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NARA [144]
3 years ago
7

Given: AR ⊥ RS , TS ⊥ RS AT=26, RS=24 AR=12 Find: TS Answer: TS =

Mathematics
2 answers:
Olenka [21]3 years ago
6 0

Answer:

Solution -

Drawing a perpendicular from AQ to TS, we get a right angle triangle AQT

Using Pythagoras Theorem,

AT² = AQ² + QT²

⇒26² = 24² + QT²                    (∵ Due to symmetry AQ = RS)

⇒QT² = 676-576 = 100

⇒QT = 10

As TS = QT + QS = 12 + 10 = 22  ( ∵ Due to symmetry AR = QS )

∴ TS = 22 (ans)

kogti [31]3 years ago
4 0

Solution -

Drawing a perpendicular from AQ to TS, we get a right angle triangle AQT

Using Pythagoras Theorem,

AT² = AQ² + QT²

⇒26² = 24² + QT²                    (∵ Due to symmetry AQ = RS)

⇒QT² = 676-576 = 100

⇒QT = 10

As TS = QT + QS = 12 + 10 = 22  ( ∵ Due to symmetry AR = QS )

∴ TS = 22 (ans)

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For the region, we first find its boundary curves' points of intersection.
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Since x > x^4 for y in [0, 1],

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\int\limits^1_0 { \int\limits_{x^4}^x {(5x+9y)} \, dy } \, dx = \int\limits^1_0 {\left[5xy+ \frac{9}{2} y^2\right]_{x^4}^{x}} \, dx  \\  \\ =\int\limits^1_0 {\left[\left(5x(x)+ \frac{9}{2} (x)^2\right)-\left(5x(x^4)+ \frac{9}{2} (x^4)^2\right)\right]} \, dx  \\  \\ =\int\limits^1_0 {\left(5x^2+ \frac{9}{2} x^2-5x^5- \frac{9}{2} x^8\right)} \, dx =\int\limits^1_0 {\left( \frac{19}{2} x^2-5x^5- \frac{9}{2} x^8\right)} \, dx \\  \\ =\left[ \frac{19}{6} x^3- \frac{5}{6} x^6- \frac{1}{2} x^9\right]^1_0

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8 0
3 years ago
Please help me with both problems thank you so much!!
motikmotik
The one on the left is -5/6
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Use the data in the table to answer the question. Citations are "speeding tickets." You may fill in the table to help you answer
Vikentia [17]

Answer:

Linear approximation is given by: y = \frac{x}{5}+2

Step-by-step explanation:

We are given the following information in the question:

Number of citations:      5      7.5      10      15       20

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The equation of line is:

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The above equation is the required linear approximation.

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3 years ago
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