Answer:
The larger number is -6, the smaller number is -15
Step-by-step explanation:
We have two numbers, a and b.
We know that one number is larger than another by 9.
Then we can write:
a = b + 9
then a is larger than b by 9 units.
If the greater number is increased by 10 (a + 10) and the lesser number is tripled (3*b), the sum of the two would be -41:
(a + 10) + 3*b = -41
So we got two equations:
a = b + 9
(a + 10) + 3*b = -41
This is a system of equations.
One way to solve this is first isolate one variable in one of the two equations:
But we can see that the variable "a" is already isolated in the first equation, so we have:
a = b + 9
now we can replace that in the other equation:
(a + 10) + 3*b = -41
(b + 9) + 10 + 3*b = -41
now we can solve this for b.
9 + b + 10 + 3b = -41
(9 + 10) + (3b + b) = -41
19 + 4b = -41
4b = -41 -19 = -60
b = -60/4 = -15
b = -15
then:
a = b + 9
a = -15 + 9 = -6
a = -6
12+6x was subtracted from the left but only 12 was subtracted from the right.
6x and 6 are not equal unless x=1
X=-11/15
The value of x in the given sector is determined as 7.1.
<h3>Area of the sector</h3>
Total area of the sector = area of the shaded + area of the unshaded
Since shaded and unshaded areas are equal, the value of x is calculated as follows;
πr²θ/360 = πx²60/360 + πx²60/360
πr²θ/360 = 2(πx²θ/360)
πr² = 2(πx²)
r² = 2x²
x² = r²/2
x² = (10²)/2
x² = 50
x = √50
x = 7.1
Thus, the value of x in the given sector is determined as 7.1.
Learn more about area of sector here: brainly.com/question/22972014
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She worked 2 and 1/2 hours yesterday, and today she did 4 and 1/4 hours.
first off, let's check how many hours she worked total, and divide 54 by it.
so, first off, let's convert the mixed fractions to "improper" and add them up.
Answer:
{t|60 <= t <= 85}
Step-by-step explanation:
The temperatures were measures at different times, but does not stop the values being real numbers (i.e. not discrete, or integer values).
So the range of the function is the set of all values between the minimum and maximum measured during the measuring interval (domain) of hours two and twenty-two.
The minimum value = 60F
The maximum value = 85F
So the interval of the range is [60,85], in interval notation.
In set-builder notation, it is
{t|60 <= t <= 85}