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3 years ago

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It was once believed that all of the bodies of the solar system were perfect spheres with absolutely nothing on their surfaces.

How did Galileo disprove this notion in the early 1600s?
A.
He used a gyroscope to measure the gravity of Jupiter.

B.
He used a rocket to measure the width of Earth's atmosphere.

C.
He used a spectroscope to observe the light from the Sun.

D.
He used a telescope to observe mountains on the Moon.

1 answer:

Lady bird [3.3K]3 years ago

6 0

Answer:

The answer is d.

Step-by-step explanation:

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Please can someone help me with this question

sasho [114]

<em>Answer:</em>

<em>r = -</em> $\frac{6a}{1-5m^2}$ <em />

<em>Step-by-step explanation:</em>

<em>Rewrite the equation as </em> $\sqrt{x} 6a+r/5r$ <em> = m</em>

<em>Remove the radical on the left side of the equation by squaring both sides of the equation.</em>

<em>(</em> $\sqrt{\frac{6a+r}{5r} )^{2}$ <em> = m^2</em>

<em>Then, you simplify each of the equation. </em>

<em>Rewrite: (</em> $\sqrt{\frac{6a+r}{5r} )^{2}$ <em> as </em> $\frac{6a+r}{5r} = m^{2}$ <em />

<em>Remove any parentheses if needed.</em>

<em>Solve for r. </em>

<em>Multiply each term by r and simplify."</em>

<em>Multiply both sides of the equation by 5.</em>

<em>6a+r= m^2r⋅(5)</em>

<em>Remove parentheses.</em>

<em>Move 5 to the left of (m ^2) r </em>

<em>6a+r=5m^2)r</em>

<em>Subtract 5m^2)r from both sides of the equation.</em>

<em>6a+r-5m^2)r=0</em>

<em>Subtract 6a from both sides of the equation.</em>

<em>r-5m^2)r=-6a</em>

<em>Factor r out of r-5m^2)r </em>

<em>r(1-5m^2)=-6a</em>

Divide each term by 1-5m^2 and simplify.

r = - $\frac{6a}{1-5m^2}$

There you go, hope this helps!

4 0

3 years ago

PLEASE HELP! WILL GIVE BRAINLIEST!!

Maslowich

Your answer is -6+3i.

6 0

3 years ago

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Which linear function has a negative slope, f(x) or g(x)?

Anit [1.1K]

Answer:

f(x) is negative and g(x) is positive

Step-by-step explanation:

f(x) = 4-x

Rewriting

f(x) = -x+4

We recognize this as

y= mx+b or the representation of a line with slope -1 and y intercept 4

f(x) is linear

g(x) as we increase x by 1 we increase g(x) by 2

The slope is 2, and the y intercept is 1

y = 2x+1

Lets check a point

(2,5)

5 = 2(2)+1

5=5 so we are correct

f(x) is negative and g(x) is positive

8 0

3 years ago

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PLSSS HELP MEEEEEEEEE

Tpy6a [65]

2^3:
a) 2 x 2 x 2
b) 8

3^2 x e^3:
a) 3 x 3 x e x e x e
b) 9e^3

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3 years ago

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Dilate the the triangle, scale factor = 3 (positive 3, ignore the

HACTEHA [7]

Answer:

The rule of dilation is $P'(x,y) = 3\cdot P(x,y)$ .

The vertices of the dilated triangle are $A'(x,y) = (-21, -18)$ , $B'(x,y) = (-15,-10)$ and $C'(x,y) = (-3,-15)$ , respectively.

Step-by-step explanation:

From Linear Algebra, we define the dilation by the following definition:

$P'(x,y) = O(x,y) + k\cdot[P(x,y)-O(x,y)]$ (1)

Where:

$O(x,y)$ - Center of dilation, dimensionless.

$k$ - Scale factor, dimensionless.

$P(x,y)$ - Original point, dimensionless.

$P'(x,y)$ - Dilated point, dimensionless.

If we know that $O(x,y) = (0,0)$ , $k = 3$ , $A(x,y) = (-7,-6)$ , $B(x,y) = (-5,-2)$ and $C(x,y) =(-1,-5)$ , then dilated points of triangle ABC are, respectively:

$A'(x,y) = O(x,y) + k\cdot [A(x,y)-O(x,y)]$ (2)

$A'(x,y) = (0,0) + 3\cdot [(-7,-6)-(0,0)]$

$A'(x,y) = (-21, -18)$

$B'(x,y) = O(x,y) + k\cdot [B(x,y)-O(x,y)]$ (3)

$B'(x,y) = (0,0) + 3\cdot [(-5,-2)-(0,0)]$

$B'(x,y) = (-15,-10)$

$C'(x,y) = O(x,y) + k\cdot [C(x,y)-O(x,y)]$ (4)

$C'(x,y) = (0,0) +3\cdot [(-1,-5)-(0,0)]$

$C'(x,y) = (-3,-15)$

The rule of dilation is:

$P'(x,y) = 3\cdot P(x,y)$ (5)

The vertices of the dilated triangle are $A'(x,y) = (-21, -18)$ , $B'(x,y) = (-15,-10)$ and $C'(x,y) = (-3,-15)$ , respectively.

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2 years ago