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Alchen [17]
3 years ago
5

I REALLY NEED HELP I FORGOT HOW TO DO THIS :(

Mathematics
1 answer:
Elena-2011 [213]3 years ago
3 0

Answer:

117.8

Step-by-step explanation:

i used heron's formula

√s(s-a)(s-b)(s-c)

s= (20+16+33)/2 = 34.5

√34.5(34.5-20)(34.5-16)(34.5-33)

= 117.8

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The numerator of fraction is 1 less then its denominator if the numerator is increased by 5 the new fraction becomes equivalent
jek_recluse [69]

Answer:

4/5

Step-by-step explanation:

the numerator of fraction is 1 less than its denominator. If the numerator is increased by 4 and the denominator is increased by 5, the new fraction becomes equivalent to 4/5. what is the original fraction?

Let n = numerator

d = denominator

n = d - 1

\frac{d - 1}{d}

\frac{d - 1 + 4}{d + 5} = \frac{4}{5}

\frac{d + 3}{d+ 5} = \frac{4}{5}

cross multiply

5(d + 3) = 4(d + 5)

5d + 15 = 4d + 20

d = 5

n = d - 1

5 - 1 = 4

original fraction 4/5

8 0
3 years ago
For all values of x,<br>f(x) = x^2 + 3x and g(x) = x - 4<br>Show that fg(x) = x^2 - 5x + 4​
Westkost [7]

Answer:

f(g(x)) =  x² - 5x +4

Step-by-step explanation:

<u><em>Explanation</em></u>

Given that the functions

         f(x) = x² + 3 x   and g(x) = x-4

     f(g(x)) = f(x -4)   (∵ g(x) = x-4)

              = (x-4)² + 3(x-4)

              = x² - 2(4)x + 4² + 3x -12

             = x² - 8 x + 16 + 3x -12

             = x² - 5x +4

∴ f(g(x)) =  x² - 5x +4

4 0
3 years ago
For the parallelogram ABCD the extensions of the angle bisectors AG and BH intersect at point P. Find the area of the parallelog
natita [175]

Answer:

The area of a parallelogram is 360 in.²

Step-by-step explanation:

Where DG = GH

GP = 12 in.

AB = 39 in.

∠DAB + ∠ABC = 180° (Adjacent angles of a parallelogram)

Whereby ∠DAB is bisected by AG and ∠ABC is bisected by BH

Therefore, ∠GAB + ∠HBA = 90°

Hence, ∠BPA = 90° (Sum of interior angles of a triangle)

cos(\angle GAB) = \dfrac{AP}{AB} = \dfrac{AP}{39} = \dfrac{GP}{GH} =\dfrac{12}{GH}

We note that ∠AGD = ∠GAB (Alternate angles of parallel lines)

∴ ∠AGD = ∠AGD since ∠AGD = ∠GAB (Bisected angle)

Hence AD = DG (Side length of isosceles triangle)

The bisector of ∠ADG is parallel to BH and will bisect AG at point Q

Hence ΔDAQ  ≅ ΔDGQ ≅ ΔGPH and AQ = QG = GP

Hence, AP = 3 × GP = 3 × 12 = 36

cos(\angle GAB) = \dfrac{AP}{AB} = \dfrac{36}{39}

\angle GAB = cos^{-1} \left (\dfrac{36}{39}  \right )

∠GAB = 22.62°

cos(\angle GAB) =  \dfrac{36}{39} = \dfrac{12}{GH}

GH =  \dfrac{39}{36} \times {12}

GH = 13 in.

∴ AD 13 in.

BP = 39 × sin(22.62°) = 15 in.

GH = √(GP² + HP²)

∠DAB = 2 × 22.62° = 45.24°

The height of the parallelogram = AD × sin(∠DAB) =  13 × sin(45.24°)

The height of the parallelogram = 120/13 =  9.23 in.

The area of a parallelogram = Base × Height = (120/13) × 39 = 360 in.²

7 0
3 years ago
Read 2 more answers
Find the sum of a finite geometric series. The sides of an equilateral triangle measure 16 inches. The midpoints of the sides of
olga2289 [7]
48 + 24 + 12 + 6 = 90 inches or <span>3*(16+8+4+2) = 90 will be the right answer.</span>
7 0
3 years ago
Read 2 more answers
Simplify.
Ugo [173]


so you take -10.5+5.3+20.2=??

-10.5+5.3=-5.2

-5.2+20.2=15

Ur Answer Is 15

Did I Help??

Hope I Did!!


7 0
3 years ago
Read 2 more answers
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