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Crazy boy [7]
3 years ago
15

Determine the solution and the reasoning that justifies the solution to the systems of equations.

Mathematics
1 answer:
Lena [83]3 years ago
8 0

Answer:

<u>Option 2</u>

(2,8), because the graph of the two equations intersects at this point

Step-by-step explanation:

Just looking at the graph, you can tell that the point (2,8) is where the two lines intersect.

Hope this helped! :)

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Work out the inverse function <br><br>y=x-2
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Answer:

x + 2

Step-by-step explanation:

Replace the x with y.

Making is x = y -2

Solve for x=y-2

and you get x+2

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16.80 is the only reasonable answer
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(2x-3)^5 please I need it fast thank you.
exis [7]
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3 years ago
Find the area of the trapezoid below to the nearest tenth.
kondor19780726 [428]

Answer:

Area of trapezoid = 67.6 square units

Step-by-step explanation:

Area of a trapezoid is given by the expression,

Area of the trapezoid = \frac{1}{2}(b_1+b_2)h

Here, b_1 and b_2 are the parallel sides of the given trapezoid.

And 'h' = Height between the parallel sides

From the given triangle ABE,

m(∠ABE) = m(∠ABC) - m(∠EBC)

m(∠ABE) = 120° - 90°

               = 30°

By applying cosine rule in the given triangle,

cos(30°) = \frac{\text{Adjacent side}}{\text{Hypotenuse}}

\frac{\sqrt{3} }{2}=\frac{BE}{AB}

\frac{\sqrt{3} }{2}=\frac{BE}{6}

BE = 3\sqrt{3} units

By applying sine rule in ΔABE,

sin(30°) = \frac{\text{Opposite side}}{\text{Hypotenuse}}

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AE = 3 units

Length of b_1=BC=10

Length of b_2=AD=(AE+EF+FD) [AE = FD, since given trapezoid ABCD is an isosceles trapezoid]

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                b_2=16

Height between the parallel sides h=3\sqrt{3}

Area of the trapezoid = \frac{1}{2}(BC+AD)BE

                                    = \frac{1}{2}(10+16)(3\sqrt{3})

                                    = 39\sqrt{3}

                                    = 67.6 square units

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3 years ago
Write the polynomial in standard form. Then name the polynomial based on its degree and number of terms.2-5x^3+6xA) 2-5x^3+6x; n
puteri [66]
The answer is B, hope this helps
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