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Igoryamba
2 years ago
14

Need help finding the measure of the arc or angle indicated

Mathematics
1 answer:
evablogger [386]2 years ago
6 0

Answer:

70 degrees

Step-by-step explanation:

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Which are solutions of x2=-11+4?
Dovator [93]

For this case we have a quadratic equation of the form:

ax ^ 2 + bx + c = 0\\x ^ 2 + 11x-4 = 0

Where:

a = 1\\b = 11\\c = -4

We find the roots:

x = \frac {-b \pm \sqrt {b ^ 2-4 (a) (c)}} {2 (a)}\\x = \frac {-11 \pm \sqrt {11 ^ 2-4 (1) (- 4)}} {2 (1)}\\x = \frac {-11 \pm \sqrt {121 + 16}} {2}\\x = \frac {-11 \pm \sqrt {137}} {2}

The roots are:

x_ {1} = \frac {-11+ \sqrt {137}} {2}\\x_ {2} = \frac {-11- \sqrt {137}} {2}

Answer:

x_ {1} = \frac {-11+ \sqrt {137}} {2}\\x_ {2} = \frac {-11- \sqrt {137}} {2}

3 0
3 years ago
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Which table does NOT represent a function?<br> A) A <br> B) B <br> C) C <br> D) D
atroni [7]

Answer: The answer is D

Step-by-step explanation: I just did the question.

7 0
3 years ago
It takes Evan 6 3/4 hours to mow 3 lawns. it takes him 2 1/3 hours to mow Mr. Smiths yard and 1 3/4 hours to mow Ms. Lee's yard.
Ratling [72]

Answer: 2\frac{2}{3}\ hours

Step-by-step explanation:

Given: The total time taken by Evan to mow the 3 lawns = 6\frac{3}{4}=\frac{27}{4}\ \text{hours}

Time taken by Evan to mow the first lawn =2\frac{1}{3}=\frac{7}{3}\ \text{hours}

Time taken by Evan to mow the second lawn =1\frac{3}{4}=\frac{7}{4}\ \text{hours}

Now,Time taken by Evan to mow the third lawn =\frac{27}{4}-\frac{7}{3}-\frac{7}{4}

⇒Time taken by Evan to mow the third lawn =\frac{81-28-21}{12}

⇒Time taken by Evan to mow the third lawn =\frac{32}{12}=\frac{8}{3}=2\frac{2}{3}\ hours

Hence, the time taken by Evan to mow the third lawn = 2\frac{2}{3}\ hours

7 0
3 years ago
Find the solution for 2-x/3=5
PtichkaEL [24]

Answer:

7

Step-by-step explanation:

6 0
3 years ago
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PLEASE HELP I’LL MAKE YOU THE BRAINIEST <br> find the area of the trapezoid to the nearest tenth
murzikaleks [220]

Answer:

2.7m

Step-by-step explanation:

1.0m

0.5m

+1.2m

2.7m - answer

5 0
2 years ago
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