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3 years ago

Help please I really can’t fail this exam

Mathematics

1 answer:

shtirl [24]3 years ago

7 0

Answer:E'F' and EF are equal in length.

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At a sale this week, a desk is being sold for $312. This is a 35% discount from the original price.

tekilochka [14]

The original price I believe would be $349.44

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3 years ago

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Express each statement using an inequality involving absolute value: The height of the plant must be within 2 inches of the stan

34kurt

Answer:

|h-13| ≤ 2

Step-by-step explanation:

The difference between the height of the plant (h) and show size (13 in) can be written as ...

h - 13

This value is allowed to be positive or negative, but its absolute value must not exceed 2 inches. Thus, the desired inequality is ...

|h -13| ≤ 2

6 0

3 years ago

In a cookie recipe 1 and 2/3 cups milk makes 16 cookies. How many cups of needed to make 24 cookies

aleksandr82 [10.1K]

Answer:

2 1/2 or 5/2 cups of milk

Step-by-step explanation:

if 16 cookies are made from 1 2/3 cups of milk which equals 5/3 cups of milk that means that you need 5/2 or 2 1/2 cups of milk to make 24 cookies because you need to multiply 5/3 by 16 and then multiply 5/48 by 24 which is 5/2 or 2 1/2 cups of milk

8 0

3 years ago

I need an answer to this math problem.

tatiyna

The greatest common factor is c^2

This is because c^2 is the most that you can divide both numbers by.

7 0

2 years ago

Read 2 more answers

A camera is mounted at a point 4000 feet away from a geyser. If the water is rising vertically at 900 ft/s when it is 3000 feet

Leno4ka [110]

Answer:

$\dot \theta = 0.144\,\frac{rad}{s}$ , $\dot \theta = 8.251\,\frac{deg}{s}$ (Option B)

Step-by-step explanation:

The trigonometric diagram is included herein as attachment. The expression is presented below:

$\tan \theta = \frac{y}{x}$

Where:

$x$ - Horizontal distance between the geyser and the camera.

$y$ - Vertical distance between the geyser and the camera.

The rate of change in terms of time is:

$\dot \theta \cdot \sec^{2}\theta = \frac{\dot y\cdot x-y\cdot \dot x}{x^{2}}$

$\dot \theta \cdot \left(\frac{1}{\cos^{2}\theta} \right) = \frac{\dot y \cdot x - y\cdot \dot x}{x^{2}}$

$\dot \theta = \left(\frac{\dot y \cdot x - y \cdot \dot x}{x^{2}} \right)\cdot \cos^{2}\theta$

$\dot \theta = \left(\frac{\dot y \cdot x - y\cdot \dot x}{x^{2}} \right)\cdot \left(\frac{x^{2}}{x^{2}+y^{2}} \right)$

$\dot \theta = \frac{\dot y \cdot x - y\cdot \dot x}{x^{2}+y^{2}}$

Finally,

$\dot \theta = \frac{\left(900\,\frac{ft}{s} \right)\cdot (4000\,ft)-(3000\,ft)\cdot \left(0\,\frac{ft}{s} \right)}{(4000\,ft)^{2}+(3000\,ft)^{2}}$

$\dot \theta = 0.144\,\frac{rad}{s}$

$\dot \theta = 8.251\,\frac{deg}{s}$

3 0

3 years ago