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Dovator [93]
4 years ago
8

P= 2/ + 2w Find The Width

Mathematics
1 answer:
Gnesinka [82]4 years ago
7 0

9514 1404 393

Answer:

  W = P/2 -L

Step-by-step explanation:

Divide by 2 and subtract L.

  P = 2L +2W

  P/2 = L +W

  P/2 -L = W

The width is ...

  W = P/2 -L

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One ounce of olive oil costs $0.23. How much does the whole bottle cost if it contains 32 ounces?
Gwar [14]

Answer:$7.36

Step-by-step explanation:

Just multiply 0.23 with 32

7 0
3 years ago
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Help me , this is my homework (Geometry) . I really need help
klemol [59]

8. Similar parts


\dfrac{6}{18} = \dfrac{AQ}{12}


AQ = 12(6)/18=12/3=4


Choice a


9. The Law of Sines says a:b:c and \sin A : \sin B \sin C are the same ratio.


The sines will go in order for acute angles, we have to convert 120 to its supplement 60 and then the order is 29,31,"60" or opposite sides JK, KL, JL


choice b


10.


DEF and FEG are supplments, FEG=180-DEF=180-114=66


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6 0
3 years ago
Write the polynomial as a product of a difference and a sum: b^2-4/9
Alenkinab [10]

Answer:

\sf \huge \boxed{ \boxed{(b  +   \frac{2}{3} )(b -  \frac{2}{3} )}}

Step-by-step explanation:

<h3>to understand this</h3><h3>you need to know about:</h3>
  • algebra
  • PEMDAS
<h3>tips and formulas:</h3>
  • {x}^{2}  -  {y}^{2}  = (x + y)(x - y)
<h3>let's solve:</h3>
  1. \sf \: rewrite : \\ (  {b})^{2}  - (  \frac{2}{3}  {)}^{2}
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4 0
3 years ago
Uranus moves in an elliptical orbit with the sun at one of the foci. The length of the half of the major axis is 2,876,769,540 k
Alina [70]

Answer:

The minimum distance (perihelion) of Uranus from the sun is 2,749,040,972.

Step-by-step explanation:

Consider the provided information.

The length of the half of the major axis is 2,876,769,540 kilometers, and the eccentricity is 0.0444.

The eccentricity (e) of an ellipse is the ratio of the distance from the center to the foci (c) and the distance from the center to the vertices (a).

e=\frac{c}{a}

Substitute a = 2,876,769,540 and e = 0.0444 in above formula and solve for c.

0.0444=\frac{c}{2,876,769,540 }

c=127728567.576

Minimum distance of Uranus from the sun is:

a-c=2,876,769,540-127728567.576\\a-c=2749040972.424\approx2749040972

Hence, the minimum distance (perihelion) of Uranus from the sun is 2,749,040,972.

6 0
3 years ago
Please help due today
marusya05 [52]

Answer:

15

Step-by-step explanation:

Let the smallest integer be x. The integers will be x, x+1, x+2, all the way up to x+6.

119 = 7x + (1+2+3+4+5+6)

119 = 7x + 21

7x = 98

x = 14

The second smallest integer will be x+1, so x+1 = 15.

5 0
3 years ago
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