Answer both questions!

Mathematics

1 answer:

vodomira [7]2 years ago

3 0

1) What is the equation for the line?
slope = 16/2 = 8

answer
equation
y = 8x

2) What is the equation of a line with a slope of 3 and a point (3, 1) on the line?

Express the equation in the form of y=mx+b where m is the slope and b is the y-intercept.

y - 1 = 3(x - 3)
y - 1 = 3x - 9
y = 3x - 8

answer
y = 3x - 8

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(Q2) Which of the following points lies on the graph of the function y = 4x ?

stepladder [879]

Answer:

Step-by-step explanation:

I think I will answer the question the your attachment because it has full information.

The function is: y = 3* $2^{x}$

a. The function decreases

Wrong, because the base number is 2 and it is greater than 0. The function will go up

b. y=intercept of (0.2)

Wrong, Let substitute x =2 into the function: y = 3* $2^{2}$ = 12

c. if x increase by 1, the value of y will double

Because the base number is 2. For example:

If x = 2, y = 3* $2^{2}$ = 12

If x =3, y =3* $2^{3}$ =(3* $2^{2}$ *)2 = 24

d. contains the point (2,4)

Wrong, Let substitute x =4 into the function: y = 3* $2^{4}$ = 48

So we choose C

5 0

2 years 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

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