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geniusboy [140]
3 years ago
6

Write an equation parallel to y=3x−1 y=3x−1 and goes through the point (-2, -13)

Mathematics
1 answer:
Lena [83]3 years ago
3 0

Answer:

y=3x-(-2,-13)

Step-by-step explanation:

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"In ABC, mBAC = 5x + 8, mABC = 6x + 22, and mBCA = 3x  25.
strojnjashka [21]
<span>In angle ABC, the measure of angle BAC = 5x + 8, </span>measure of angle ABC = 6x + 22, and measure of angle BCA = 3x  25. You know that the sum of the angles of a triangle is 180 degrees. And since the angles BAC, ABC and BCA are the corners of the triangle then just add them and equate to 180 degrees.

5x + 8 + 6x + 22 + 3x + 25 = 180
14x + 55 = 180
14x = 125
x = 8.93
5 0
3 years ago
Read 2 more answers
a random sample of size 100 is taken from a population described by the proportion P equals 0.60 using the appropriate normal tr
Nataliya [291]

Answer:

767

Step-by-step explanation:

5 0
3 years ago
I NEED HELP IDENTIFYING THE SLOPE PLEASE !!! Pic attatched !!!!
Alexus [3.1K]

Answer:

see explanation

Step-by-step explanation:

Calculate the slope m using the slope formula

m = (y₂ - y₁ ) / (x₂ - x₁ )

with (x₁, y₁ ) = 2,4 ) and (x₂, y₂ ) = (5, 1)

m = \frac{1-4}{5-2}  = \frac{-3}{3} = - 1 ← negative slope

Repeat

with (x₁, y₁ ) = - 7,8) and (x₂, y₂ ) = (- 7, 0)

m = \frac{0-8}{-7+7} = \frac{-8}{0}

Division by zero is undefined, thus slope is undefined.

Repeat with

(x₁, y₁ ) = (6, - 3) and (x₂, y₂ ) = (- 4, - 3)

m = \frac{-3+3}{-4-6} = \frac{0}{-10} = 0 ← zero slope

Repeat with

(x₁, y₁ ) = (3, 5) and (x₂, y₂ ) = (- 1, 2)

m = \frac{2-5}{-1-3} = \frac{-3}{-4} = \frac{3}{4} ← positive slope

7 0
3 years ago
Tell whether this sequence is arithmatic. If it is, what is the common difference. 10, 15, 21, 28
Vlad [161]

Answer:

No this is not arithmetic this is geometric

Step-by-step explanation:

10 to 15 is a increase of 5

15 to 21 is a increase of 6

21 to 28 is a increase of 7 arithmetic have to have steady increases

8 0
3 years ago
How do you do this question?
Gennadij [26K]

Step-by-step explanation:

Find the next term of the Taylor series.

f⁽⁰⁾(x) = x^⅓

f⁽¹⁾(x) = ⅓ x^-⅔

f⁽²⁾(x) = -²/₉ x^-⁵/₃

f⁽³⁾(x) = ¹⁰/₂₇ x^-⁸/₃

f⁽⁴⁾(x) = -⁸⁰/₈₁ x^-¹¹/₃

So the fourth term would be:

(-⁸⁰/₈₁ z^-¹¹/₃) (x − 1)⁴ / 4!

For 0.8 ≤ z ≤ 1.2, │f⁽⁴⁾(z)│is a maximum at z = 0.8.  Therefore:

│R₃(x)│≤ │(-⁸⁰/₈₁ (0.8)^-¹¹/₃) (0.8 − 1)⁴ / 4!│

│R₃(x)│≤ 0.00014923

Looks like you accidentally wrote an extra zero.

7 0
3 years ago
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