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3 years ago

A line passes through the points (5, –7.5) and has a y-intercept 10.

Mathematics

1 answer:

andrezito [222]3 years ago

4 0

Answer: y = -3.5x + 10 so C.

Step-by-step explanation: We’re in same class and I got that right :) lmk if you need help or answers .

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4. Using the geometric sum formulas, evaluate each of the following sums and express your answer in Cartesian form.

nikitadnepr [17]

Answer:

$\sum_{n=0}^9cos(\frac{\pi n}{2})=1$

$\sum_{k=0}^{N-1}e^{\frac{i2\pi kk}{2}}=0$

$\sum_{n=0}^\infty (\frac{1}{2})^n cos(\frac{\pi n}{2})=\frac{1}{2}$

Step-by-step explanation:

$\sum_{n=0}^9cos(\frac{\pi n}{2})=\frac{1}{2}(\sum_{n=0}^9 (e^{\frac{i\pi n}{2}}+ e^{\frac{i\pi n}{2}}))$

$=\frac{1}{2}(\frac{1-e^{\frac{10i\pi}{2}}}{1-e^{\frac{i\pi}{2}}}+\frac{1-e^{-\frac{10i\pi}{2}}}{1-e^{-\frac{i\pi}{2}}})$

$=\frac{1}{2}(\frac{1+1}{1-i}+\frac{1+1}{1+i})=1$

2nd

$\sum_{k=0}^{N-1}e^{\frac{i2\pi kk}{2}}=\frac{1-e^{\frac{i2\pi N}{N}}}{1-e^{\frac{i2\pi}{N}}}$

$=\frac{1-1}{1-e^{\frac{i2\pi}{N}}}=0$

3th

$\sum_{n=0}^\infty (\frac{1}{2})^n cos(\frac{\pi n}{2})==\frac{1}{2}(\sum_{n=0}^\infty ((\frac{e^{\frac{i\pi n}{2}}}{2})^n+ (\frac{e^{-\frac{i\pi n}{2}}}{2})^n))$

$=\frac{1}{2}(\frac{1-0}{1-i}+\frac{1-0}{1+i})=\frac{1}{2}$

What we use?

We use that

$e^{i\pi n}=cos(\pi n)+i sin(\pi n)$

and

$\sum_{n=0}^k r^k=\frac{1-r^{k+1}}{1-r}$

6 0

3 years ago

Find the solution to the inequality -0.3x-5<6.7.

I am Lyosha [343]

Answer:

x > -39

Step-by-step explanation:

-0.3x < 6.7 + 5

-0.3x < 11.7

divide by -0.3

Answer is x > -39

6 0

3 years ago

Tyler’s brother works in a shoe store. The store was selling a pair of shoes for $80. They now are on sale for $50. What is the

Sedbober [7]

Answer:

0.3%

Step-by-step explanation:

80-50=30

30÷100

=0.3%

5 0

2 years ago

Calculate pt3 such that a line from pt1 to pt3 is perpendicular to the line from pt1 to pt2, and the distance between pt1 and pt

Leni [432]

Let the point_1 = p₁ = (1,4)
and      point_2 = p₂ = (-2,1)
and      Point_3 = p₃ = (x,y)

The line from point_1 to point_2 is L₁ and has slope = m₁
The line from point_1 to point_3 is L₂ and has slope = m₂
m₁ = Δy/Δx = (1-4)/(-2-1) = 1
m₂ = Δy/Δx = (y-4)/(x-1)
L₁⊥L₂ ⇒⇒⇒⇒ m₁ * m₂ = -1
∴ (y-4)/(x-1) = -1 ⇒⇒⇒ (y-4)= -(x-1)
(y-4) = (1-x) ⇒⇒⇒⇒⇒⇒⇒⇒⇒⇒⇒⇒⇒⇒ equation (1)

The distance from point_1 to point_2 is d₁
The distance from point_1 to point_3 is d₂
d = $\sqrt{Δx^2+Δy^2}$
d₁ = $\sqrt{(-2-1)^2+(1-4)^2}$
d₂ = $\sqrt{(x-1)^2+(y-4)^2}$
d₁ = d₂
∴ $\sqrt{(-2-1)^2+(1-4)^2} = \sqrt{(x-1)^2+(y-4)^2}$ ⇒⇒ eliminating the root
∴(-2-1)²+(1-4)² = (x-1)²+(y-4)²
(x-1)²+(y-4)² = 18
from equatoin (1) y-4 = 1-x
∴(x-1)²+(1-x)² = 18            ⇒⇒⇒⇒⇒ note: (1-x)² = (x-1)²
2 (x-1)² = 18
(x-1)² = 9
x-1 = $\pm \sqrt{9} = \pm 3$
∴ x = 4 or x = -2
∴ y = 1 or y = 7

Point_3 = (4,1) or (-2,7)

8 0

3 years ago

Can someone please help me .?

MariettaO [177]

Answer:

Step-by-step explanation:

5+7

4 0

3 years ago