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3 years ago

2/3x+1/3y=2 in graph

Mathematics

1 answer:

Nady [450]3 years ago

6 0

Solve for y:

$\dfrac{2}{3}x+\dfrac{1}{3}y=2\ \ \ \ |\cdot3\\\\2x+y=6\ \ \ \ |-2x\\\\y=-2x+6$

It's the slope intercept form of a straight line.

We need only two points to draw a straight line.

$for\ x=0\to y=-2(0)+6=0+6=6\to(0;\ 6)\\for\ x=3\to y=-2(3)+6=-6+6=0\to(3;\ 0)$

Look at the picture.

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What are the explicit equation and domain for a geometric sequence with a first term of 4 and a second term of −12?

Lerok [7]

$a_1=4;\ a_2=-12\\\\r=\dfrac{a_2}{a_1}\to r=\dfrac{-12}{4}=-3$

The formula of a geometric sequence:

$a_n=a_1\cdot r^{n-1}$

substitute

$a_n=4\cdot(-3)^{n-1}$

$The\ domain:\ \text{all integers where}\ n\geq1$

Answer: $\boxed{a_n=4(-3)^{n-1};\ \text{all integers where}\ n\geq1}$

4 0

3 years ago

Determine if the sequence below is arithmetic or geometric and determine the

asambeis [7]

Answer:

Step-by-step explanation:

This is an arithmetic series.

3 , 5 , 7 ,......

Common difference = second term - first term

= 5 - 3

= 2

4 0

2 years ago

Please help can't figure this out to save the life of me.

bogdanovich [222]

An equation written in slope-intercept form is y=mx+b, where m is a constant equal to the slope. Parallel lines have the same slope. So a line parallel to y=-2x+3 is y=-2x+5. Perpendicular lines have slopes which are negative reciprocals of each other. So a line perpendicular to y=-2x+3 is y=1/2x+7. y=2x-1 is neither parallel of perpendicular to y=-2x+3.

3 0

3 years ago

Read 2 more answers

The Department of Transportation would like to test the hypothesis that the average age of cars on the road is less than 12 year

puteri [66]

Answer:

$z=\frac{10.6-12}{\frac{4.1}{\sqrt{45}}}=-2.29$

Step-by-step explanation:

information given

$\bar X=10.6$ represent the sample mean

$\sigma=4.1$ represent the population standard deviation

$n=45$ sample size

$\mu_o =12$ represent the value that we want to test

$\alpha=0.05$ represent the significance level for the hypothesis test.

z would represent the statistic

$p_v$ represent the p value for the test

Hypothesis to test

We want to test if the true mean is less than 12, the system of hypothesis would be:

Null hypothesis: $\mu \geq 10$

Alternative hypothesis: $\mu < 10$

The statistic is given by:

$z=\frac{\bar X-\mu_o}{\frac{\sigma}{\sqrt{n}}}$ (1)

Replacing the info given we got:

We can replace in formula (1) the info given like this:

$z=\frac{10.6-12}{\frac{4.1}{\sqrt{45}}}=-2.29$

8 0

3 years ago

Given: KLMN is a trapezoid m∠N = m∠KML ME ⊥ KN , ME = 3√5 , KE = 8, LM:KN = 3:5 Find: KM, LM, KN, Area of KLMN

lora16 [44]

Q1)Find KM
As ME is perpendicular to KN, ∠KEM is a right angle
Therefore ΔKEM is a right angled triangle
KE is given and and ME is also given, we need to find KM
for this we can use Pythogoras' theorem where the square of hypotenuse is equal to the sum of the squares of the adjacent sides.
KM² = KE² + ME²
KM² = 8² + (3√5)²
= 64 + 9x5
KM = √109
KM = 10.44

Q2)Find LM
It is said that ratio of LM:KN is 3:5
Therefore if we take the length of one unit as x
length of LM is 3x
and the length of KN is 5x
KN is greater than LM by 2 units
If we take the figure ∠K and ∠N are equal.
Since the angles on opposite sides of the bases are equal then this is called an isosceles trapezoid. So if we draw a line from L that cuts KN perpendicularly at D, ΔKEM and ΔLDN are congruent therefore KE = DN
since KN is greater than LM due to KE and DN , the extra 2 units of KN correspond to 16 units
KN = LM + 2x
2x = KE + DN
2x = 8+8
x = 8
LM = 3x = 3*8 = 24

Q3)Find KN
Since ∠K and ∠N are equal, when we take the 2 triangles KEM and LDN, they both have;
same height ME = LD perpendicular distance between the 2 parallel sides
same right angle when the perpendicular lines cut KN
∠K = ∠N
when 2 angles and one side of one triangle is equal to two angles and one side on another triangle then the 2 triangles are congruent according to AAS theorem (AAS). Therefore KE = DN
the distance ED = LM
Therefore KN = KE + ED + DN
since ED = LM = 24
and KE + DN = 16
KN = 16 + 24 = 40
another way is since KN = 5x and x = 8
KN = 5 * 8 = 40

Q4)Find area KLMN
Area of trapezium can be calculated using the following general equation
Area = 1/2 x perpendicular height between parallel lines x (sum of the parallel sides)
where perpendicular height - ME
2 parallel sides are KN and LM
substituting values into the general equation
Area = 1/2 * ME * (KN+ LM)
= 1/2 * 3√5 * (40 + 24)
= 1/2 * 3√5 * 64
= 3 x 2.23 * 32
= 214.66 units²

8 0

2 years ago

Read 2 more answers