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mestny [16]
3 years ago
9

What is the slope? 25 points + brainly

Mathematics
1 answer:
uysha [10]3 years ago
8 0

Answer:

-5/4  or -1.25 is the slope

Step-by-step explanation:

Slope (gradient)  = rise/run or (y1 - y2/x1 - x2)

We will take the points so that we have a whole number:

Therefore the two points are:

(10,0) and (50, -30)

Now the gradient;

(y1 - y2/x1 - x2)

0 - 50/10 - (-30)

-50/40

-5/4

-1.25

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Wavelength varies inversely with frequency. Let k be the product of wavelength and frequency. Complete the table using the inver
Lerok [7]

Answer:

..

Step-by-step explanation:

a =  

⇒ 670

b =  

⇒ 301600

c =  

⇒ 560

8 0
3 years ago
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A hemisphere fake stone made of clay has a diameter of 16 inches. The density of clay is 0.0035 pounds per cubic inch. The clay
Rufina [12.5K]

Answer:

$137.82

Step-by-step explanation:

The volume of a hemisphere stone is given by the volume of a half-sphere:

Volume = (2/3) * pi * radius^3

If the diameter is 16 inches, the radius is 16/2 = 8 inches, so the volume is:

Volume = (2/3) * pi * 8^3 = 1072.33 in3

We need a total of 8 stones, so the total volume is:

Total Volume = 8 * 1072.33 = 8578.64 in3

The density of clay is 0.0035 pounds per cubic inch, so the total weight of the 8 stones is:

Weight = Total Volume * density = 8578.63 * 0.0035 = 30.0252 pounds

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Price = 30.0252 * 4.59 = $137.82

8 0
4 years ago
Someone help me please
gogolik [260]

x/5 = 6/15
15x = 30
    x = 2

answer
x = 2 ft

y/10 = 6/15
15y = 60
y = 4

answer
y = 4 ft


3 0
3 years ago
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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
PLEASE HELP I WILL GIVE BRAINLIEST
MA_775_DIABLO [31]

Answer:

The answer is 1\frac{1}{2}

Step-by-step explanation:

4 0
3 years ago
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