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Anna [14]
3 years ago
6

Please hElp!! I don't know how to do this!!!

Mathematics
1 answer:
pishuonlain [190]3 years ago
7 0
150 more male bats need to be tagged

if you want 2/5 to be female you divide the 260 (<160+100) by 2 to get 130
then you multiply 130 by 3 to get the total amount of male bats that need to be tagged
130*3=390
then you subtract 240 from the 390 to find out how many more bats need to be tagged
390-240=150
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This for math. Rewriting Fractions <br> How do you decide which number to use in the Giant One?
ASHA 777 [7]

Answer:

factor of hundred is

100 - 1, 2,4,5,10,100

4 0
3 years ago
I am once again asking the mathematical geniuses of the world to solve a rather simple problem that my small brain can't compreh
Nitella [24]
Ooh ok these are fun. So these triangles are similar, meaning they are in proportion. In triangle LMN, line LM is 21. To find the corresponding line on the other triangle, we need to find the corresponding letters in the names. Because L and M are the first two letters, we need to use the first two letter s in the other name, so FG. Line FG is 9, so our first proportion is 9/21
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3 years ago
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the sum of nine times a number and fifteen is less than or equal to the sum of twenty four and ten times the number
lana66690 [7]
9n + 15 < = 10n + 24
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8 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
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Someone pleasee help me with this freaking IXL.
Norma-Jean [14]
The answer is 1761 1/2 (1761.5 in decimal form)
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