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3 years ago

1) Order the numbers from least to greatest.

Mathematics

2 answers:

Cerrena [4.2K]3 years ago

6 0

I think 1 I hope I help

masha68 [24]3 years ago

3 0

1. -0.8, -2, -3, -1/10, 1/2, 0.8, 7 i think

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$300 at 6% for 3 years

Charra [1.4K]

Its I=prt
so plug it in
I=300(.06)3 i got .06 because you have 6% you have to move the decimal the left twice 0.06 is what you get
now its basic multiplying
300 times .06 times 3
so it would be 54
I=$54
the balence you get from adding the P to the I so the balence would be 356
Bal=$356

7 0

3 years ago

Please help!! 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

He found a quotient of 7 which division problem was he using?

marshall27 [118]

The answer is 3 1/2 divided by 1 1/2 because those equal 7.
~Silver

6 0

3 years ago

a bottle of orange juice holds for 4/6 quarts of juice. a bottle of pineapple juice holds 2/3 quarts of juice. which bottle has

earnstyle [38]

Answer:

the orange juice has more ounces than the pineepple juice

Step-by-step explanation:

3 0

3 years ago

PLS HELP ME WITH THIS!!!!!! HOW DID THEY GET 80ft^2

noname [10]

$\huge \boxed{\mathfrak{Question} \downarrow}$

Determine the surface area of the right square pyramid.

$\large \boxed{\mathfrak{Answer \: with \: Explanation} \downarrow}$

The formula for finding the surface area of a right square pyramid is ⇨ b² + 2bl, where

b = base of the right square pyramid
l = slant height of the right square pyramid.

In the given figure,

base (b) = 4 ft.
slant height (l) = 8 ft.

Now, let's substitute the values of b & l in the formula & solve it :-

$\sf \: {b}^{2} + 2bl \\ = \sf \: {4}^{2} + 2 \times 4 \times 8 \\ = \sf \: 16 + 8 \times 8 \\ = \sf \: 16 + 64 \\ = \huge\boxed{\boxed{ \bf 80 \: ft ^{2} }}$

So, the surface area of the right square pyramid is 80 ft².

7 0

3 years ago