During a solar eclipse, what is causing the earth’s view of the sun to be darkened?

2 answers:

8 0

A solar eclipse occurs when the moon gets between Earth and the sun, and the moon casts a shadow over Earth. A solar eclipse can only take place at the phase of new moon, when the moon passes directly between the sun and Earth and its shadows fall upon Earth's surface.

Serggg [28]3 years ago

7 0

Answer:

the moon blocks light from the sun by coming between the earth and the sun

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Write an equation in point-slope form of the line that passes through the point 4,-1and has slope of -5

Basile [38]

Answer:

y=-5+19

Step-by-step explanation:

First of all, remember what the equation of a line is:

y = mx+b

Where:

m is the slope, and

b is the y-intercept

To start, you know what m is; it's just the slope, which you said was -5. So you can right away fill in the equation for a line somewhat to read:

y=-5x+b.

Now, what about b, the y-intercept?

To find b, think about what your (x,y) point means:

(4,-1). When x of the line is 4, y of the line must be -1.

Because you said the line passes through this point, right?

Now, look at our line's equation so far: . b is what we want, the -5 is already set and x and y are just two "free variables" sitting there. We can plug anything we want in for x and y here, but we want the equation for the line that specfically passes through the the point (4,-1).

So, why not plug in for x the number 4 and for y the number -1? This will allow us to solve for b for the particular line that passes through the point you gave!.

(4,-1). y=mx+b or -1=-5 × 4+b, or solving for b: b=-1-(-5)(4). b=19.

6 0

3 years ago

MATH HELP!!! 100PTS!!!

Lostsunrise [7]

Answer:

$x =\dfrac{20 +\sqrt{454}}{18}, \quad \dfrac{20 -\sqrt{454}}{18}$

Step-by-step explanation:

Given functions:

$p(x)=2x-4$

$r(x)=\dfrac{6x-1}{9x+1}$

Solve for p(x) = r(x):

$\begin{aligned}p(x) & = r(x)\\\implies 2x-4 & = \dfrac{6x-1}{9x+1}\\(2x-4)(9x+1)&=6x-1\\18x^2+2x-36x-4&=6x-1\\18x^2-40x-3&=0\end{aligned}$

As the found quadratic equation cannot be factored, use the Quadratic Formula to solve for x:

Quadratic Formula

$x=\dfrac{-b \pm \sqrt{b^2-4ac} }{2a}\quad\textsf{when }\:ax^2+bx+c=0$

Therefore:

$a=18, \quad b=-40, \quad c=-3$

Substitute the values of a, b and c into the quadratic formula and solve for x:

$\begin{aligned}\implies x & =\dfrac{-(-40) \pm \sqrt{(-40)^2-4(18)(-3)} }{2(18)}\\\\& =\dfrac{40 \pm \sqrt{1816}}{36}\\\\& =\dfrac{40 \pm \sqrt{4 \cdot 454}}{36}\\\\& =\dfrac{40 \pm \sqrt{4}\sqrt{454}}{36}\\\\& =\dfrac{40 \pm2\sqrt{454}}{36}\\\\& =\dfrac{20 \pm\sqrt{454}}{18}\end{aligned}$