Use the Pythagorean theorem to find z:
z = √(6^2 + 9^2)
z= √(36 + 81)
z = √117
z = 3√13 = 10.8
Use leg rule to find x:
x/6 = 6/9
Cross multiply:
9x = 36
Divide both sides by 9:
x = 36/9
x = 4
Use Pythagorean theorem to find y:
y = √(6^2 + 4^2)
y= √(36 + 16)
y = √52
y = 2√13 = 7.2
Answer:
(3x+4)(5x+7)
Step-by-step explanation:
15x^2
+41x+28
Factor the expression by grouping. First, the expression needs to be rewritten as 15x^2
+ax+bx+28. To find a and b, set up a system to be solved.
a+b=41
ab=15×28=420
Since ab is positive, a and b have the same sign. Since a+b is positive, a and b are both positive. List all such integer pairs that give product 420.
1,420
2,210
3,140
4,105
5,84
6,70
7,60
10,42
12,35
14,30
15,28
20,21
Calculate the sum for each pair.
1+420=421
2+210=212
3+140=143
4+105=109
5+84=89
6+70=76
7+60=67
10+42=52
12+35=47
14+30=44
15+28=43
20+21=41
The solution is the pair that gives sum 41.
a=20
b=21
Rewrite 15x^2
+41x+28 as (15x^2
+20x)+(21x+28).
(15x^2
+20x)+(21x+28)
Factor out 5x in the first and 7 in the second group.
5x(3x+4)+7(3x+4)
Factor out common term 3x+4 by using distributive property.
(3x+4)(5x+7)
Answer:
C) 42.14 cm²
Step-by-step explanation:
<u>Recall the area formulas for a square and circle:</u>
Area of square: A=s²
Area of circle: A=πr²
<u>Given:</u>
π=3.14
s=14
r=s/2=14/2=7
Therefore, the area of the square is A=14²=196 cm²
The area of the two semicircles is A=3.14(7)²=3.14(49)=153.86 cm²
Find the difference between the two areas to get the shaded area:
196 - 153.86 = 42.14
Therefore, the shaded area is 42.14 cm², making C the correct choice.
<u>Given </u><u>:</u><u>-</u>
- Length :- 8m
- Width :- 10
- Height :- 10m
<u>To </u><u>find</u><u> </u><u>:</u><u>-</u>
<u>Solution</u><u> </u><u>:</u><u>-</u>
formula for Surface Area = 2(LW + WH + HL)
- putting the known values,
Surface Area = 2(8×10 + 10×10 + 10×8) m²
<u>Surface</u><u> Area</u><u> </u><u>=</u><u> </u><u>2</u><u>(</u><u>2</u><u>6</u><u>0</u><u>)</u><u> </u><u>m²</u>
<u>Surface</u><u> Area</u><u> </u><u>=</u><u> </u><u>5</u><u>2</u><u>0</u><u>m</u><u>²</u><u> </u><u>(</u><u> </u><u>Option</u><u> </u><u>C</u><u>)</u>