Answer:
Step-by-steplanation:
Add: 1
2
+ 2
5
= 1 · 5
2 · 5
+ 2 · 2
5 · 2
= 5
10
+ 4
10
= 5 + 4
10
= 9
10
For adding, subtracting, and comparing fractions, it is suitable to adjust both fractions to a common (equal, identical) denominator. The common denominator you can calculate as the least common multiple of the both denominators - LCM(2, 5) = 10. In practice, it is enough to find the common denominator (not necessarily the lowest) by multiplying the denominators: 2 × 5 = 10. In the next intermediate step the fraction result cannot be further simplified by cancelling.
In words - one half plus two fifths = nine tenths.
Conversion a mixed number 6 2
7
to a improper fraction: 6 2/7 = 6 2
7
= 6 · 7 + 2
7
= 42 + 2
7
= 44
7
To find new numerator:
a) Multiply the whole number 6 by the denominator 7. Whole number 6 equally 6 * 7
7
= 42
7
b) Add the answer from previous step 42 to the numerator 2. New numerator is 42 + 2 = 44
c) Write previous answer (new numerator 44) over the denominator 7.
Six and two sevenths is forty-four sevenths
Add: the result of step No. 1 + 44
7
= 9
10
+ 44
7
= 9 · 7
10 · 7
+ 44 · 10
7 · 10
= 63
70
+ 440
70
= 63 + 440
70
= 503
70
For adding, subtracting, and comparing fractions, it is suitable to adjust both fractions to a common (equal, identical) denominator. The common denominator you can calculate as the least common multiple of the both denominators - LCM(10, 7) = 70. In practice, it is enough to find the common denominator (not necessarily the lowest) by multiplying the denominators: 10 × 7 = 70. In the next intermediate step the fraction result cannot be further simplified by cancelling.
In words - nine tenths plus forty-four sevenths = five hundred three seventieths.
14. 1.5, 10 <- Answer
15. 5,1 <- Answer
Proof 14
Solve the following system:
{2 x - y = -7 | (equation 1)
4 x - y = -4 | (equation 2)
Swap equation 1 with equation 2:
{4 x - y = -4 | (equation 1)
2 x - y = -7 | (equation 2)
Subtract 1/2 × (equation 1) from equation 2:
{4 x - y = -4 | (equation 1)
0 x - y/2 = -5 | (equation 2)
Multiply equation 2 by -2:
{4 x - y = -4 | (equation 1)
0 x+y = 10 | (equation 2)
Add equation 2 to equation 1:
{4 x+0 y = 6 | (equation 1)
0 x+y = 10 | (equation 2)
Divide equation 1 by 4:
{x+0 y = 3/2 | (equation 1)
0 x+y = 10 | (equation 2)
Collect results:
Answer: {x = 1.5
y = 10
Proof 15.
Solve the following system:
{5 x + 7 y = 32 | (equation 1)
8 x + 6 y = 46 | (equation 2)
Swap equation 1 with equation 2:
{8 x + 6 y = 46 | (equation 1)
5 x + 7 y = 32 | (equation 2)
Subtract 5/8 × (equation 1) from equation 2:{8 x + 6 y = 46 | (equation 1)
0 x+(13 y)/4 = 13/4 | (equation 2)
Divide equation 1 by 2:
{4 x + 3 y = 23 | (equation 1)
0 x+(13 y)/4 = 13/4 | (equation 2)
Multiply equation 2 by 4/13:
{4 x + 3 y = 23 | (equation 1)
0 x+y = 1 | (equation 2)
Subtract 3 × (equation 2) from equation 1:
{4 x+0 y = 20 | (equation 1)
0 x+y = 1 | (equation 2)
Divide equation 1 by 4:
{x+0 y = 5 | (equation 1)
0 x+y = 1 | (equation 2)
Collect results:
Answer: {x = 5 y = 1
(-7,3) here is the answer, easy one
Answer:
16
Step-by-step explanation:
h(x) × h(x) = (6 - x)²
(h × h)(10) = (6 - 10)² = (- 4)² = 16
Answer:
and 
.
Step-by-step explanation:
Please find the attachment.
We have been given that a norman window is constructed by adjoining a semicircle to the top of an ordinary rectangular. The total perimeter is 38 feet.
The perimeter of the window will be equal to three sides of rectangle plus half the perimeter of circle. We can represent our given information in an equation as:
We can see that diameter of semicircle is W. We know that diameter is twice the radius, so we will get:
Let us find area of window equation as:
Now, we will solve for L is terms W from perimeter equation as:
Substitute this value in area equation:
Since we need the area of window to maximize, so we need to optimize area equation.
Let us find derivative of area equation as:
To find maxima, we will equate first derivative equal to 0 as:
Upon substituting in equation , we will get:
Therefore, the dimensions of the window that will maximize the area would be and .
