What is the apparent solution to the system of equations graphed above?

Simplify the expression, Please help!!

Serhud [2]

For the first one it would be C.

3 0

2 years ago

The function is defined by f(x)=x^2+6

jeka57 [31]

Answer: $9x^2 +6$

Step-by-step explanation:

$f(3x)=(3x)^2 +6=9x^2 +6$

8 0

1 year ago

part 3. Find the value of the trig function indicated, use Pythagorean theorem to find the third side if you need it.

Nat2105 [25]

Answer: $\bold{9)\ \sin \theta=\dfrac{1}{3}\qquad 10)\ \sin \theta = \dfrac{4}{5}\qquad 11)\ \cos \theta = \dfrac{\sqrt{11}}{6}\qquad 12)\ \tan \theta = \dfrac{17\sqrt2}{26}}$

Step-by-step explanation:

Pythagorean Theorem is: a² + b² = c² , where "c" is the hypotenuse

$9)\ \sin \theta=\dfrac{\text{side opposite of}\ \theta}{\text{hypotenuse of triangle}}=\dfrac{4}{12}\quad \rightarrow \large\boxed{\dfrac{1}{3}}$

Note: 4² + (8√2)² = hypotenuse² → hypotenuse = 12

$10)\ \sin \theta=\dfrac{\text{side opposite of}\ \theta}{\text{hypotenuse of triangle}}=\dfrac{16}{20}\quad \rightarrow \large\boxed{\dfrac{4}{5}}$

Note: 12² + opposite² = 20² → opposite = 16

$11)\ \cos \theta=\dfrac{\text{side adjacent to}\ \theta}{\text{hypotenuse of triangle}}=\dfrac{\sqrt{11}}{6}\quad =\large\boxed{\dfrac{\sqrt{11}}{6}}$

Note: adjacent² + 5² = 6² → adjacent = √11

$12)\ \tan \theta=\dfrac{\text{side opposite of}\ \theta}{\text{side adjacent to}\ \theta}=\dfrac{17}{13\sqrt2}\quad =\large\boxed{\dfrac{17\sqrt2}{26}}$

Note: adjacent² + 7² = (13√2)² → adjacent = 17

5 0

3 years ago

Two piecewise functions are shown below. What is the value of

gizmo_the_mogwai [7]

Step-by-step explanation:

I am not sure what your problem here is.

you understand the inequality signs ?

anyway, to get

6×f(-2) + 3×g(1)

we can calculate every part of the expression separately, and then combine all the results into one final result.

f(-2)

we look at the definition.

into what category is -2 falling ? the one with x<-2, or the one with x>=-2 ?

is -2 < -2 ? no.

is -2 >= -2 ? yes, because -2 = -2. therefore, it is also >= -2.

so, we have to use

1/3 x³

for x = -2 that is

1/3 × (-2)³ = 1/3 × -8 = -8/3

g(1)

again, we look at the definition.

into what category is 1 falling ? the one with x > 2 ? or the one with x <= 1 ?

is 1 > 2 ? no.

is 1 <= 1 ? yes, because 1=1. therefore it is also <= 1.

so we have to use

2×|x - 1| + 3

for x = 1 we get

2×0 + 3 = 3

6×f(-2) = 6 × -8/3 = 2× -8 = -16

3×g(1) = 3× 3 = 9

and so in total we get

6×f(-2) + 3×g(1) = -16 + 9 = -7

3 0

1 year ago

PLEASE I NEED THE SLOPE OF THE RED LINE AND THE BLUE LIKE BY NOW!!!!

netineya [11]

Red line slope=3/4
Blue line slope=2/3

3 0

2 years ago

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