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jeka94
3 years ago
10

Find the slope of the line graphed below

Mathematics
2 answers:
Illusion [34]3 years ago
8 0
The slope of the line is -2/3
taurus [48]3 years ago
3 0
Hello,

A=(3,-4)
B=(-2,-1)

m=(-4+1)/(3+2)=-3/5
==============
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If a+b+c=0 then find the value of a^2+b^2+c^2/a^2-bc pls help me
VladimirAG [237]

a+b+c=0

[(a+b+c)^2=a^2+b^2+c^2+2ab+2ac+2bc]

[a^2+b^2+c^2+2ab+2ac+2bc=0]

[a^2+b^2+c^2=-(2ab+2ac+2bc)]

[a^2+b^2+c^2=-2(ab+ac+bc)] (i)

also

[a=-b-c]

[a^2=-ab-ac] (ii)

[-c=a+b]

[-bc=ab+b^2] (iii)

adding (ii) and (iii) ,we have

[a^2-bc=b^2-ac] (iv)

devide (i) by (iv)

[(a^2+b^2+c^2)/(a^2-bc)=(-2(ab+bc+ca))/(b^2-ac)]
8 0
3 years ago
A worker is paid $0.05 on the first day, $0.10 on the second day, $0.20 on the third day, $0.40 on the fourth day, and so on. Ho
Novay_Z [31]
This is a geometric sequence with a common ratio of 2

an = a1 * r^(n - 1)
n = term to find = 21
a1 = first term = 0.05
r = common ratio = 2
now we sub
a21 = 0.05 * 2^(21 - 1)
a21 = 0.05 * 2^20
a21 = 0.05 * 1048576
a21 = 52428.80 <==== after working 21 days
7 0
3 years ago
Mixed fractions: What is the value of 2 1/3 + 6 3/5 *
Studentka2010 [4]

Answer:

134/15 or

8 14/15

Step-by-step explanation:

3 0
3 years ago
Suppose a student started with 139.0 mg of trans-cinnamic acid, 463 mg of pyridinium tribromide, and 2.45 mL of glacial acetic a
Sliva [168]

Answer:

0.2889 g  brominated product

64.6 %

Step-by-step explanation:

This is a bromination chemical reaction of an alkene  and we are asked to calculate the theoretical and percent yield. Thus to solve it we have to perform a calculation based on the balanced chemical reaction.

The pyridinium tribromide is used as a generator of molecular bromine in situ, and bromine will add to the double bond in trans-cinnamic acid so we know the reaction occur in a one to one mole fashion.

trans-cinnamic acid + pyridinium  tribromide ⇒ 2,3-dibromo-3-                          

                                                                             phenylpropanoic acid

Molar weight  trans-cinnamic acid  =  148.16 g/mol

mass trans-cinnamic acid  = 139.0 mg x  1g/1000 mg = 0.139 g

# mol trans-cinnamic acid =  0.139 g / 148 g/mol = 9.38 x 10⁻⁴ mol

Since our reaction is 1 mol trans-cinnamic acid  produces 1 mol 2,3-dibromo-3-phenylpropanoic acid, it follows that the theoretical yield is:

1 mol 2,3-dibromo-3-phenylpropanoic acid /  trans-cinnamic acid x 9.38 x 10⁻⁴ mol trans-cinnamic acid  

=  9.38 x 10⁻⁴ mol  2,3-dibromo-3-phenylpropanoic acid

In grams the the theoretical yield is:

molar mass  2,3-dibromo-3-phenylpropanoic acid = 307.97 g/mol

The theoretical mass  2,3-dibromo-3-phenylpropanoic acid:

=   9.38 x 10⁻⁴ mol  2,3-dibromo-3-phenylpropanoic acid x 307.97 g/mol

=   0.2889 g

% yield = mass experimental/mass theoretical

= 0.1866 g /  0.2889 g x 100 = 64.6 %

4 0
2 years ago
Need help on these thank you if you help
navik [9.2K]
First box: 7 to the power of 2
second box: 7
third box: 7 to the power of four

first box: m to the power of 2
second box: m to the power of 5
5 0
2 years ago
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