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2 years ago

Which is the line shown in the figure? Oline XY Oline XZ Oline WX Oline WZ

Mathematics

1 answer:

Anika [276]2 years ago

3 0

Answer:

line Xz

Step-by-step explanation:

because xz is

trasnveral

to XYZ

to XWZ

ok make

as brain list

thanks

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I could really use your help

nydimaria [60]

Answer:

x = $\frac{c-a}{5y}$

Step-by-step explanation:

Given

5xy + a = c ( isolate 5xy by subtracting a from both sides )

5xy = c - a ( isolate x by dividing both sides by 5y )

x = $\frac{c-a}{5y}$

8 0

4 years ago

For this graph, mark the statements that are true.

lord [1]

Given:

The graph of function.

To find:

The correct statement for the domain and range of the given graph.

Solution:

Domain: The set of input values.

Range: The set of output value.

From the given graph it is clear that the graph represents a polynomial function.

The given graph of a polynomial function is defined for all real values of x. So, the domain is the set of all real numbers.

The graph of the function approaches from negative infinite to positive infinite. So, the output the given graph is set of all real numbers. It means the range is the set of all real numbers.

Therefore, the correct options are B and D.

6 0

3 years ago

From a window 20 feet above the ground, the angle of elevation to the top of a building across

Nikitich [7]

Answer: The answer is 381.85 feet.

Step-by-step explanation: Given that a window is 20 feet above the ground. From there, the angle of elevation to the top of a building across the street is 78°, and the angle of depression to the base of the same building is 15°. We are to calculate the height of the building across the street.

This situation is framed very nicely in the attached figure, where

BG = 20 feet, ∠AWB = 78°, ∠WAB = WBG = 15° and AH = height of the bulding across the street = ?

From the right-angled triangle WGB, we have

$\dfrac{WG}{WB}=\tan 15^\circ\\\\\\\Rightarrow \dfrac{20}{b}=\tan 15^\circ\\\\\\\Rightarrow b=\dfrac{20}{\tan 15^\circ},$

and from the right-angled triangle WAB, we have'

$\dfrac{AB}{WB}=\tan 78^\circ\\\\\\\Rightarrow \dfrac{h}{b}=\tan 15^\circ\\\\\\\Rightarrow h=\tan 78^\circ\times\dfrac{20}{\tan 15^\circ}\\\\\\\Rightarrow h=361.85.$