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2 years ago

To measure the distance across a pond, Sandy and Garrett mark a spot at on

Mathematics

1 answer:

DerKrebs [107]2 years ago

4 0

Answer:

distance across the pond: 13 yd

distance walked by Garret: 12 yd

Step-by-step explanation:

Data:

distance across the pond: x yd

distance walked by Sandy: 5 yd

distance walked by Garret: x-1 yd

A right triangle is formed where the distance across the pond is the hypotenuse. From Pythagorean theorem:

x² = 5² + (x-1)²

x² = 25 + x² - 2x + 1

0 = 26 - 2x

x = 13

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Help me find the slope of this imagine. it a quizizz question and i cant figure it out

lidiya [134]

Answer:

slope = $\frac{5}{2}$

Step-by-step explanation:

Select any 2 ordered pairs from the given table, the use the slope formula to calculate the slope m

m = (y₂ - y₁ ) / (x₂ - x₁ )

with (x₁, y₁ ) = (2, 5) and (x₂, y₂ ) = (4, 10) ← 2 ordered pairs from table

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3 years ago

Is the point 3/5 and 4/5 corresponds to an angle in the unit circle what is the csc

Mariulka [41]

<u>Answer:</u>

csc = 5/4

<u>Step-by-step explanation:</u>

Assuming that the two given points 3/5 and 4/5 are the two sides of a triangle and a right angle 0 with legs x and y and z as its hypotenuse.

Then using Pythagoras Theorem:

$z^2=x^2+y^2$

$z^2=(\frac{3}{5} )^2+(\frac{4}{5} )^2$

$z^2 = \frac{9}{25} + \frac{16}{25}$

$z^2=\frac{25}{25}$

$z^2=1$

Taking square root on both the sides:

$z=1$

Now, $sin\alpha=\frac{4}{5}$

and we know that $csc = \frac{1}{sin\alpha }$

Therefore, csc = 5/4.

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3 years ago

For the given function f and g, complete parts (a)-(h). For parts (a)-(d), also find the domain.

kakasveta [241]

Answer:

a) (f + g)(x) = 9x + 1

Domain: x ε R

b) (f - g)(x) = (-5x + 13)

Domain: x ε R

c) (f.g)(x) = 14x² + 37x - 42

Domain: x ε R

d) (f/g)(x) = (2x + 7)/(7x -6)

Domain: (x ε R except x = 6/7)

e) (f + g)(7) = 64

f) (f - g)(2) = 3

g) (f.g)(3) = 195

h) (f/g)(x) = 1/3

Step-by-step explanation:

f(x) = 2x + 7, g(x) = 7x - 6

a) (f + g)(x) = f(x) + g(x) = (2x + 7) + (7x - 6)

(f + g)(x) = 9x + 1

Since x is defined for functions f & g for all real numbers, the domain of (f + g)(x) is x ε R

b) (f - g)(x) = f(x) - g(x) = (2x + 7) - (7x - 6)

(f - g)(x) = (-5x + 13)

Since x is defined for functions f & g for all real numbers, the domain of (f - g)(x) is x ε R

c) (f.g)(x) = f(x) × g(x) = (2x + 7)(7x - 6)

(f.g)(x) = 14x² - 12x + 49x - 42 = 14x² + 37x - 42

Since x is defined for functions f & g for all real numbers, the domain of (f.g)(x) is x ε R

d) (f/g)(x) = f(x)/g(x) = (2x + 7)/(7x -6)

x is defined for functions f & g for all real numbers, the domain of (f/g)(x) will be x ε R except when the denominator vanishes (that is, goes to zero). This will cause the function to take up values of ∞.

This will happen when 7x - 6 = 0, x = 6/7.

Therefore, the domain of (f/g)(x) is x ε R except the point, x = 6/7.

e) (f + g)(7) = 9(7) + 1 = 64

f) (f - g)(2) = -5(2) + 13 = 3

g) (f.g)(3) = 14(3²) + 37(3) - 42 = 195

h) (f/g)(27) = (2(27) + 7)/(7(27) - 6) = 61/183 = 1/3

Hope this Helps!!!

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2 years ago