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Xelga [282]
3 years ago
6

How do you Solve 2|x|=3

Mathematics
1 answer:
HACTEHA [7]3 years ago
8 0
If yo have |a|=b solve for a such that a=b or a=-b
divideboth sides by 2

|x|=3/2

x=3/2 or -3/2
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A car starts with a dull tank of gas. After driving around 5he city, 1/7 of the gas has been used. With the rest of the gas in t
Sati [7]

Answer:

\frac{2}{7}

Step-by-step explanation:

Given:

A car starts with a dull tank of gas

1/7 of the gas has been used around the city.

With the rest of the gas in the car, the car can travel to and from Ottawa three times.

Question asked:

What fractions of a tank of gas does each complete trip to Ottawa use?

Solution:

Fuel used around the city = \frac{1}{7}

Remaining fuel after driving around the city = 1 - \frac{1}{7}

                                                                         =    \frac{7 - 1}{7}  = \frac{6}{7}

According to question:

As from the rest of the gas in the car that is \frac{6}{7}, the car can complete 3 trip to Ottawa  which means,

By unitary method:

The car can complete 3 trip by using = \frac{6}{7} tank of gas.

The car can complete 1 trip by using =  \frac{6}{7} \div 3

                                                             =\frac{6}{7} \times\frac{1}{3}

                                                             =  \frac{6}{21}

                                                             = \frac{2}{7} tank of gas

Thus, \frac{2}{7} tank of gas used for each complete trip to Ottawa.

5 0
3 years ago
Please solve all i will mark brainliest and 15 points.​
OleMash [197]

Answer:

Hope that will help you get

3 0
3 years ago
Make x the subject of y=alog(5x)
Bond [772]

Answer:

x = 0.2(10^y/a).

Step-by-step explanation:

y = alog(5x)

y = log(5x)^a)   By the definition of a log:

(5x)^a  = 10^y

5x = (10^y) ^ 1/a

x = 0.2(10^y/a)

4 0
3 years ago
HELPP ME<br> s + 3 Five6= 6 17
Oduvanchick [21]

Answer:The crystal structures of five 6-mercaptopurine derivatives, viz. 2-[(9-acetyl-9H-purin-6-yl)sulfan­yl]-1-(3-meth­oxy­phen­yl)ethan-1-one (1), C16H14N4O3S, 2-[(9-acetyl-9H-purin-6-yl)sulfan­yl]-1-(4-meth­oxy­phen­yl)ethan-1-one (2), C16H14N4O3S, 2-[(9-acetyl-9H-purin-6-yl)sulfan­yl]-1-(4-chloro­phen­yl)ethan-1-one (3), C15H11ClN4O2S, 2-[(9-acetyl-9H-purin-6-yl)sulfan­yl]-1-(4-bromo­phen­yl)ethan-1-one (4), C15H11BrN4O2S, and 1-(3-meth­oxy­phen­yl)-2-[(9H-purin-6-yl)sulfan­yl]ethan-1-one (5), C14H12N4O2S. Compounds (2), (3) and (4) are isomorphous and accordingly their mol­ecular and supra­molecular structures are similar. An analysis of the dihedral angles between the purine and exocyclic phenyl rings show that the mol­ecules of (1) and (5) are essentially planar but that in the case of the three isomorphous compounds (2), (3) and (4), these rings are twisted by a dihedral angle of approximately 38°. With the exception of (1) all mol­ecules are linked by weak C—H⋯O hydrogen bonds in their crystals. There is π–π stacking in all compounds. A Cambridge Structural Database search revealed the existence of 11 deposited compounds containing the 1-phenyl-2-sulfanyl­ethanone scaffold; of these, only eight have a cyclic ring as substituent, the majority of these being heterocycles.

Keywords: crystal structure, mercaptopurines, supra­molecular structure

Go to:

Chemical context  

Purines, purine nucleosides and their analogs, are nitro­gen-containing heterocycles ubiquitous in nature and present in biological systems like man, plants and marine organisms (Legraverend, 2008 ▸). These types of heterocycles take part of the core structure of guanine and adenine in nucleic acids (DNA and RNA) being involved in diverse in vivo catabolic and anabolic metabolic pathways.

6-Mercaptopurine is a water insoluble purine analogue, which attracted attention due to its anti­tumor and immunosuppressive properties. The drug is used, among others, in the treatment of rheumathologic disorders, cancer and prevent

Step-by-step explanation:

6 0
3 years ago
Read 2 more answers
Please help me again
gulaghasi [49]

Answer:

65 degrees

Step-by-step explanation:

<u>Step 1:  Find what the measure of JLK is</u>

JLM + JLK = 180

140 + JLK - 140 = 180 - 140

JLK = 40

<u>Step 2:  Find what the measure of KJL is</u>

KJL + JLK + JKL = 180

KJL + 40 + 75 = 180

KJL + 115 - 115 = 180 - 115

KJL = 65

Answer:  65 degrees

5 0
3 years ago
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