Answer:
1. z = 128
2. x = 4.2
3. c = 10
4. w = 100
5. a = 95.2
Step-by-step explanation:
1. Solve for z:
z/16 = 8
Multiply both sides of z/16 = 8 by 16:
(16 z)/16 = 16×8
(16 z)/16 = 16/16×z = z:
z = 16×8
16×8 = 128:
Answer: z = 128
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2. Solve for x:
3.5 x = 14.7
Divide both sides of 3.5 x = 14.7 by 3.5:
(3.5 x)/3.5 = 14.7/3.5
3.5/3.5 = 1:
x = 14.7/3.5
14.7/3.5 = 4.2:
Answer: x = 4.2
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3. Solve for c:
32 = 3.2 c
32 = 3.2 c is equivalent to 3.2 c = 32:
3.2 c = 32
Divide both sides of 3.2 c = 32 by 3.2:
(3.2 c)/3.2 = 32/3.2
3.2/3.2 = 1:
c = 32/3.2
32/3.2 = 10:
Answer: c = 10
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4. Solve for w:
(2 w)/5 = 40
Multiply both sides of (2 w)/5 = 40 by 5/2:
(5×2 w)/(2×5) = 5/2×40
5/2×2/5 = (5×2)/(2×5):
(5×2)/(2×5) w = 5/2×40
5/2×40 = (5×40)/2:
(5×2 w)/(2×5) = (5×40)/2
(5×2 w)/(2×5) = (2×5)/(2×5)×w = w:
w = (5×40)/2
2 | 2 | 0
| 4 | 0
- | 4 |
| | 0
| - | 0
| | 0:
w = 5×20
5×20 = 100:
Answer: w = 100
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5. Solve for a:
a/14 = 6.8
Multiply both sides of a/14 = 6.8 by 14:
(14 a)/14 = 14×6.8
(14 a)/14 = 14/14×a = a:
a = 14×6.8
14×6.8 = 95.2:
Answer: a = 95.2
Answer:
First, find tan A and tan B.
cosA=35 --> sin2A=1−925=1625 --> cosA=±45
cosA=45 because A is in Quadrant I
tanA=sinAcosA=(45)(53)=43.
sinB=513 --> cos2B=1−25169=144169 --> sinB=±1213.
sinB=1213 because B is in Quadrant I
tanB=sinBcosB=(513)(1312)=512
Apply the trig identity:
tan(A−B)=tanA−tanB1−tanA.tanB
tanA−tanB=43−512=1112
(1−tanA.tanB)=1−2036=1636=49
tan(A−B)=(1112)(94)=3316
kamina op bolte
✌ ✌ ✌ ✌
Ok, First you have to find the GCF of 120 and 90. When I got the answer I got a GCF of 30. I got 30 by multiplying all the number prime numbers 120 and 90 had, 2×3×5=30 What I did next was divide thirty by the number of cracker and also by the number of can of juice.
120÷30= 4
90÷30= 3
So as you can see there could be a maximum 30 of kid's. Each of the kids would receive 4 cans of juice and 3 packs of cheese crackers.
I hope you enjoyed my explanation. :)
Step-by-step explanation:
2. answer is RS 150
follow unitary method
If one angle is d° then it's supplement is
(180-d)°