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dybincka [34]
3 years ago
8

Identify the slope and y-intercept of the line. y = 4x − 11 slope= y-intercept (x, y) =

Mathematics
1 answer:
MArishka [77]3 years ago
5 0

Step-by-step explanation:

Equation of a line is y = mx + c

where

m is the slope

c is the y intercept

From the question the equation is

y = 4x − 11

Comparing with the general equation above

<h3>Slope = 4</h3><h3>y - intercept = - 11 or ( 0 , - 11)</h3>

Hope this helps you

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1. If &lt;1 is 120 degrees , what is the measure of &lt;2?
natali 33 [55]

Answer:

240

Step-by-step explanation:

2 is 1 + 1 so 120 + 120 must be the equivalent of 2.

5 0
3 years ago
Pls, can someone help me with this problem?
ExtremeBDS [4]

Answer:

0.\overline{33}

Step-by-step explanation:

If you divide number 1 by number 3, you will notice that number 3 will constantly repeat at the decimal part.

So, the answer is 0.\overline{33}.

5 0
2 years ago
Hey yall like dat sweet algeba? then 5ab - 2c where a = 2 b = 3 and c is 0. gud luck daddio
nordsb [41]

Answer:

30

Step-by-step explanation:

5ab-2c

5(2)(3)-2(0)

5(6)-2(0)

30-0

30

7 0
3 years ago
If the discriminant is less than zero than the equation has ______________ solutions.
aleksley [76]

Answer:

The discriminant is the part under the square root in the quadratic formula, b²-4ac. If it is more than 0, the equation has two real solutions. If it's less than 0, there are no solutions. If it's equal to 0, there is one solution.

Brainliest please

Step-by-step explanation:

7 0
3 years ago
100 points , please help. I am not sure if I did this correct if anyone can double-check me thanks!
Nookie1986 [14]

Step-by-step explanation:

\lim_{n \to \infty} \sum\limits_{k=1}^{n}f(x_{k}) \Delta x = \int\limits^a_b {f(x)} \, dx \\where\ \Delta x = \frac{b-a}{n} \ and\ x_{k}=a+\Delta x \times k

In this case we have:

Δx = 3/n

b − a = 3

a = 1

b = 4

So the integral is:

∫₁⁴ √x dx

To evaluate the integral, we write the radical as an exponent.

∫₁⁴ x^½ dx

= ⅔ x^³/₂ + C |₁⁴

= (⅔ 4^³/₂ + C) − (⅔ 1^³/₂ + C)

= ⅔ (8) + C − ⅔ − C

= 14/3

If ∫₁⁴ f(x) dx = e⁴ − e, then:

∫₁⁴ (2f(x) − 1) dx

= 2 ∫₁⁴ f(x) dx − ∫₁⁴ dx

= 2 (e⁴ − e) − (x + C) |₁⁴

= 2e⁴ − 2e − 3

∫ sec²(x/k) dx

k ∫ 1/k sec²(x/k) dx

k tan(x/k) + C

Evaluating between x=0 and x=π/2:

k tan(π/(2k)) + C − (k tan(0) + C)

k tan(π/(2k))

Setting this equal to k:

k tan(π/(2k)) = k

tan(π/(2k)) = 1

π/(2k) = π/4

1/(2k) = 1/4

2k = 4

k = 2

8 0
3 years ago
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