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yawa3891 [41]
3 years ago
8

t=" \: \: \: \: " align="absmiddle" class="latex-formula">
hi pls someone help me

{ \tt{write  \:  \: the  \:  \: equation  \:  \: of  \:  \: variation  \:  \: and  \:  \: solve  \:  \: for  \:  \: the \:  \:  constant  \:  \: of  \:  \: variation \:  \: }}
​

Mathematics
2 answers:
Alborosie3 years ago
5 0

\large \gray{ \boxed{ { \colorbox{black}{answer}}}}

<h2><u>1</u><u>.</u><u>)</u></h2>

\tt \: q = 7 \times  {p}^{2}

  • for the given values of <u>p,</u><u>q </u>the equation of variation is q=7×p²

  • for constant of variation is <u>7</u>

<h2><u>2</u><u>.</u></h2>

<u>variation</u>

p = 48/q

<u>constant</u>

<u>4</u><u>8</u>

<h2><u>3</u><u>.</u><u>)</u></h2>

<u>variation</u>

p= 48/q

<u>constant</u>

<u>4</u><u>8</u>

<h2><u>4</u><u>.</u><u>)</u></h2>

<u>variation</u>

q=kp

<u>constant</u>

k=⅓

hope it helps

Slav-nsk [51]3 years ago
4 0

Answer:

  1. q = 7p²
  2. q = 48/p
  3. q = 48/p . . . same table as 2
  4. q = 1/3p

Step-by-step explanation:

When faced with identifying sequences or forms of variation, the usual ones we look for are ...

  • directly proportional (y = kx)
  • inversely proportional (y = k/x)
  • proportional to the square (y = kx²)
  • related by some polynomial function (y = ax³ +bx² +cx +d)
  • exponential (y = a·b^x)

Proportional relations will have a constant ratio between y and x. That ratio is the value of k.

Inversely proportional relations will have a constant product of x and y. That product is the value of k.

Relations that are proportional to a square will have ratios of adjacent terms that are ratios of square numbers. (It is helpful to be familiar with the first few square numbers, at least.)

Exponential relations will have a constant ratio between terms, the value of b.

Polynomial relations will have constant n-th differences when the polynomial is of degree n. (First differences of a linear sequence are constant. Second differences of a "proportional to square" sequence are constant.)

In these tables, we assume q is the output and p is the input. The constant in each equation is the "constant of variation."

__

<h3>1. </h3>

Ratios of q-values are 28/7 = 4, 63/28 = 9/4, 112/63 = 16/9. These are ratios of consecutive squares. The constant of variation is the multiplier of 1² that results in the first term: 7

  q = 7p²

__

<h3>2, 3.</h3>

The products of table values are pq = 48. The equation is ...

 q = 48/p

__

<h3>4. </h3>

Ratios q/p are all the same, 2/6 = 4/12 = ... = 1/3. This is the constant of proportionality:

  q = 1/3p

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