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3 years ago

15 POINTS! What is the correct answer for Dean's problem? Explain where he went wrong.

Mathematics

1 answer:

prohojiy [21]3 years ago

8 0

Answer:

The slope is 5/-2.

Step-by-step explanation:

Slope is y1-y2 over m1-m2 (rise over run). The first ordered pair is -2 (m1) and 11 (y1). We then subtract the second ordered pair (4 (m2) and -4 (y2)) from the first.

11 - (-4) = 11 + 4 = 15

-2 - 4 = -6

Remember, slope is rise over run (y over x), so the slope is 15/-6. Now, we must simplify. 15/-6 = 5/-2

Dean went wrong because he thought that slope was run over rise (x over y). If he had switched the two numbers, his answer would have been correct.

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Answer: =23

Step-by-step explanation:

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2 years ago

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Use the model below to estimate the average annual growth rate of a certain country's population for 1950, 1988, and 2010, where

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Answer:

The average annual growth rate of a certain country's population for 1950, 1988, and 2010 are 2.398, 0.9985 and 0.2236 respectively.

Step-by-step explanation:

The given equation is

$Y=-0.0000084x^3+0.00211x^2-0.205x+8.423$

Where Y is the annual growth rate of a certain country's population and x is the number of years after 1900.

Difference between 1950 and 1900 is 50.

Put x=50 in the given equation.

$Y=-0.0000084(50)^3+0.00211(50)^2-0.205(50)+8.423$

$Y=2.398$

Therefore the estimated average annual growth rate of the country's population for 1950 is 2.398.

Difference between 1988 and 1900 is 88.

Put x=88 in the given equation.

$Y=-0.0000084(88)^3+0.00211(88)^2-0.205(88)+8.423$

$Y=0.9984752\approx 0.9985$

Therefore the estimated average annual growth rate of the country's population for 1988 is 0.9985.

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Put x=110 in the given equation.

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2 years ago

Let v1=(3,5) and v2=(-4,7).

Nat2105 [25]

PART A

The given vectors are,

$v_1 = \: < \: 3 , \: 5 \: >$

$v_2 = \: < \: - 4 , \: 7 \: >$

The magnitude of the vector

$v= \: < \: x , \: y \: > \:$

is given by:

$|v| = \sqrt{ {x}^{2} + {y}^{2} }$

This implies that,

$|v_1| = \sqrt{ {3}^{2} + {5}^{2} }$

$|v_1| = \sqrt{9 + 25}$

$|v_1| = \sqrt{34}$

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$|v_2| = \sqrt{ 16+ 49}$

$|v_2| = \sqrt{65}$

PART B

To find the unit vector in the direction of a given vector, we divide by the magnitude of that vector.

$^{ - } _{v_1} = \: < \: \frac{3}{ \sqrt{34} } , \: \frac{5}{ \sqrt{34} } \: >$

Rationalize the denominator.

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Also,

$^{ - } _{v_2} = \: < \: \frac{ - 4}{ \sqrt{65} } , \: \frac{7}{ \sqrt{65} } \: >$

$^{ - } _{v_2} = \: < \: \frac{ - 4\sqrt{65}}{ 65 } , \: \frac{7\sqrt{65}}{ 65 } \: >$

PART C

The sketch of the given vectors as well as their unit vectors are shown in the attachment.

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