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sergiy2304 [10]
2 years ago
6

Please help I need the answer quick this assignment is almost due!

Mathematics
2 answers:
Svetach [21]2 years ago
3 0
The correct answer is x= -3
Lilit [14]2 years ago
3 0

Answer:

x = -3

Step-by-step explanation:

The two angles shown are <em>corresponding angles</em>, meaning that they have the same value. Because of this, we can create an equation to solve for the value of x.

2x + 9 = 7x + 24

Remember, those two angles are similar, and that's why we can make the two values equal to each other. First, let's remove 7x from both sides.

2x - 7x + 9 = 7x - 7x + 24

-5x + 9 = 24

Then, we'll remove 9 from both sides.

-5x + 9 - 9 = 24 - 9

-5x = 15

Finally, we'll divide both sides by -5 to isolate x and figure out what it is equivalent to.

-5x/-5 = 15/-5

x = -3

And there it is... the value of x. Hopefully that's helpful! :)

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The 4 sides of a square are usually the same and the side length of squares 4 and 5 is a+b.

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From the diagram, each side is called the side length. Therefore,

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Let f= {(1,2),(2,3),(3,5),(4,7)}<br><br> What is f ⁻¹ ∘ f
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Test of a horizontal line.Suppose f is a function.F does not have an inverse if any horizontal line crosses the graph of f more than once.F does have an inverse if no horizontal line crosses the graph of f more than once

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f= {(1,2),(2,3),(3,5),(4,7)}

Dom(gof)=dom(f)={1,3,4}.

(gof)(1)=g{f(1)}=g(2)=3,

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If there’s is an increase in price at which a product is sold what is most likely going to occur
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Find the given value of x 2x + 3 for x = 3
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4 0
4 years ago
Read 2 more answers
An airplane leaves an airport and flies due west 150 miles and then 170 miles in the direction S 49.17°W. How far is the plane f
Sphinxa [80]

Answer:

Distance between plane and airport is 134.4 miles.

Step-by-step explanation:

Given : An airplane leaves an airport and flies due west 150 miles and then 170 miles in the direction S 49.17°W.

To find : How far is the plane from the airport.

Solution : Distance from airport to west is 150 miles and then 170 miles in the direction south  and angle form is S 49.17° W

Refer the attached picture for clearance.

Applying law of cosines

c^2=a^2+b^2-2ab Cos(C)

c=\sqrt{a^2+b^2-2ab Cos(C)}

where a= 150 miles

b=170 miles

C=   49.17° angle in degree

c = distance between plane from the airport

Put values in the formula,

c=\sqrt{a^2+b^2-2ab Cos(C)}

c=\sqrt{150^2+170^2-2(150)(170) Cos(49.17^{\circ})}

c=\sqrt{22500+28900-51000(0.653)}

c=\sqrt{51400-33303}

c=\sqrt{18057}

c=134.37

Therefore, Distance between plane and airport is 134.4 miles.

5 0
3 years ago
Read 2 more answers
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