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3 years ago

What is the equation of the line that is perpendicular to the

Mathematics

1 answer:

Bas_tet [7]3 years ago

7 0

Answer:

y=3x-5

Step-by-step explanation:

If the new line is perpendicular to the original line, they the slope for the new line should be the reciprocal and opposite of the original line. Since the original line's slope is -1/3, the slope of the new line is 3, which is the coefficient in the new equation representing the new line. Then, you create an equation to solve for "b" in the equation y=mx+b where "b" is the y-intercept. You plug in 3 for x, 4 for y, and 3 for m (you know that 3 is x and 4 is y because it says that the line must past (3,4). So the equation to solve for b should be "4=3(3)+b". Then for b you should get -5.

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The given line passes through the points (0, −3) and (2, 3).

Alex

Answer:

The equation would be y + 1 = 3(x + 1)

Step-by-step explanation:

In order to find the equation of this line, we have to use the slope formula to find the slope.

m (slope) = (y2 - y1)/(x2 - x1)

m = (3 - -3)/(2 - 0)

m = (3 + 3)/(2 - 0)

m = 6/2

m = 3

Now that we have that, we can use it along with the second point to put into the point-slope equation.

y - y1 = m(x - x1)

y - -1= 3(x - -1)

y + 1 = 3(x + 1)

6 0

3 years ago

A city's population increases by 5% every year. If its population today is 300,000, write a function that models the population

ratelena [41]

Answer:

f(x) = 300,000(1.05)^x

Step-by-step explanation:

Use the exponential growth equation:

f(x) = a(1 + r)^x

Plug in 300,000 as a, since it is the original population:

f(x) = 300,000(1 + r)^x

The city's population is growing at 5% each year, so plug in 0.05 as r:

f(x) = 300,000(1 + 0.05)^x

f(x) = 300,000(1.05)^x

So, the function for the population growth is f(x) = 300,000(1.05)^x

6 0

3 years ago

ILL GIVE MEDAL AND BRAINLIST AWNSER TO whoEVER IN THE FIRST!!!!Roland's heart beats 495 times in 6 minutes. At that rate, how ma

Sonbull [250]

Answer,

C. 1650

Explanation,

495/6 = 82.5

82.5 x 20 = 1650

Hope this helps :-)

7 0

3 years ago

Show work and explain with formulas:

Marina86 [1]

17 Answer: 205

Step-by-step explanation:

$\{1\dfrac{2}{3}+1\dfrac{5}{6}+2+...+8\dfrac{1}{3}\}\implies a_1=1\dfrac{2}{3},\ d=\dfrac{1}{6}\\\\\\a_n=a_1+d(n-1)\qquad solve\ for\ n\\\\8\dfrac{1}{3}=1\dfrac{2}{3}+\dfrac{1}{6}(n-1)\\\\\\\dfrac{25}{3}=\dfrac{5}{3}+\dfrac{1}{6}n-\dfrac{1}{6}\\\\\\\dfrac{50}{6}=\dfrac{10}{6}+\dfrac{1}{6}n-\dfrac{1}{6}\\\\\\\dfrac{41}{6}=\dfrac{1}{6}n\\\\\\\dfrac{41}{6}\cdot 6=n\\\\41=n$

$\text{Now use the sum formula:}\\\\S_n=\dfrac{a_1+a_n}{2}\cdot n\\\\\\S_{41}=\dfrac{1\dfrac{2}{3}+8\dfrac{1}{3}}{2}\cdot 41\\\\\\.\quad =\dfrac{10}{2}\cdot 41\\\\\\.\quad =5\cdot 41\\\\.\quad =\large\boxed{205}$

18 Answer: 1968

Step-by-step explanation:

$a_1=-6,\ d=2,\ n=48, \quad \text{solve for }a_{48}\\\\a_{n}=a_1+d(n-1)\\\\a_{48}=-6+2(48-1)\\\\.\quad =-6+2(47)\\\\.\quad =-6+94\\\\.\quad =88$

$\text{Now use the sum formula:}\\\\S_n=\dfrac{a_1+a_n}{2}\cdot n\\\\\\S_{48}=\dfrac{-6+88}{2}\cdot 48\\\\\\.\quad =\dfrac{82}{2}\cdot 48\\\\\\.\quad =41\cdot 48\\\\.\quad =\large\boxed{1968}$

19 Answer: -116

Step-by-step explanation:

$\{3, -2, -7, ...\}\\a_1=3,\ d=-5,\ n=8, \quad \text{solve for }a_{8}\\\\a_{n}=a_1+d(n-1)\\\\a_{8}=3-5(8-1)\\\\.\quad =3-5(7)\\\\.\quad =3-35\\\\.\quad =-32\\\\\text{Now use the sum formula:}\\\\S_8=\dfrac{a_1+a_8}{2}\cdot 8\\\\\\S_{8}=\dfrac{3-32}{2}\cdot 8\\\\\\.\quad =\dfrac{-29}{2}\cdot 8\\\\\\.\quad =-29\cdot 4\\\\.\quad =\large\boxed{-116}$

6 0

3 years ago

The graph shows a probability distribution. What is P(1.5 ≤ X ≤ 3)? Enter your answer, as a decimal, in the box.

Bumek [7]

The probability from 1.5 ≤ x ≤ 3 can be calculated by dividing the Area from x=1.5 to x=3 by the total Area of the distribution.

The given distribution is rectangular shaped, so its Area will be = Length x Width = 1 x 3 = 3 square units

From x = 1.5 to x = 3, the length is 1.5 and width is 1. So the area between these two intervals = 1.5 square units.

Thus, P(1.5 ≤ X ≤ 3) = 1.5/3 = 0.5

5 0

3 years ago