Answer:
The correct answer is - 625.36.
Step-by-step explanation:
Given:
Fixed mothly plan - 616
free minutes - 150
Used minutes - 170
Free text - 150
Used text - 182
Charging amount over free limit - 18p
Solution:
Number of minutes over the limit - 170 - 150 = 20
Number of text over the limit - 182 - 150 = 32
So the extra amount that will be add to the monthly charge would be -
(20*0.18) +(32*0.18)
and the total charge of the month would be -
= 616+(20*0.18) +(32*0.18)
= 616+3.60+5.76
= 625.36
Thus, the correct answer would be - 625.36
Answer:
A. 29
B. 11
C. 10
D. 12
Step-by-step explanation:
A. 180 - 122 = 58
58 divided by 2 = 29
-
B. 180 - 37 = 143
x = 11
5 x 11 = 55
8 x 11 = 88
55 + 88 = 143 + 37 - 180
-
C. the small box in a triangle means right angle = 90
90 - 68 = 22
(2x - 2) we subtract before finding out what x is
22 - 2 = 20
2x = 20 divide by 2 = 10
x is equal to 10
-
D. to find the total, you must figure out what each equation is to add them up to make 180.
(8x + 3) + (2x + 9) + 4x = 180
add up the numbers with x and the numbers without them
14x + 12 = 180
subtract 180 - 12 = 168
168 divided by 14 = 12
x = 12
Hope this helped !!
Answer:
Step-by-step explanation:
This is a proportion problem. I think you just lost a zero.
12/100 = 9/x Cross multiply
12x = 100*9 Combine the right.
12x = 900 Divide by 12
12x/12 = 900/12
x = 75 grams
So if you put 75 grams of tasty Oats in a bowl and covered it with milk the cereal would give you 9 grams of Protein.
2.
1 gram of Tasty Oats cereal contains 12/100 = 0.12 grams of protein.
1 gram of Cornbits contains 5/45 = 0.11 grams of protein.
By the narrowest of margins the Tasty Oats has more protein in it.
Answer:
Option B
Step-by-step explanation:
Looking at the options, option B is correct because when multiplying it by matrix A, it yields the matrix AB as follows;
First row of A multiplied by first column of matrix in option D;
(1 × -1) + (0 × 0) + (0 × 0) = -1 which corresponds to the first number on the first row of Matrix AB
Since majority of matrix AB are zero, I will just prove the ones that are not zero.
Thus;
Second row of matrix A is multiplied by second column of matrix in option D;
(0 × 0) + (-1 × -1) + (0 × 0) = 1 which is same as 2nd number on second row in matrix AB
Lastly, third row of matrix A is multiplied by third column of matrix in option D;
(0 × 0) + (0 × 0) + (1 × -1) = -1 which is same as third number in third row in matrix AB
