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Grace [21]
2 years ago
5

In a new video game, Franky had -16 1/2 points. He was able to earn 3/4 of the points he lost back. How many points did he have?

Mathematics
1 answer:
lyudmila [28]2 years ago
7 0

Answer:

slightly confused on the wording if he got back 3/4 from 16.5 then he earned back 12.375 points

if -16.5 is the 1/4 he didnt get back then he had 66 points

Step-by-step explanation:

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Two trains leave the station at the same time, one heading east and the other west. The eastbound train travels 18 miles per hou
blagie [28]

Formula: Distance = rate * time

570 = (R+(R-18))*3

570 = (2R - 18)*3

190 = 2R-16

208 = 2R

R = 104


Eastbound Train: 86mph

Westbound Train: 104mph

3 0
3 years ago
What is the sum of the geometric sequence -4, 24, -144, ... if there are 6 terms?
kondor19780726 [428]

Answer:

1556963

Step-by-step explanation:

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6 0
4 years ago
Cidnee is weighing combinations of geometric solids. She found that 4 cylinders and 5 prisms weigh 32 ounces and that 1 cylinder
vitfil [10]

Answer:

The weights of each cylinder and prism are 3 and 4 ounces, respectively.

Step-by-step explanation:

Let be x and y the masses of a cylinder and a prism, measured in ounces, respectively. After a careful reading of the statement we get the following linear equations by interpretation:

i) <em>She found that 4 cylinders and 5 prisms weigh 32 ounces:</em>

4\cdot x +5\cdot y = 32\,oz (Eq. 1)

ii) <em>And that 1 cylinder and 8 prisms weigh 35 ounces:</em>

x + 8\cdot y = 35\,oz (Eq. 2)

Now we solve the system of linear equations algebraically:

From (Eq. 2):

x = 35 - 8\cdot y

(Eq. 2) is (Eq. 1):

4\cdot (35-8\cdot y) + 5\cdot y = 32

140-32\cdot y +5\cdot y = 32

32\cdot y - 5\cdot y = 140-32

27\cdot y = 108

y = 4\,oz

From (Eq. 2):

x = 35 - 8\cdot (4)

x = 35 - 32

x = 3\,oz

The weights of each cylinder and prism are 3 and 4 ounces, respectively.

4 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
Plz help me i'm lost badly
Annette [7]

Answer:

Im not sure but i think the answer is 0

3 0
3 years ago
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