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3 years ago

like terms -5b2 considered like to 5b2 or because of the negative they would be unlike or is it the b2 that makes them like?

1 answer:

7 0

-5b2 is still considered to be like 5b2 because they share terms (the variable)
so it is the b that makes them alike the exponent and the negatives dont matter

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$\qquad \qquad\huge \underline{\boxed{\sf Answer}}$

Here's the solution ~

Let's find the measure of hypotenuse first, by using Pythagoras theorem ;

$\qquad \sf \dashrightarrow \: h {}^{2} = {8}^{2} + {6}^{2}$

$\qquad \sf \dashrightarrow \: h {}^{2} = {36}^{} + {64}^{}$

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$\qquad \sf \dashrightarrow \: \sin(x) = \dfrac{opposite \: side}{hypotenuse}$

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$\qquad \sf \dashrightarrow \: \sin(x) = \dfrac{3}{5} \: or \: 0.6 \: units$

For Cos y :

$\qquad \sf \dashrightarrow \: \cos(y) = \dfrac{adjcant \: side}{hypotenuse}$

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2 years ago

Question 9 of 25

tresset_1 [31]

Answer:

One approach to this problem is to obtain the graph for the given equation.

We need to find every intersection those functions have with the axis 'x' and 'y'

starting with g(x)

g(x=0)=0-3, first point (0,-3) it iis the crossing point with 'x' axis

g(x)=0=x-3, second point (3,0) it iis the crossing point with 'y' axis

Lets do the same for f(x)

g(x=0)=0, this leads to the first point (0,0) it iis the crossing point with 'x' axis and also, with the 'y' axis

We dont need to find any other, since always y=x

By plotting we have the attached picture

Now you can see that g(x) differs from its parent function in that is shifted 3 units to the right, and also 3 units down.

Step-by-step explanation:

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