Let x represent the worth of the professional basketball player's autograph a year ago. Since his autograph's worth increased by 40% now, $364 is 1.40x.
364 = 1.40x
Solving for x gives x = 260.
Thus, the autograph of the basketball player was worth only $260 last year.
As <em>x</em> → 10 from above, <em>h(x)</em> is steadily approaching 18.5.
it would be 70/100 then it would go down to 7/10 so 1/10 of 70kg would be 7kg because you change 70 you make it a decimal witch is .70 or .7 now make that into a fraction and it would be 70/100 or 7/10 then you take your 7 because its 1/10 of 70. Or you could do it a whole lot easier way and just divide 70 by 10 and get 7. There you have it your answer is 7.
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Answer:
The center is (-10,10) and the radius is 4sqrt(3)
Step-by-step explanation:
(x + 10)^2 + (y - 10)^2 = 48
We can write the equation of a circle as
(x -h)^2 + (y - k)^2 = r^2 where (h,k) is the center and r is the radius
(x- -10)^2 + (y - 10)^2 = (sqrt(16*3) )^2
(x- -10)^2 + (y - 10)^2 = (4sqrt(3)) ^2
The center is (-10,10) and the radius is 4sqrt(3)
Answer:
a. 3
Step-by-step explanation:
We can get to know if a number is divisible by 3 if after adding all the digits of that number we get a number which is completely divisible by 3 or we can say the number appears in the table of 3.
For example: In the given number if we will add all the digits the we will get 21 and we all know that 21 is divisible by 3.
10,534,341 = 1 + 0 + 5 + 3 + 4 + 3 + 4 + 1 = 21
21 = 21/3 = 7
Other options are incorrect.
Divisibility rule for 2 : A number is said to be divisible by 2 if it's ones place ends in 0, 2, 4, 6 or 8 i.e. an even number. The given number does not satisfy this condition because it ends in 1 which is an odd number.
Divisibility rule for 9 : A number is said to be divisible by 9 if after adding all the digits of that number we get a number which is completely divisible by 9. Since the sum total of the given number is 21 which is not completely divisible by 9 we can easily infer that 9 is not the answer for this question.
