They must have the same solution because they are doubled from eachother
Q55.
UTA+ATS=UTS
x+18+120=12x+17
x+138=12x+17
12x-x=138-17
11x=121
x=121/11
x=11
mUTA=x+18
put the value of x
11+18
mUTA=29
Q57.
AQP+RQA=RQP
9x+2+75=1+28x
9x+77=1+28x
28x-9x=77-1
19x=76
x=76/19
x=4
mRQP=1+28x
put the value of x
1+28(4)
1+112
mRQP=113
to write 98 as a product of its prime factors we have to first find the prime factors of 98
prime factors are prime numbers by which the given number can be divided by.
98 we have to keep dividing it by prime numbers
98 is an even number so we can first divide by 2
98 / 2 = 49
49 is a multiple of 7 which too is a prime number so we can divide 49 by 7
49/7 = 7
7 can be divided again by 1
7/7 = 1
98 is divisible by 2 and 7
so 98 written as a product of prime factors is
98 = 2 x 7 x 7
We know that
cos²(theta)=0.21
sin²(theta)+cos²<span>(theta)=1
</span>sin²(theta)=1-cos²(theta)------> sin²(theta)=1-(0.21)-----> 0.79
sin(theta)=√0.79
sin(theta)=0.89
the answer is
the value of sin(theta) is 0.89
Answer:
subtract 7.13 and 6.2 and you will get your answer
Step-by-step explanation:
Im doing the assignment on Edgenuity
and i passed hope this helps:)
