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maw [93]
3 years ago
10

Rita collected and stored beads in 4 different bags.the table below shows the number and color of beeds in each bag.

Mathematics
2 answers:
bezimeni [28]3 years ago
8 0
Hey may we please see the table? :))
Yuki888 [10]3 years ago
4 0

Answer:

Add the table and i might answer correctly

Step-by-step explanation:

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*PLEASE HELP ME* (math attachment) IT'S PAST MIDNIGHT I'M TIRED AND DESPERATE
Luda [366]
<h3><u>Answer:</u></h3>

D

<h3><u>Step-by-step explanation:</u></h3>

let's rewrite this

x + y = 11                  <em>- equation 1</em>

   - x = - y - 9           <em>- equation 2</em>

solve the first equation for x

x + y = 11            (add -y to both sides to get x by itself)

     x = -y + 11          <em> - equation 3</em>

now substitute -y + 11 for x in the second equation

 y - 11 = - y - 9        (2 negative signs next to each other equals positive)

2y - 11 = -9              (add y to both sides)

      2y = 2               (add 11 to both sides)

        y = 1                (divide both sides by 2)

now substitute y = 1 into equation 3

x = -1 + 11

x = 10

therefore, x = 10 and y = 1. in coordinates, x goes before y

therefore, the coordinates are (10, 1)

therefore the answer is D

good luck my friend! hope this helped a lot! also, i found a website if you don't fully understand this, its called MathPapa, i didn't use it to work this out so my working might be different but it definitely helps me sometimes. for an equation like this, type it likes this: x+y=11; -x=-y-9

-cheesetoasty

3 0
3 years ago
Read 2 more answers
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
HELP ME PLEASE I'LL PICKA BRAINLIEST PERSON!!!
Kamila [148]
Its C! Lets say x=30
|30-323|=50
This could technically be true because it would show if the temp is within range
6 0
3 years ago
Read 2 more answers
Can someone help? I’ll give brainliest!
Shkiper50 [21]

Answer:

V=65.94 in^3 (inches cubed)

Step-by-step explanation:

V=πr^2h/3

h = 7

r = 3

V = (3.14)(3^2)(7/3)

V= (3.14)(9)(2.333)

V=65.94 in^3

4 0
3 years ago
Read 2 more answers
!!!Help me please :)!!!
Paraphin [41]

Answer:

1/8

Step-by-step explanation:

Chance of getting heads: 1/2

Chance of getting orange marble; 1/4

OMG this is so hard to explain.

basically, you have a 1/2 chance of getting Heads. from there, you have to multiply your chance of getting heads by 1/4. This gives us 1/8. We have a 1/8 chance of getting both heads and the orange marble.

           H                                                         T

     B  P  G  O                                           B  P  G  O

These are the charts that you use to solve this.

Hopefully this helped. Good luck!

8 0
3 years ago
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