Instructions:Type the correct answer in each box. Use numerals instead of words. If necessary, use / for the fraction bar(s).James bought two T-shirts and one pair of jeans at an online store and paid $40, not including taxes, for his purchase. A month later, the same store sold the T-shirts and jeans at a 50% discount from their original prices. James bought two T-shirts and five pairs of jeans for $60, not including taxes.
Assuming the base prices of the T-shirts and the jeans are the same on both occasions, and ignoring the taxes, the price of a T-shirt is $ and the price of a pair of jeans is $.
Answers
Initially (without discount)
2T + P = 40
One month later (half prices)
2 (T/2) + 5(P/2) = 60
T +5P/2 = 60
To solve the system of equations multiply the second equation by 2 and substract it from the first equation
2 T + 5P = 120
- (2 T + P = 40 )
________________
4P = 80
P = 80/4
P = 20
From 2T + P = 40
T = (40 - P) / 2 = (40 -20) / 2 = 20/2 = 10.
The price of a T-shirt is $10 and the price of a pair of jeans is $20.
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(a) Find the approximate length of the plank. Round to the nearest tenth of a foot.
(b) Find the height above the ground where the plank touches the wall. Round to the nearest tenth of a foot.
2. A support wire is strung from a tree. The bottom of the wire is 24 ft from the tree. The length of the wire is 30 ft. Find each of the following values, and show all your work.
(a) Find the value of x. Round to the nearest tenth of a degree.
(b) Find the approximate height where the wire touches the tree. Round to the nearest foot.
Answers
(a) Using sine rule
ground/Sin 49 = plank/Sin 90
But sine 90 = 1, ground = 3 ft
Then,
3/Sin 49 = plank
Length of the plank = 3/Sin 49 ≈ 4.0 ft (rounded to nearest tenth)
(b) Height where the plank touches the wall
wall = Sqrt (plank^2 - ground^2) = Sqrt (4.0^2 - 3.0^2) ≈ 2.6 ft (Rounded to nearest tenth)
Question 2:
(a) Angle x
ground = 24 ft
support wire = 30 ft
Applying sine rule
support wire/Sin 90 = ground/Sin (90-x) ----- but Sin 90 = 1
Then,
Support wire = ground/Sin (90-x)
Sin (90-x) = ground/support wire = 24/30 = 0.8
90-x = Sin^-1(0.8) = 53.13 => x = 90-53.13 = 36.9°
(b) Height where the wire touches the tree (tree)
tree = Sqrt (support wire^2 - ground^2) = Sqrt (30^2 - 24^2) = 18 ft
Answers
Answer:
x = 90 - 2α
Step-by-step explanation:
Solution:-
- Consider the right angled triangle " ABD ". The sum of angles of an triangle is always "180°".
< BAD > + < ADB > + < ABD > = 180°
< ABD > = 180 - 90° - α
< ABD > = 90° - α
- Then we look at the figure for the triangle "ABE". Where " E " is the midpoint and intersection point of two diagonals " AC and BD ".
- We name the foot of the perpendicular bisector as " F ": " BF " would be the perpendicular bisector. The angle < BAE > is equal to < ABD >.
< ABD > = < BAE > = 90° - α ... ( Isosceles triangle " BEA " )
Where, sides ( BE = AE ).
- Use the law of sum of angles in a triangle and consider the triangle " BFA " as follows:
< ABF> + < BFA > + < BAF > = 180°
< ABF > = 180 - (90° - α) - 90°
< ABF > = α
Where, < BAF > = < BAE >
- The angle < ABD > = < ABE > is comprised of two angles namely, < ABF > and < FBE > = x.
< ABD > = < ABE > = < ABF > + x
90° - α = α + x
x = 90 - 2α ... Answer
Answer:
Step-by-step explanation:
The length of this triangle is 10 squares
and the width is 4 squares
The diagonals divide the rectangle into four triangles
These traingles are isoceles
Each two triangles facing each others are identical
<B = 90 degree
B = alpha + Beta
Let Beta be the angle next alpha
The segment that is crossing Beta is its bisector since it perpendicular to the diagonals wich means that:
Beta = 2x
Then B = alpha + 2x
90 = alpha +2x
90-alpha = 2x
x = (90-alpha)/2
B. $2.80
C. $10.00
D. $7.20
Answers
Answer:
C. $10.00
Step-by-step explanation:
She has to pay which ever is greater.
To get whichever is greater you need to check what 2% of her last billing cycle. That would be done by multiplying .02 and 360. That equals $7.20.
10>7.20 and therefore the answer would be 10 dollars aka C.
Answers
324 is exactly three times 108. Thus, the product of 324 and 4 is exactly three times the product of 108 and 4.
What is product?
In mathematics, a product is the outcome of multiplying two or more numbers together.
The outcome of multiplication is a product, or a factor, which is an expression that specifies the objects (numbers or variables) to be multiplied.
For instance, the result of multiplying 2 and 5 together is 10, or the product. Math requires a lot of multiplication.
Explanation:
Since, 324 is exactly three times 108,
324 = 3 x 108.
Therefore,
324 x 4 = (3 x 108) x 4
324 x 4 = 3 x (108 x 4)
Solving,
108 x 4 = 432
therefore,
324 x 4 = 3 x 432
324 x 4 = 1296.
As, shown above we can use the product of 108 and 4 to find the product of 324 and 4.
Thus, you multiply the product of 108*4 by thee to get 324*4
Therefore, 324 is exactly three times 108. Thus, the product of 324 and 4 is exactly three times the product of 108 and 4.
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Answers
67.935=67+935/1000=67+187/200 = 67|187/200
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Answers
Answer:
Sun is cm away from alpha century.
Sun is m.
Sun is 1.391016 time the size of an atom at this scale.
Step-by-step explanation:
Light year is a measure of distance. It is the distance light travels in an year.
Light year = km
So 4.4 light years = km
km
Lets scale this down to the level of
km
= km
Change the units to centimeters:
cm
= cm
= cm
Therefore on the new scale sun is cm away from alpha century.
Diameter of the sun is 1.391016 million km
Lets change Sun's diameter to the new scale:
km
=km
Lets change kilometers in to meters:
m
=m
Therefore, sun is m
and an atom is
Therefore the sun is 1.391016 time the size of an atom at this scale.
Final answer:
On a 1-to-10^19 scale, the distance from the Sun to the Alpha Centauri is about 44 cm. On this same scale, the Sun itself would have a diameter of about 150 picometers, which is larger than a typical atom.
Explanation:
The 1-to-10^19 scale means for every actual meter in space, we represent 10^19 meters on our model. The Alpha Centauri is 4.4 light-years away from the sun. Considering 1 light-year equals to approximately 9.46x10^15 meters, the real distance from the sun to Alpha Centauri is about 4.16x10^16 meters. So, on the scale, this is about 0.44 meters or 44 cm.
The Sun's real size, with a diameter of 1.5 million kilometers or 1.5x10^9 meters, is represented as 1.5x10^-10 meters or 150 picometers on the scale. This is much bigger than an actual atom, which has a diameter of 0.1 to 0.5 nanometers or 100 to 500 picometers. Hence, on this scale, the Sun would be larger than a typical atom.
Learn more about Scale Representation here:
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