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yKpoI14uk [10]
3 years ago
6

the birth rate in a certain country in 1994 was 14,6 birthd per thousand population. in 2004 the birth rate was 14.32 per thousa

nd. let x represent years after 1994 and y represent the birth rate . assume that the relationship between x and y is linear over this period . write a linear equation that relates y in terms of x
Mathematics
1 answer:
amm18123 years ago
4 0

Answer:

y(x)=-0.028x+14.6

Step-by-step explanation:

Birth Rate in 1994 =  14.6 births per thousand population.

Birth Rate in 2004 =  14.32 births per thousand population.

A linear equation is of the form: y=mx+c

Where:

  • x=years after 1994
  • y=the birth rate
  • m=Yearly Birth Rate

First, we determine the birth rate per year

m=\dfrac{14.32-14.6}{10-0}\\=\dfrac{-0.28}{10}\\m=-0.028

Therefore, a linear equation that relates y in terms of x is of the form:

y(x)=-0.028x+c

When x=0, y=14.6

14.6=-0.028(0)+c

c=14.6

Therefore, the linear equation that relates y in terms of x is:

y(x)=-0.028x+14.6

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Step-by-step explanation:

Hi this is a simple mathematics issue.

First, let's imagine what is 3%. '3 percents' means 3/100. If your family have 100 apples and you have 3 apples, that means you have 3% of your family apples.

Now, get back to this problem. We need to find the price before increase. Let's say the price was £ X. Now we will find the X number.

It's said that the price was increased by 3%. That means they took 3% of X and added that amount to X to have the new price : £ 650. That means we have this equation :

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So we can see 650 is equaled to X multipled by 103/100.

To find the correct answer, simply divide 650 by 103/100.

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The correct options are;

EFGH has 4 congruent sides

Diagonal FH bisects angles EFG and EHG

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Step-by-step explanation:

1) Given that for a reflection, we have;

The distance of the reflected preimage from the line of reflection = The distance of the reflected image from the line of reflection

Therefore;

The distance of the point E from the line HF = The distance of the point G from the line HF

Also the reflection of an preimage (x, y) about the x-axis, gives an image (x, -y)

We can show that from the length of a line given by the equationl = \sqrt{\left (y_{2}-y_{1}  \right )^{2}+\left (x_{2}-x_{1}  \right )^{2}}, that the length EH ≅ GH and EF ≅ GF

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Similarly, ∠GFH ≅ ∠EFH

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Therefore, diagonal FH bisects angles EFG and EHG

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Angle FEH is congruent to angle FGH, by Congruent Parts of Congruent Triangles are Congruent

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