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3 years ago

How many solutions does the system of equations below have? y = 6x − 7 y = 5 3 x + 1 3

Mathematics

1 answer:

True [87]3 years ago

4 0

Answer:

There will be 1 solution to the system of equations.

Step-by-step explanation:

y = 6x - 7

y = 53x + 13

Since the two equations have different slopes and different y-intercepts, there will be 1 solution to the system of equations.

tadah!

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Vicki started jogging the first time she ran she ran 3/16 mile the second time she ran 3/8 mile and the third time she ran 9/16

IrinaK [193]

Answer:

Jogging 6th time.

Step-by-step explanation:

We have been given that Vicki started jogging the first time she ran she ran 3/16 mile the second time she ran 3/8 mile and the third time she ran 9/16 mile.

We can see that the distance Vicki covers each time forms a arithmetic sequence, where 1st term is 3/16.

We know that an arithmetic sequence is in form $a_n=a_1+(n-1)d$ , where,

$a_n$ = nth term of sequence,

$a_1$ = 1st term of sequence,

n = Number of terms in sequence,

d = Common difference.

Let us find common difference of our given sequence as:

$\frac{3}{8}-\frac{3}{16}\Rightarrow \frac{6}{16}-\frac{3}{16}=\frac{3}{16}$

Since Vicki needs to cover more than 1 mile, so we nth term of sequence should be greater than 1.

$1$

Let us solve for n.

$1$

$1\cdot \frac{16}{3}$

$5.333$

$n>5.333$

We can also write next terms of our sequence as:

$\frac{3}{16},\frac{6}{16}, \frac{9}{16},\frac{12}{16},\frac{15}{16},\frac{18}{16}$

Therefore, Vicki will run more than 1 mile when she is jogging for 6th time.

7 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

Points J and K lie in plane H. Plane H contains two points. Point J is to the lower left and point K is to the uppercase right.

xxMikexx [17]

Answer:

Step-by-step explanation:

There's a theorem that says "through 2 points it only cross one and only one line". Having this as our start point, we can guarantee that there's only one line that can be drawn through those points

4 0

3 years ago

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Translate the triangle left 1 units and up 5 units. What are the coordinates of point C after the translation?

Step2247 [10]

Answer:

b++nfr

Step-by-step explanation:

jjkiki

5 0

3 years ago

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A brand new filled jar of salsa is 11 cm tall and a has a radius of 6 cm. Kelly eats some of the salsa and the salsa in the jar

Sonbull [250]

Before we begin please note that a jar is always cylindrical in shape if not mentioned otherwise. Thus, we will take a cylindrical jar in this case.

Since Kelly eats some of the salsa and the salsa in the jar is now 6 cm high, we know that Kelly ate the amount of Salsa whose volume is missing from the jar.

To find the missing volume on the Salsa jar all that we need to do is to find the height of the missing volume which is given to us as h= 6 cm. We already know that the radius of the jar, r=6 cm. Thus, the amount of Salsa Kelly has eaten is:

$V=\pi(6)^2\times 6\approx678.6\approx679$ $cm^3$

Thus, the third option is the correct answer

8 0

3 years ago

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