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irina [24]
3 years ago
9

Need help please show how you did it.

Mathematics
1 answer:
jolli1 [7]3 years ago
5 0

Answer:

(0, 1).

Method 1 (Substitution):

Substituting our two y's, we get the following:

x^2 + 5x + 1 = x^2 + 2x + 1 \Rightarrow 3x = 0 \Rightarrow x = 0

Thus, the only set of solutions is (0, 1). A quick sketch (either by hand or on Desmos) can confirm this.

Method 2 (Elimination):

We have two equations. We'll let the top one be equation 1 and the bottom one be equation 2. Eliminating as many variables as we can, we subtract (2) from (1) to get:

0 = 3x => x = 0.

So the only set of solutions is (0, 1).

Method 3 (Gaussian elimination):

We can place this in an augmented matrix and row reduce.

\left[\begin{array}{cccc}1&5&1 & 1\\1&2&1 & 1\end{array}\right]

Row reducing this gives us:

\left[\begin{array}{cccc}1&5&1 & 1\\0&3&0 & 0\end{array}\right]

This tells us that the only solution for x is x = 0 (since we read this as "3x = 0") and thus, the only solution we get is (0, 1).

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3 years ago
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Step-by-step explanation:

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3 years ago
. Use the quadratic formula to solve each quadratic real equation. Round
Liono4ka [1.6K]

Answer:

A. No real solution

B. 5 and -1.5

C. 5.5

Step-by-step explanation:

The quadratic formula is:

\begin{array}{*{20}c} {\frac{{ - b \pm \sqrt {b^2 - 4ac} }}{{2a}}} \end{array}, with a being the x² term, b being the x term, and c being the constant.

Let's solve for a.

\begin{array}{*{20}c} {\frac{{ 5 \pm \sqrt {5^2 - 4\cdot1\cdot11} }}{{2\cdot1}}} \end{array}

\begin{array}{*{20}c} {\frac{{ 5 \pm \sqrt {25 - 44} }}{{2}}} \end{array}

\begin{array}{*{20}c} {\frac{{ 5 \pm \sqrt {-19} }}{{2}}} \end{array}

We can't take the square root of a negative number, so A has no real solution.

Let's do B now.

\begin{array}{*{20}c} {\frac{{ 7 \pm \sqrt {7^2 - 4\cdot-2\cdot15} }}{{2\cdot-2}}} \end{array}

\begin{array}{*{20}c} {\frac{{ 7 \pm \sqrt {49 + 120} }}{{-4}}} \end{array}

\begin{array}{*{20}c} {\frac{{ 7 \pm \sqrt {169} }}{{-4}}} \end{array}

\begin{array}{*{20}c} {\frac{{ 7 \pm 13 }}{{-4}}} \end{array}

\frac{7+13}{4} = 5\\\frac{7-13}{4}=-1.5

So B has two solutions of 5 and -1.5.

Now to C!

\begin{array}{*{20}c} {\frac{{ -(-44) \pm \sqrt {-44^2 - 4\cdot4\cdot121} }}{{2\cdot4}}} \end{array}

\begin{array}{*{20}c} {\frac{{ 44 \pm \sqrt {1936 - 1936} }}{{8}}} \end{array}

\begin{array}{*{20}c} {\frac{{ 44 \pm 0}}{{8}}} \end{array}

\frac{44}{8} = 5.5

So c has one solution: 5.5

Hope this helped (and I'm sorry I'm late!)

4 0
3 years ago
Please help me out....................
maxonik [38]

Answer:

x = 180 - 2 * 35 = 110°

y = (180 - 82)/2 = 49°


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The answer to your question is d 354
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