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3 years ago

Need help ASSAAAPPPPPPP

Mathematics

1 answer:

lilavasa [31]3 years ago

7 0

Answer:

Help with what... there is no question

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What is the conclusion in this conditional statement?

Trava [24]

Answer:

Yes

Step-by-step explanation:

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3 years ago

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. 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

Put the following equation of a line into slope-intercept form, simplifying all

Kipish [7]

Answer:

y=1/4x−6

Step-by-step explanation:

Write in slope-intercept form, y = m x + b .

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3 years ago

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What is 22/9×.13 repeating?

astraxan [27]

0.31777777777. I am pretty sure. I used my calculator

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3 years ago

Unit 10: circles homework 2: central angles & arc measures

tresset_1 [31]

The central angle (127 degrees) is the angle at point K

The measures of JL and JML are 127 and 233 degrees, respectively

<h3>How to determine the measures of angles JL and JML?</h3>

From the complete question, we have:

JL = 127 degrees.

The sum of angles at a point is 360 degrees

So, we have:

JML + 127 = 360

Subtract 127 from both sides

JML = 233

Hence, the measures of JL and JML are 127 and 233 degrees, respectively

Read more about circles and arcs at:

brainly.com/question/25305793

4 0

2 years ago