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dimulka [17.4K]
3 years ago
10

Can someone please help me :)

Mathematics
1 answer:
Bas_tet [7]3 years ago
3 0

Answer:

2\sqrt{13}

Step-by-step explanation:

12x12 = 144

14x14 = 196

144 + ? = 196

196 - 144 = 52

\sqrt{52} = 2\sqrt{13}

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I need help due in 5 min
Serhud [2]

Answer:

7

Step-by-step explanation:

3 0
3 years ago
Find the value of the discriminant for <img src="https://tex.z-dn.net/?f=7x%5E%7B2%7D%20%2B5x%2B1%3D0" id="TexFormula1" title="7
kondor19780726 [428]

Answer:

No real roots

Step-by-step explanation:

Given

7x² + 5x + 1 = 0 ← in standard form

with a = 7, b = 5, c = 1

To determine the nature of the roots use the discriminant

Δ = b² - 4ac

• If b² - 4ac > 0 then roots are real and distinct

• If b² - 4ac = 0 then roots are real and equal

• If b² - 4ac < 0 then the roots are not real

Here

b² - 4ac = 5² - (4 × 7 × 1) = 25 - 28 = - 3

Thus the 2 roots are not real

7 0
4 years ago
Hey someone who is good in math please help me with this question
frosja888 [35]
Your answer is 10% hoped this helped
3 0
3 years ago
Read 2 more answers
Question 5 of 10 What is the domain of the function in this table? х y 1 2 2 6 3 10 4 14 O A. (1,2,3,4) B. (1.2).(2.6), (3,10),
Mrac [35]
The answer is C: (1,2)
6 0
3 years ago
Which equation in slope-intercept form represents a line that passes through the point (6,−1) and is perpendicular to the line y
damaskus [11]

For this case we have that by definition, the equation of a line of the slope-intersection form is given by:

y = mx + b

Where:

m: It is the slope of the line

b: It is the cut point with the y axis

By definition, if two lines are perpendicular then the product of their slopes is -1.

If we have: y = 2x-7

m_ {1} = 2\\2 * m 2 = - 1\\m_ {2} = - \frac {1} {2}

Thus, the equation is of the form:

y = - \frac {1} {2} x + b

We substitute the point:

-1 = - \frac {1} {2} (6) + b\\-1 = - \frac {1} {2} (6) + b\\-1 = -3 + b\\-1 + 3 = b\\b = 2

Finally, the equation is:

y = - \frac {1} {2} x + 2

Answer:

y = - \frac {1} {2} x + 2

6 0
3 years ago
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