Directions: Measure each angle and write the measure on the line.
1 answer:
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Answer:
The factors are: (3a+2b +ab-6)(3a+2b -ab+6)
Step-by-step explanation:
We need to solve the above expression using factorization.
Multiplying (a^2-4)(9-b^2)
9(a^2-4)-b^2(a^2-4) + 24ab
9a^2 -36 -a^2b^2+4b^2 + 24ab
Rearranging:
9a^2 + 4b^2 +24ab -36 -a^2b^2
We try to make perfect square of the form a^2+2ab-b^2
We have 24ab that can be written as 12ab + 12ab
Now, we can arrange the above equation:
9a^2 +12ab+ 4b^2 -(a^2b^2-12ab +36)
(3a)^2 +2(3a)(2b) + (2b)^2 -((ab)^2 -2(ab)(6)+(6)^2)
The perfect square will be:
(3a+2b)^2 - (ab-6)^2
Now We know a^2 - b^2 = (a+b)(a-b)
Here a = 3a+2b , b=ab-6
So,
(3a+2b +(ab-6))(3a+2b - (ab-6))
(3a+2b +ab-6)(3a+2b -ab+6)
So, the factors are: (3a+2b +ab-6)(3a+2b -ab+6)
Answer:
(1)0.0138
(2)A. If the depth of the dive is increased by one meter, it adds 0.0138 minutes to the time spent under water.
Nos 3-12: See Explanation
Step-by-step explanation:
Given the regression equation for the relation of dive duration (DD) to depth (D).
(1)The slope of the regression lie =0.0138
(2)
When D=1, DD = 2.69 + 0.0138(1)=2.7038
When D=2, DD = 2.69 + 0.0138(2)=2.7176
2.7176-2.7038=0.0138
Therefore, If the depth of the dive is increased by one meter, it adds 0.0138 minutes to the time spent under water.
(3) When depth, D =200 meters
DD = 2.69 + 0.0138(200)=5.45 Minutes
(4) When depth, D =210 meters
DD = 2.69 + 0.0138(210)=5.588 Minutes
(5) When depth, D =220 meters
DD = 2.69 + 0.0138(220)=5.726 Minutes
(6) When depth, D =230 meters
DD = 2.69 + 0.0138(230)=5.864 Minutes
(7) When depth, D =240 meters
DD = 2.69 + 0.0138(240)=6.002 Minutes
(8) When depth, D =150 meters
DD = 2.69 + 0.0138(150)=4.76 Minutes
(9) When depth, D =160 meters
DD = 2.69 + 0.0138(160)=4.898 Minutes
(10) When depth, D =170 meters
DD = 2.69 + 0.0138(170)=5.036 Minutes
(11) When depth, D =180 meters
DD = 2.69 + 0.0138(180)=5.174 Minutes
(12) When depth, D =190 meters
DD = 2.69 + 0.0138(190)=5.312 Minutes