Mathematics, 26.06.2019 13:00 ian2006huang

For δabc, ∠a = 4x + 7, ∠b = 2x + 3, and ∠c = 6x - 10. if δabc undergoes a dilation by a scale factor of 2 to create δa'b'c' with ∠a' = 5x - 8, ∠b' = 3x - 12, and ∠c' = 7x - 25, which confirms that δabc∼δa'b'c by the aa criterion?

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Answer from: reyamukhtar

Bizzare problem. What are the choices?

The triangle angles add to 180.

4x + 7 + 2x +3 + 6x - 10 = 180

12x = 180

x = 15

Dilation doesn't change the angles so we must have

4x + 7 = 5x - 8 and 2x + 3 = 3x - 12 and 6x - 10 = 7x - 25

Fortunately x=15 is the solution to all of these so it's consistent.

Answer from: carterkelley12oz0kp1

The value of x is 15.

Step-by-step explanation:

It is given that in ΔABC, ∠A = 4x + 7, ∠B = 2x + 3, and ∠C = 6x - 10.

According to angle sum property of triangle, the sum of all interior angles of a triangle is 180°.

In ΔABC,

$\angle A+\angle B+\angle C=180$

(4x+7)+(2x+3)+(6x-10)=180

Combine like terms.

(4x+2x+6x)+(7+3-10)=180

12x=180

Divide both sides by 12.

x=15

The value of x is 15.

$\angle A=4x+7\Rightarrow 4(15)+7=67^{\circ}$

$\angle B=2x+3\Rightarrow 2(15)+3=33^{\circ}$

$\angle C=6x-10\Rightarrow 6(15)-10=80^{\circ}$

$\angle A'=5x-8\Rightarrow 5(15)-8=67^{\circ}$

$\angle B'=3x-12\Rightarrow 3(15)-12=33^{\circ}$

$\angle C'=7x-25\Rightarrow 7(15)-25=80^{\circ}$

We get

$\angle A\cong \angle A'$

$\angle B\cong \angle B'$

Two corresponding angles of triangles are congruent. By AA criterion

$\triangle ABC\cong \triangle A'B'C'$

Hence proved.

Answer from: Quest

part a

answer: a reasonable domain is $0 \le n \le 10$ where n is a real number. so n can be between 0 and 10. both endpoints are included.

work shown:

n is the number of days after the experiment starts. the smallest n can be is n = 0 which means that 0 days have gone by, and we're at the start. to find out how large n should be, then replace f(n) with 16.13 and solve for n. use logarithms to isolate the exponent.

f(n) = 12*(1.03)^n

16.13 = 12*(1.03)^n

16.13/12 = (1.03)^n

1.34416666666667 = (1.03)^n

(1.03)^n = 1.34416666666667

log[ (1.03)^n ] = log[ 1.34416666666667 ]

n*log[ 1.03 ] = log[ 1.34416666666667 ]

n = log[ 1.34416666666667 ]/log[ 1.03 ]

n = 10.0062999823929

this rounds to n = 10 which is fairly close but not 100% perfect. so this is the largest n can be.

note: if you plug n = 10 into f(n), you'll get roughly 16.127 which rounds to 16.13 (this comes up again in part c)

=====================================

part b

answer: the y intercept is 12. it represents the starting height of the plant in cm.

work shown:

plug n = 0 into the f(n) function. simplify

f(n) = 12*(1.03)^n

f(0) = 12*(1.03)^0

f(0) = 12*(1)

f(0) = 12

on day n = 0, aka the starting point, the height f(n) is 12 cm

=====================================

part c

answer: the average rate of change is approximately 0.43061036458991 (round however you need to). this represents the average growth rate from day n = 3 to day n = 10. so the plant grew roughly 0.43 cm per day during this timespan, assuming you round to 2 decimal places.

work shown:

compute f(3)

f(n) = 12*(1.03)^n

f(3) = 12*(1.03)^3

f(3) = 13.112724 < we'll use this later

compute f(10)

f(n) = 12*(1.03)^n

f(10) = 12*(1.03)^10

f(10) = 16.1269965521294 < we'll use this later

now use the formula below with a = 3 and b = 10

aroc = average rate of change

aroc = [ f(b) - f(a) ]/[ b - a ]

aroc = [ f(10) - f(3) ]/[ 10 - 3 ]

aroc = (16.1269965521294 - 13.112724)/(10 - 3)

aroc = 3.0142725521294/7

aroc = 0.43061036458991

round this however you need to

note: the plant grew approximately 3.01 cm over 7 days, so roughly 0.43 cm per day is the average growth rate (if you were to round to 2 decimal places).

Answer from: Quest

First, think of what you know. you know that the number t the end of the equation will need to be subtracted because it is the money he lost from buying the lawn mower. and your first two weeks have negative profit, so that tells you that c and d are wrong. now you can plug any of the weeks into a and b to see which one works. i'll start with plugging week 1 into a. $y = 60(1) - 90 \\ y = 60 - 90 \\ y = - 30$ this isn't right because it says your answer for week one should be -90.so now we need to try the same thing with b. $y = 60(1) - 150 \\ y = 60 - 150 \\ y = - 90$ this one got you the right answer, (sometimes if you only check with one of the weeks multiple will work so you need to check with more of the weeks, but it worked out fine here)this means that b is the correct answer.let me know if you need more clarification. i hope this : )

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For δabc, ∠a = 4x + 7, ∠b = 2x + 3, and ∠c = 6x - 10. if δabc undergoes a dilation by a scale factor...

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