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3 years ago

What is the common ratio for the sequence (-100, 20, -4....)

Mathematics

2 answers:

hammer [34]3 years ago

8 0

Answer:

divided by -5

Step-by-step explanation:

kakasveta [241]3 years ago

8 0

Answer:

divide by 5

Step-by-step explanation:

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The measure of an angle is twenty-four times the measure of its complementary angle. What is the measure of each angle?

Bogdan [553]

Answer:

Step-by-step explanation:

24x - measure of an angle

x - the measure of a complementary angle

Complementary angles adds to 90°

x + 24x = 90°

25x = 90°

x = 90°÷25

x = 3.6°

24x = 24·3.6° = 86.4°

7 0

3 years ago

Which of the following correctly translates "quotient"?

tatiyna

Answer:

A quotient is division.

Step-by-step explanation:

Addition is sum

Division is quotient

muultiplication is product

subtraction is difference

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2 years ago

1. At time t=0,500 bacteria are in a petting dish, and this amount triples every 15 days.

atroni [7]

Answer:

4 better inbox

Step-by-step explanation:

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2 years ago

Which figure shows a reflection of pre-image DEFG over the x-axis?

Sergio [31]

The image that shows the reflection of DFEG over the x-axis is the FIRST option. The x-axis is the mirror line. The distance from the mirror line to the reflected shape equals to the distance from the original image to the mirror line.

3 0

3 years ago

Find the points on the curve where the tangent is horizontal or vertical. If you have a graphing device, graph the curve to chec

klasskru [66]

Answer:

There is a horizontal tangent at (0,-4)

The tangent is vertical at (-2,-3) and (2,-3).

Step-by-step explanation:

The given function is defined parametrically by the equations:

$x=t^3-3t$

and

$y=t^2-4$

The tangent function is given by:

$\frac{dy}{dx}=\frac{\frac{dy}{dt} }{\frac{dx}{dt} }$

$\implies \frac{dy}{dx}=\frac{2t}{3t^2-3}$

The tangent is vertical at when $\frac{dx}{dt}=0$

$\implies \frac{3t^2-3}{2t}=0$

$\implies 3t^2-3=0$

$\implies 3t^2=3$

$\implies t^2=1$

$\implies t=\pm1$

When t=1,

$x=1^3-3(1)=-2$ and $y=1^2-4=-3$

When t=-1,

$x=(-1)^3-3(-1)=2$ and $y=(-1)^2-4=-3$

The tangent is vertical at (-2,-3) and (2,-3).

The tangent is horizontal, when $\frac{dy}{dx}=0$ or $\frac{dy}{dt}=0$

$\implies 2t=0$

$\implies t=0$

When t=0,

$x=0^3-3(0)=0$ and $y=0^2-4=-4$

There is a horizontal tangent at (0,-4)

5 0

3 years ago