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son4ous [18]
4 years ago
11

Is the relation a function? {(14, 15), (5, 7), (3, 10), (11, 1), (5, 8)} a. yes. b. no

Mathematics
1 answer:
Monica [59]4 years ago
8 0
In this relation we have two ordered pairs:
( 5, 7 ) and ( 5, 8 )
For x = 5 :  f ( x ) = 7 and also for x = 5,  f ( x ) = 8. This is not possible for a function.
Answer:
b ) No. The relation is not a function.
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suppose you got 8 mangoes and 3 apples for $18 and 3 mangoes and 5 apples for $14.50. how mush does each mango and each apple co
Zanzabum
Suppose a mango cost x dollar and a apple cost y dollar

there will be an binary system of equation

8*x+3*y=18
3*x+5*y=14.5

so soultions have to x=48/31     y=16/31

6 0
3 years ago
Read 2 more answers
If the m∠2 = 40, what is the m∠5?
77julia77 [94]

Answer: The answer is 100.

Step-by-step explanation: 40 divided by 2 = 20

x=20

20x5=100

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3 years ago
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I’m kind of having troublehelp?
Bezzdna [24]

Answer:

13

Step-by-step explanation:

a^2 + b^2 = c^2

11^2 + 7 ^2 = c^2

121 + 49 = 170

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5 0
3 years ago
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Amy is pulling a wagon with a force of 30 pounds up a hill at an angle of 25°. Give the force exerted on the wagon as a vector a
Fofino [41]

Answer:

Vector (ordered pair - rectangular form)

\vec F = (27.189,12.679)\,[lbf]

Vector (ordered pair - polar form)

\vec F = (30\,lbf, 25^{\circ})

Sum of vectorial components (linear combination)

\vec F = 27.189\cdot \hat{i} + 12.679\cdot \hat{j}\,[N]

Step-by-step explanation:

From statement we know that force exerted on the wagon has a magnitude of 30 pounds-force and an angle of 25° above the horizontal, which corresponds to the +x semiaxis, whereas the vertical is represented by the +y semiaxis.

The force (\vec F), in pounds-force, can be modelled in two forms:

Vector (ordered pair - rectangular form)

\vec F =  \left(\|\vec F\|\cdot \cos \theta, \|\vec F\|\cdot \sin \theta\right) (1)

Vector (ordered pair - polar form)

\vec F = \left(\|\vec F\|, \theta\right)

Sum of vectorial components (linear combination)

\vec {F} = \left(\|\vec F\|\cdot \cos \theta\right)\cdot \hat{i} + \left(\|\vec F\|\cdot \sin \theta \right)\cdot \hat{j} (2)

Where:

\|\vec F\| - Norm of the vector force, in newtons.

\theta - Direction of the vector force with regard to the horizontal, in sexagesimal degrees.

\hat{i}, \hat{j} - Orthogonal axes, no unit.

If we know that \|\vec F\| = 30\,lbf and \theta = 25^{\circ}, then the force exerted on the wagon is:

Vector (ordered pair - rectangular form)

\vec F= \left(30\cdot \cos 25^{\circ}, 30\cdot \sin 25^{\circ}\right)\,[lbf]

\vec F = (27.189,12.679)\,[lbf]

Vector (ordered pair - polar form)

\vec F = (30\,lbf, 25^{\circ})

Sum of vectorial components (linear combination)

\vec F = (30\cdot \cos 25^{\circ})\cdot \hat{i} + (30\cdot \sin 25^{\circ})\cdot \hat{j}\,[N]

\vec F = 27.189\cdot \hat{i} + 12.679\cdot \hat{j}\,[N]

8 0
3 years ago
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