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1 year ago

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Neptune’s average distance from the sun is 4.503 x 10e9 km. Mercury’s average distance from the sun is 5.791 x10e9 km.about how

many times farther from the sun is Neptune than mercury?

1 answer:

Igoryamba1 year ago

4 0

Answer:

77.76 times

Step-by-step explanation:

The average distance of Neptune from the sun

= 4.503 × 10 ⁹ k m .

and Mercury = 5.791 × 10 ⁷ k m .

Hence neptune is ( 4.503 × 10 ⁹) ÷ (5.791 × 10 ⁷ ) times farther from the sun than mercury

i.e.( $\frac{4.503}{5.791}$ ) × 10⁹⁻⁷ times

= 0.7776 × 10 ² times

= 77.76 times.

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Let P2 be the vector space of all polynomials of degree 2 or less, and let H be the subspace spanned by 10x2+4xâ1, 3xâ4x2+3, and

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I suppose

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The vectors that span $H$ form a basis for $P_2$ if they are (1) linearly independent and (2) any vector in $P_2$ can be expressed as a linear combination of those vectors (i.e. they span $P_2$ ).

Independence:

Compute the Wronskian determinant:

$\begin{vmatrix}10x^2+4x-1&3x-4x^2+3&5x^2+x-1\\20x+4&3-8x&10x+1\\20&-8&10\end{vmatrix}=-6\neq0$

The determinant is non-zero, so the vectors are linearly independent. For this reason, we also know the dimension of $H$ is 3.

Span:

Write an arbitrary vector in $P_2$ as $ax^2+bx+c$ . Then the given vectors span $P_2$ if there is always a choice of scalars $k_1,k_2,k_3$ such that

$k_1(10x^2+4x-1)+k_2(3x-4x^2+3)+k_3(5x^2+x-1)=ax^2+bx+c$

which is equivalent to the system

$\begin{bmatrix}10&-4&5\\4&3&1\\-1&3&-1\end{bmatrix}\begin{bmatrix}k_1\\k_2\\k_3\end{bmatrix}=\begin{bmatrix}a\\b\\c\end{bmatrix}$

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2 years ago

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2 years ago

What additional information could you use to show that ΔSTU ≅ ΔVTU using SAS? Check all that apply.

GalinKa [24]

Options

A. UV = 14 ft and m∠TUV = 45°

B. TU = 26 ft

C. m∠STU = 37° and m∠VTU = 37°

D. ST = 20 ft, UV = 14 ft, and m∠UST = 98°

E. m∠UST = 98° and m ∠TUV = 45°

Answer:

A. UV = 14 ft and m∠TUV = 45°

D. ST = 20 ft, UV = 14 ft, and m∠UST = 98°

Step-by-step explanation:

Given

See attachment for triangle

Required

What proves that: ΔSTU ≅ ΔVTU using SAS

To prove their similarity, we must check the corresponding sides and angles of both triangles

First:

$\angle UST$ must equal $\angle UVT$

So:

$\angle UST = \angle UVT = 98$

Next:

UV must equal US.

So:

$UV = US = 14$

Also:

ST must equal VT

So:

$ST = VT = 20$

Lastly

$\angle TUV$ must equal $\angle TUS$

So:

$\angle TUV = \angle TUS = 45$

Hence: Options A and D are correct

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Lyrx [107]

Answer:

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Step-by-step explanation:

We have the following function given:

$f(x) = 500(1.1)^x$

And we want to know what repreent the value 500 for this equation. If we see the general expression for an exponential function we have:

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1 year ago