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mario62 [17]
2 years ago
12

Can somebody help me? -2-7p<26

Mathematics
1 answer:
LekaFEV [45]2 years ago
4 0

Answer: p>−4

Step-by-step explanation:

Let's solve your inequality step-by-step.

−2−7p<26

Step 1: Simplify both sides of the inequality.

−7p−2<26

Step 2: Add 2 to both sides.

−7p−2+2<26+2

−7p<28

Step 3: Divide both sides by -7.

−7p

−7

<

28

−7

p>−4

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Answer:

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3x3+4x2−2x−7 = 10-2x

3x3+4x4−2x2−7 = 14

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Step-by-step explanation:

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2 years ago
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#9. Find the slope of the line in the graph.
jekas [21]
The answer to this problem is -2/2
7 0
2 years ago
Evaluate the expression: |8| - 2 x |-3| + 4<br><br> 22<br> 10<br> 42<br> 6
JulsSmile [24]

Answer:

−6x+12

Step-by-step explanation:

3 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
Find the number which when divided by 4 and increased by 12 is the same as when it is divided by 3 and decreased by 5
MatroZZZ [7]

Step-by-step explanation:

Let's make the number we're finding to be a. So, it says that it will be divided by 4 and then added by 12 if we want this in algebraic expression it will be \frac{a}{4} +12\\. It's also telling us that that expression is the same thing as our number divided by 3 and subtracted by 5. If we want an expression out of it it well be \frac{a}{3} -5\\. Since they are the same, we have the equation below.

\frac{a}{4} +12 = \frac{a}{3} -5\\

All we have to do now is to find a.

\frac{a}{4} +12 = \frac{a}{3} -5 \\ \frac{a +48}{4} = \frac{a -15}{3} \\ (3)(a +48) = (a - 15)(4) \\ 3a +144 = 4a -60 \\ (60 -3a) + 3a +144 = 4a -60 +(60 -3a) \\ a = 204

<h3>Answer:</h3>

Our number must be 204. I think

5 0
3 years ago
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