All categories

2 years ago

I don’t understand please help

Mathematics

1 answer:

kow [346]2 years ago

7 0

The anwser to this is 6.1

You might be interested in

If p and q vary inversely and p is 17 when q is 4, determine q when p is equal to 2.

Maslowich

Answer: what the hec... what kind of question-

Step-by-step explanation:

6 0

3 years ago

15. Robin ordered pizza and opted for delivery.

Ivenika [448]

Answer: No she'll need atlease 9 dollars

Step-by-step explanation:

48.95 x 18% is 8.811 which is greater than

4 0

2 years ago

Alice says that all the factors of 8 are even.

Rom4ik [11]

1 is not even yet it is a factor of 8. Alice is wrong.

6 0

2 years ago

20 points!! Please help

Bas_tet [7]

Answer:

If the statement is true, lines a and b are parallel

Step-by-step explanation:

This is due to lines intersecting with line m at identical angles.

8 0

3 years ago

You are looking at the New York ball drop on New Year’s Eve at a distance of 100 m away from the base of the structure. If the b

Fofino [41]

The question is an illustration of related rates.

The rate of change between you and the ball is 0.01 rad per second

I added an attachment to illustrate the given parameters.

The representations on the attachment are:

$\mathbf{x = 100\ m}$

$\mathbf{\frac{dy}{dt} = 2\ ms^{-1}}$ ---- the rate

$\mathbf{\theta = \frac{\pi}{4}}$

First, we calculate the vertical distance (y) using tangent ratio

$\mathbf{\tan(\theta) = \frac{y}{x}}$

Substitute 100 for x

$\mathbf{y = 100\tan(\theta) }$

$\mathbf{\tan(\theta) = \frac{y}{100}}$

Differentiate both sides with respect to time (t)

$\mathbf{ \sec^2(\theta) \cdot \frac{d\theta}{dt} = \frac{1}{100} \cdot \frac{dy}{dt}}$

Substitute values for the rates and $\mathbf{\theta }$

$\mathbf{ \sec^2(\pi/4) \cdot \frac{d\theta}{dt} = \frac{1}{100} \cdot 2}$

This gives

$\mathbf{ (\sqrt 2)^2 \cdot \frac{d\theta}{dt} = \frac{1}{100} \cdot 2}$

$\mathbf{ 2 \cdot \frac{d\theta}{dt} = \frac{1}{100} \cdot 2}$

Divide both sides by 2

$\mathbf{ \frac{d\theta}{dt} = \frac{1}{100} }$

$\mathbf{ \frac{d\theta}{dt} = 0.01 }$

Hence, the rate of change between you and the ball is 0.01 rad per second