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3 years ago

Determine the standard form of the equation of the line that passes through (-7,8) and (0,2)

Mathematics

1 answer:

SpyIntel [72]3 years ago

6 0

B is the right answer according to the result of the equation. "Thanks me" if i've been helpful

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Which polynomial correctly combines the like terms and puts the given polynomial in standard form?

Minchanka [31]

Answer: $-5x^3y^3+8x^4y^2-7xy^5+x^2y^4+8x^6$

Step-by-step explanation:

$-5x^3y^3+8x^4y^2-xy^5-2x^2y^4+8x^6+3x^2y^4-6xy^5$

$-5x^3y^3+8x^4y^2+(-xy^5-6xy^5)+(-2x^2y^4+3x^2y^4)+8x^6$

$-5x^3y^3+8x^4y^2+(-7xy^5)+(x^2y^4)+8x^6$

$-5x^3y^3+8x^4y^2-7xy^5+x^2y^4+8x^6$

8 0

3 years ago

Read 2 more answers

Question 6 (1 point)

vlada-n [284]

Answer:

Zero

Step-by-step explanation:

3x - 7 = 3x - 9 + 2

3x - 7 = 3x - 7

Add 7 on both sides

3x = 3x

Subtract 3x you get 0.

Hope this helps.

4 0

3 years ago

In the interval 0º < x < 360°, find the values of x for which tan x= -0.6745

SSSSS [86.1K]

Tan(θ) = 3 tan(θ), 0° ≤ θ ≤ 360°

Solve for θ to the nearest degree.

tan(θ) = 3 tan(θ)

Subtract tan(θ) from both sides:

0 = 2 tan(θ)

Divide by 2 both sides:

tan(θ) = 0

If (x,y) is a point on the terminal ray of θ,

then tan(θ) = y/x = 0, and y = 0.

y = 0 ==> θ = 0°, 180°, or 360° in the interval 0° ≤ θ ≤ 360°.

7 0

3 years ago

A Cepheid variable star is a star whose brightness alternately increases and decreases. For a certain star, the interval between

sattari [20]

Answer:

$B'(t) = \dfrac{0.9\pi}{4.4}\cos\bigg(\dfrac{2\pi t}{4.4}\bigg)$

b) 0.09

Step-by-step explanation:

We are given the following in the question:

$B(t) = 4.2 +0.45\sin\bigg(\dfrac{2\pi t}{4.4}\bigg)$

where B(t) gives the brightness of the star at time t, where t is measured in days.

a) rate of change of the brightness after t days.

$B(t) = 4.2 +0.45\sin\bigg(\dfrac{2\pi t}{4.4}\bigg)\\\\B'(t) = 0.45\cos\bigg(\dfrac{2\pi t}{4.4}\bigg)\times \dfrac{2\pi}{4.4}\\\\B'(t) = \dfrac{0.9\pi}{4.4}\cos\bigg(\dfrac{2\pi t}{4.4}\bigg)$

b) rate of increase after one day.

We put t = 1

$B'(t) = \dfrac{0.9\pi}{4.4}\cos\bigg(\dfrac{2\pi t}{4.4}\bigg)\\\\B'(1) = \dfrac{0.9\pi}{4.4}\bigg(\cos(\dfrac{2\pi (1)}{4.4}\bigg)\\\\B'(t) = 0.09145\\B'(t) \approx 0.09$

The rate of increase after 1 day is 0.09

8 0

3 years ago

After the strand of measurement why do you think that graphing and data analysis are easier? Please help ASAP!!! :(

Vikki [24]

It’s organized and you can see it more visually , gives you a way to double check and write down the data properly.

8 0

3 years ago