To evaluate the expression A-7x2 for a -15, we need to substitute -15 for a and then simplify the expression.
A-7x2 = A-7(x^2)
Substituting -15 for a, we get:
-15 - 7(x^2)
Therefore, the value of the expression A-7x2 for a -15 is -7(x^2) - 15 12.
Hope this helped :)
10 + 2³ ∙ 4 - 1
10 + 2³ ∙ (4 - 1)
(10 + 2³) ∙ 4 - 1
10 + (2³ ∙ 4) - 1
Hello there!
10+2^3*4-1
10+8*4-1
10+32-1
10+31
41
10+2^3*(4-1)
10+8*(3)
10+24
34
(10+8)*4-1
18*4-1
72-1
71
10+(2^3*4)-1
10+(8*4)-1
10+32-1
10+31
41
Hope this helps you!
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For this case the generic equation of the line is given by:
y = mx + b
Where,
m: slope of the line
b: intersection with the y axis.
To graph a line it is necessary to know the values of m and b.
Answer:
y = mx + b
You need to know m and b to graph the function completely.
For this reason, someone would choose to use the y-intercept and the slope
Answer:
0.75 years.
Step-by-step explanation:
Assuming 360 days in a year or 12 months.
So, we get days each month.
In 9 months there will be = days
So, convert this period to years = = 0.75 years.
Hence, the answer is 0.75 years.
Answer: x 0
Step-by-step explanation:
The domain of a function consists of the x-values that have a y-value in a function. The +1 shifts the graph upwards 1 unit. This only affects the range, the y-values. The x is in the denominator of a function, and the denominator of a function can never be 0, so we can say x 0. There are no other restrictions on the domain; x can be negative, x can be positive, but x cannot be 0.
If we graph this function, we'll see that there are no y-values when x = 0. We say that there is a vertical asymptote at x = 0, an imaginary line at an x-value that does not have any y-values.
I hope this helps! :)