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guapka [62]
2 years ago
15

ac{1}{3} + \frac{7}{15} " alt=" \frac{1}{3} + \frac{7}{15} " align="absmiddle" class="latex-formula">
​
Mathematics
2 answers:
luda_lava [24]2 years ago
6 0

Answer:

Step-by-step explanation:

\frac{5}{15}  + \frac{7}{15y}  = \frac{12}{15}  = \frac{4}{5}

Inessa05 [86]2 years ago
6 0
1/3+ + 7/15 = 0.8 as a decimal and 4/5 as a fraction
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What is the vertex of a parabola having the equation y = –2(x – 5)2 + 7? Which direction does the parabola open?
Vika [28.1K]

y = -2(x-5)^2 + 7

That's vertex form for a parabola y=a(x-p)^2+q and we read off vertex (p,q) as

Answer: Vertex (5,7)

The negative <em>a</em> tells us this is a downward opening parabola (upside down from the usual y=x^2. I remember CUP - Concave Up Positive; here <em>a </em>is negative so not a cup, instead concave <em>down.</em> The same rule applies to second derivatives in calculus so memorize it now and use it later.

Answer: downward

7 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
Answer all questions shown
FromTheMoon [43]
SOH CAH TOA

—>

Sin D = 13 sqrt3/26 = 0.76
Sin E = 13/26 = 0.47
Cos D = 13/26 = 0.87
Cos E = 13 sqrt3/26 = 0.64
7 0
2 years ago
About how much farther is it to drive than to walk directly from building A to building B? Round to the nearest whole number.
Sav [38]

Answer:

A. 183 meters

Step-by-step explanation:

Building A and building B are 500 meters apart. There is no road between them, so to drive from building A to building B, it is necessary to first drive to building C and then to building B. About how much farther is it to drive than to walk directly from building A to building B? Round to the nearest whole number. A) 183 meters B) 250 meters C) 366 meters D) 683 meters

Find distance BC

Cos (60°)=BC / AB (Adjacent divided by the hypotenuse)

Cos (60°)=1/2

BC=a

AB=500

Cos (60°)=BC / AB

1/2=a/500

1/2 * 500=a

250=a

a=250m

Find distance AC

Sin(60°)=AC/AB (opposite side divided by hypotenuse)

Sin(60°)=√3/2

AC=b

AB=500

Sin(60°)=AC/AB

√3/2=b/500

√3/2 * 500=b

250√3=b

b=433m

Distance AC and BC=AC+BC

433m+250m=683m

Subtract the distance AB from AC+BC

= 683m - 500m

=183m

Answer is A. 183 meters

8 0
3 years ago
Peaches come in large and small cylindrical cans. The larger can has a radiusand height that are both four times longer than the
Alik [6]

Answer:

D. 2058.24 in³

Step-by-step explanation:

Let r,h be the radius and height of smaller can and R,H be the radius and height of larger can

Then, R=4r  and H=4h

Volume of smaller can=32.16 in³

        ⇒                  πr²h=32.16 in³

Volume of larger can= πR²H

                                  = π(4r²)×4h

                                 = 64πr²h

                                 = 64×32.16 in³

                                =  2058.24 in³

Hence, correct option is:

D. 2058.24 in³

7 0
2 years ago
Read 2 more answers
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