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3 years ago

The graph represents function 1 and the equation represents function 2:

Mathematics

1 answer:

Grace [21]3 years ago

8 0

A horizontal line has a slope(rate of change) of 0

y = 6x + 1....slope(rate of change) = 6

so function 2 is 6 higher then function 1

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I need help with this review question

WINSTONCH [101]

Answer:

Δ MAN

reason: ASA

Step-by-step explanation:

if you look at the congruency marks you can see that ∡F ≅ ∡A and we know that ∡N ≅ ∡N

markings also show us that sides UF ≅ MA

therefore, we know two angles and one side for each triangle

6 0

3 years ago

SA scuba diver reached a depth of -14 feet (below

shtirl [24]

Answer:

+2 ft

Step-by-step explanation:

He descends -12 feet (negative descend is an ascend) so he ascends 12 feet

-14 + 12 = -2 so he needs to travel 2ft

3 0

3 years ago

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Will someone please explain how to do this?

VladimirAG [237]

The answer you have selected is correct

3 0

3 years ago

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Which shows all factors pairs of 45.

maks197457 [2]

1, 3, 5, 9, 15, and 45

4 0

3 years ago

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Vector u has a magnitude of 5sq rt 3 feet and is directed due west. Vector v has a magnitude

Vikki [24]

The magnitude and the direction angle for the resultant vector are 10N and 120degrees respectively

The formula for calculating the resultant vector is expressed as:

$R^2 = u^2 + v^2\\R =\sqrt{u^2+v^2}$

Given the following distances:

u = -5√3 feet (west direction)

v = 5 feet

Substitute the given values into the formula above;

$R=\sqrt{(5\sqrt{3} )^2+5^2} \\R=\sqrt{(25 \times 3)+25}\\R=\sqrt{75 + 25}\\R=\sqrt{100}\\R = 10 feet$

The magnitude of the resultant vector is 10feet

Get the direction:

$\theta = tan^{-1}\frac{y}{x} \\\theta = tan^{-1}\frac{5}{-5\sqrt{3} }\\\theta =tan^{-1}\frac{-1}{\sqrt{3} }\\\theta = -60^0$

Since tan is negative in the second quadrant, hence;

$\theta = 180 - 60\\\theta = 120^0$

Hence the magnitude and the direction angle for the resultant vector are 10N and 120degrees respectively.

Learn more here: brainly.com/question/5177683

7 0

3 years ago