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ivanzaharov [21]
3 years ago
14

Triangle FGH is equilateral. The midpoints of the sides are connected to form triangle XYZ. Line segment XY is parallel to line

segment FH.
Triangle F G H is an equilateral triangle. Equilateral triangle X Y Z is inside of triangle F G H and the points are the midpoints of triangle F G H.

What type of figure is quadrilateral FXYH?
Mathematics
1 answer:
Hatshy [7]3 years ago
3 0

Answer:

Isosceles trapezoid

Step-by-step explanation:

The figure will look like one large triangle on the outside and four smaller triangles on the inside. you are only concerned with the inside triangle that is pointing to the line FH. Since XY and FH are parallel and the distance between FX and YH are equivalent (the midpoints of each side should be the same for all three sides of the larger triangle) the shape FXYH is an isosceles trapezoid.

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Step-by-step explanation:

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Answer the question below. Be sure to show your work
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ANSWER:

We have to find new side-lengths of PQR triangle.

Original sides are.

\begin{gathered} PQ\text{ = 8cm} \\ QR=17\operatorname{cm} \\ RP=15\operatorname{cm} \end{gathered}

After multiplying by 2.5 we get

\begin{gathered} PQ^{\prime}=2.5\times PQ=(8\times2.5)cm=20\operatorname{cm} \\ QR^{\prime}=(17\times2.5)cm=42.5\operatorname{cm} \\ RP^{\prime}=(15\times2.5)cm=37.5\operatorname{cm} \end{gathered}

These are the new sides of the P'Q'R' triangle.

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1 year ago
Differentiating a Logarithmic Function in Exercise, find the derivative of the function. See Examples 1, 2, 3, and 4.
Leni [432]

Answer:  \dfrac{2x^2-1}{x(x^2-1)}

Step-by-step explanation:

The given function : y=\ln(x(x^2 - 1)^{\frac{1}{2}})

\Rightarrow\ y=\ln x+\ln (x^2-1)^{\frac{1}{2}}    [\because \ln(ab)=\ln a +\ln b]

\Rightarrow y=\ln x+\dfrac{1}{2}\ln (x^2-1)}  [\because \ln(a)^n=n\ln a]

Now , Differentiate both sides  with respect to x , we will get

\dfrac{dy}{dx}=\dfrac{1}{x}+\dfrac{1}{2}(\dfrac{1}{x^2-1})\dfrac{d}{dx}(x^2-1) (By Chain rule)

[Note : \dfrac{d}{dx}(\ln x)=\dfrac{1}{x}]

\dfrac{1}{x}+\dfrac{1}{2}(\dfrac{1}{x^2-1})(2x-0)

[ \because \dfrac{d}{dx}(x^n)=nx^{n-1}]

=\dfrac{1}{x}+\dfrac{1}{2}(\dfrac{1}{x^2-1})(2x) = \dfrac{1}{x}+\dfrac{x}{x^2-1}\\\\\\=\dfrac{(x^2-1)+(x^2)}{x(x^2-1)}\\\\\\=\dfrac{2x^2-1}{x(x^2-1)}

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8 0
3 years ago
The junior and senior students at Mathville High School are going to present an exciting musical entitled, "Math, What is it Goo
xeze [42]

Some parts are missing in the queston. Find attached the picture with the complete question

Answer:

                   \large\boxed{\large\boxed{161}}

Explanation:

Let's put the information in a table step-by step.

                                                (number of remaining students)

                                                        Juniors          Seniors

Condition

  • Initially                                           J                     S
  • 15 seniors left                                                   S - 15
  • Twice juniors as seniors         2(S - 15)
  • 3/4 of the juniors left              1/4×2(S - 15)
  • 1/3 of seniors left                                             2/3×(S - 15)

At the end, there were 8 more seniors than juniors:

  • 2/3×(S - 15) -  1/4×2(S - 15) = 8

Now you have obtained one equation, which you can solve to find S, the number of senior students, and then the number of junior students.

Solve the equation:

2/3\times (S - 15) -  1/4\times 2(S - 15) = 8

  • Mutilply all by 12:

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  • Distribution property:

8S-120-6S-90=96

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8S-6S=96+120+90

  • Add like terms:

2S=306

  • Division property of equalities:

S=306/2=153

That is the number of senior students that came out to the information meeting, but the number of students remaining to perform in the school musical is (from the table above):

2/3\times (S-15)+1/4\times 2(S-15)

Just substitute S with 153 fo find the number of students that remained to perfom in the musical:

          2/3\times (153-15)+1/4\times 2(153-15)\\ \\ 2/3(138)+1/2(138)

          161

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olasank [31]

Answer:

See image

Step-by-step explanation:

You put One-place decimals over ten and then simplify.

You put Two-place decimals over 100 and then simplify. See image.

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