CAN ANYONE HELP ME WITH THIS !!!!??? *****Theres 2 questions *****

In general, the y intercept of the function F(x)= a•b^x is the point

serg [7]

The y intercept of this function is always (0, a).

This is because when we place 0 in for x (which is the only way it'll be on the y-axis, we get 'a' as a result. This is because of the rule that raising anything to the 0th power will result in the number 1 and multiplying anything by 1 gives us the same number. See the work below for the example.

F(x) = a*b^x

F(0) = a*b^0

F(0) = a*1

F(0) = a

And for an example with a random number, we'll use a = 5 and b = 3

F(x) = 5*3^x

F(0) = 5*3^0

F(0) = 5*1

F(0) = 5

No matter what a and b equal, the intercept will be the a value.

4 0

2 years ago

Solve using the quadratic formula 3x^2+11x+5=0

storchak [24]

Answer:

$3x^2 + 11x + 5 = 0 : x = \frac{-11 + \sqrt{61} }{6} , x = \frac{-11 - \sqrt{61} }{6}$

Decimal:

$(x = -0.53162..., x = -3.13504...)$

Step-by-step explanation:

$3x^2+11x+5=0$

Solve with the quadratic formula:

For a quadratic equation of the form $ax^2+bx+c=0$ the solutions are:

$x_{1,\:2}=\frac{-b\pm \sqrt{b^2-4ac}}{2a}$

For $a=3,\:b=11,\:c=5:\quad x_{1,\:2}=\frac{-11\pm \sqrt{11^2-4\cdot \:3\cdot \:5}}{2\cdot \:3}$

$x=\frac{-11 + \sqrt{11^2-4\cdot \:3\cdot \:5}}{2\cdot \:3} : \frac{-11 + \sqrt{61} }{6}$

$x=\frac{-11 - \sqrt{11^2-4\cdot \:3\cdot \:5}}{2\cdot \:3} : \frac{-11 - \sqrt{61} }{6}$

The solutions to the quadratic equation are:

$x=\frac{-11+\sqrt{61}}{6},\:x=\frac{-11-\sqrt{61}}{6}$

Hope I helped. If so, may I get brainliest and a thanks?

Thank you, have a good day! =)

8 0

3 years ago

Is 3 x 5/6: A: 15/6 B:15/18 ????

kiruha [24]

Answer:

3 x 5/6 is 2.5, so 15/6

Step-by-step explanation:

3/1 x 5/6

6 0

2 years ago

Read 2 more answers

ZA Round your answer to the nearest hundredth.

Ahat [919]

Answer:

A = 48.81°

Step-by-step explanation:

Angles in a Right Triangle

When the side lengths of a right triangle are known and an angle must be determined, we can use the trigonometric ratios that relate angles and sides.

The tangent ratio is defined as:

$\displaystyle \tan A=\frac{\text{opposite leg}}{\text{adjacent leg}}$