

Saw a board 5 ft 1 in. Long into six equal pieces. If the loss per cut is 1/5 in., how long will each piece be?

Mathematics

1 answer:

Pie3 years ago

5 0

Answer:

The length of each piece is 0.833 ft.

Step-by-step explanation:

Length of board, L = 5 ft 1 inch

loss per cut = 1/5 inch

The board should be cut by 5 times to get the 6 pieces.

The loss in 5 cuts = 5 x 1/5 = 1 inch

So, the length of each piece is

$L' =\frac{5 ft 1 in - 1in }{6}\\\\L' =\frac{5}{6}\\\\L' = 0.833 ft$

You might be interested in

Given the following equation of the hyperbola, what are the x-values of the vertices? x^2/4 - y^2/49 = 1 is it 2 and -2?

Jet001 [13]

Answer:

yes, ±2

Step-by-step explanation:

The x-intercepts are found by setting y=0 and solving for x:

x^2/4 = 1

x^2 = 4

x = ±√4

x = ±2

The x-values of interest are -2 and +2.

3 0

3 years ago

Charlie knows that 3 blocks are needed to make a tower that is 0.5 foot high.

iogann1982 [59]

Answer:

24 blocks will be needed to make the tower that is 4 feet high

Step-by-step explanation:

Let us use the ratio method to solve the question

∵ Charlie knows that 3 blocks are needed to make a tower that is

0.5 foot high

∵ He needs to make a tower that is 4 feet high

→ By using the ratio method

→ Blocks : height (ft)

→ 3 : 0.5

→ x : 4

→ By using the cross multiplication

∵ x × 0.5 = 3 × 4

∴ 0.5x = 12

→ Divide both sides by 0.5 to find x

∵ $\frac{0.5x}{0.5}$ = $\frac{12}{0.5}$

∴ x = 24

∵ x represents the number of blocks

∴ 24 blocks will be needed to make the tower that is 4 feet high

7 0

2 years ago

In a triangle , the measures of the angles are x,x + 20 & 2x . What is the value of x ?

Alexus [3.1K]

X+x+20+2x=180
4x+20=180
4x=160
x=40

5 0

3 years ago

I need help on this question

Kitty [74]

What is the question? i think that’s important

8 0

3 years ago

Please help!!<br> Write a matrix representing the system of equations

frozen [14]

Answer:

$(4, -1, 3)$

Step-by-step explanation:

We have the system of equations:

$\left\{ \begin{array}{ll} x+2y+z =5 \\ 2x-y+2z=15\\3x+y-z=8 \end{array} \right.$

We can convert this to a matrix. In order to convert a triple system of equations to matrix, we can use the following format:

$\begin{bmatrix}x_1& y_1& z_1&c_1\\x_2 & y_2 & z_2&c_2\\x_3&y_2&z_3&c_3 \end{bmatrix}$

Importantly, make sure the coefficients of each variable align vertically, and that each equation aligns horizontally.

In order to solve this matrix and the system, we will have to convert this to the reduced row-echelon form, namely:

$\begin{bmatrix}1 & 0& 0&x\\0 & 1 & 0&y\\0&0&1&z \end{bmatrix}$

Where the (x, y, z) is our solution set.

Reducing:

With our system, we will have the following matrix:

$\begin{bmatrix}1 & 2& 1&5\\2 & -1 & 2&15\\3&1&-1&8 \end{bmatrix}$

What we should begin by doing is too see how we can change each row to the reduced-form.

Notice that R₁ and R₂ are rather similar. In fact, we can cancel out the 1s in R₂. To do so, we can add R₂ to -2(R₁). This gives us:

$\begin{bmatrix}1 & 2& 1&5\\2+(-2) & -1+(-4) & 2+(-2)&15+(-10) \\3&1&-1&8 \end{bmatrix}\\\Rightarrow\begin{bmatrix}1 & 2& 1&5\\0 & -5 & 0&5 \\3&1&-1&8 \end{bmatrix}$

Now, we can multiply R₂ by -1/5. This yields:

$\begin{bmatrix}1 & 2& 1&5\\ -\frac{1}{5}(0) & -\frac{1}{5}(-5) & -\frac{1}{5}(0)& -\frac{1}{5}(5) \\3&1&-1&8 \end{bmatrix}\\\Rightarrow\begin{bmatrix}1 & 2& 1&5\\ 0 & 1 & 0& -1 \\3&1&-1&8 \end{bmatrix}$

From here, we can eliminate the 3 in R₃ by adding it to -3(R₁). This yields:

$\begin{bmatrix}1 & 2& 1&5\\ 0 & 1 & 0& -1 \\3+(-3)&1+(-6)&-1+(-3)&8+(-15) \end{bmatrix}\\\Rightarrow\begin{bmatrix}1 & 2& 1&5\\ 0 & 1 & 0& -1 \\0&-5&-4&-7 \end{bmatrix}$

We can eliminate the -5 in R₃ by adding 5(R₂). This yields:

$\begin{bmatrix}1 & 2& 1&5\\ 0 & 1 & 0& -1 \\0+(0)&-5+(5)&-4+(0)&-7+(-5) \end{bmatrix}\\\Rightarrow\begin{bmatrix}1 & 2& 1&5\\ 0 & 1 & 0& -1 \\0&0&-4&-12 \end{bmatrix}$

We can now reduce R₃ by multiply it by -1/4:

$\begin{bmatrix}1 & 2& 1&5\\ 0 & 1 & 0& -1 \\ -\frac{1}{4}(0)&-\frac{1}{4}(0)&-\frac{1}{4}(-4)&-\frac{1}{4}(-12) \end{bmatrix}\\\Rightarrow\begin{bmatrix}1 & 2& 1&5\\ 0 & 1 & 0& -1 \\0&0&1&3 \end{bmatrix}$

Finally, we just have to reduce R₁. Let's eliminate the 2 first. We can do that by adding -2(R₂). So:

$\begin{bmatrix}1+(0) & 2+(-2)& 1+(0)&5+(-(-2))\\ 0 & 1 & 0& -1 \\0&0&1&3 \end{bmatrix}\\\Rightarrow\begin{bmatrix}1 & 0& 1&7\\ 0 & 1 & 0& -1 \\0&0&1&3 \end{bmatrix}$

And finally, we can eliminate the second 1 by adding -(R₃):

$\begin{bmatrix}1 +(0)& 0+(0)& 1+(-1)&7+(-3)\\ 0 & 1 & 0& -1 \\0&0&1&3 \end{bmatrix}\\\Rightarrow\begin{bmatrix}1 & 0& 0&4\\ 0 & 1 & 0& -1 \\0&0&1&3 \end{bmatrix}$

Therefore, our solution set is $(4, -1, 3)$

And we're done!

3 0

3 years ago