Kalino earned 85, 95, 93, and 80 points on the 4 tests, each worth 100 points, given so far this term. how many points must he earn on his fifth test, also worth 100 points, to average 90 points for the 5 tests given this term?

Kalino would need to earn at least a 97 on his fifth test.

Step-by-step explanation:

The average is calculated by adding up the amounts earned on each test and dividing by the total number of tests.  In this case, the score for the last test is not known, but the expected average of 90 points is, so we can set up an equation to solve for the unknown test value:

'x' represents the score of the fifth test.  Using inverse (opposite) operations and doing the same operation to both sides of the equation, we can first multiply both sides by 5:

x 5 = 90 x 5

353 + x = 450, now subtract 353 from both sides: 353 -353 + x = 450 - 353

x = 97

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step-by-step explanation:

9) i can't see the angle so whatever it is just do this: cos(angle) x hypotenuse and then use that answer and they hypotenuse and plug it into this: (answer you just got)^2 + b^2 = (hypotenuse)^2. the first answer will be the answer of the adjacent side and the second answer after this equation will be the opposite side.

10) y = 14 x = √147 or 12.12435565

11) x = 4.8

12) x = 4.2

13) x = 9.3

14) pq: 13.2

qr : 17.6

angle p: 53º

15) angle c: 54.3º

angle a: 35.7º

16) h = 21040

v = 1841

step-by-step explanation:

10) 7/sin(30) = 14 = y

7^2 + b^2 = 14^2

49 + b^2 = 196

b^2 = 196 - 49

√b^2 = √147

x = √147

11) cos(70) x 14 = 4.788282007

x = 4.8

12) tan(25) x 9 = 4.196768923

x = 4.2

13) sin(31) x 18 = 9.270685348

x = 9.3

14) pq: sin(37) x 22 = 13.23993051

pq: 13.2

qr: cos(37) x 22 = 17.56998122

qr : 17.6

angle p: 180 - (37 + 90) = 53º

15) cos-1(7/12) = 54.31466529

angle c: 54.3º

angle a: 180 - (90 + 54.31466529) = 35.68533471

angle a: 35.7º

16) 4 x 5280 = 21120

cos(5) x 21120

h = 21039.63202

h = 21040

sin(5) x 21120

v = 1840.729287

v = 1841