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Cloud [144]
3 years ago
5

Which of the following integrals cannot be evaluated using simple substitution?

Mathematics
1 answer:
adell [148]3 years ago
8 0

the second and the last,

since quadratic inside radical without multiplication with linear function

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If 3(r+300)=6, then what would be the value of r+300-2?
saveliy_v [14]

Solve for r in the first equation:

3(r+300) = 6

Use the distributive property:

3r + 900 = 6

Subtract 900 from both sides:

3r = -894

Divide both sides by 3:

r = -894 / 3

r = -298

Now you have r, replace r in the second equation and solve:

r +300 -2 =

-298 + 300 - 2 = 0

The answer is 0.

6 0
3 years ago
Area of rectangle 12ft 6ft
Brilliant_brown [7]
12ft×6ft= area of 72ft
8 0
4 years ago
Find the system of inequalities which represents the shaded region on the coordinate plane.
dybincka [34]

Answer:

given, y≥2x+1 and

y>

2

1

x−1

first, draw the graph for equations y=2x+1 and y=

2

1

x−1

for y=2x+1

substitute y=0 we get, 2x+1=0⟹x=−0.5

substitute x=0 we get, y=1

therefore, y=2x+1 line passes through (0.5,0) and (0,1) as shown in fig.

Hence, y≥2x+1 includes the region above the line.

for y=

2

1

x−1

substitute y=0 we get,

2

1

x−1=0⟹x=2

substitute x=0 we get, y=−1

therefore, y=2x+1 line passes through (2,0) and (0,-1) as shown in fig.

Hence, y>

2

1

x−1 includes the region above the line.

the intersection region is the shaded region as shown in above figure which includes I, II and III quadrants.

Therefore, quadrant IV has no solution please make as brainlelist

7 0
3 years ago
Add the following complex numbers:
Fittoniya [83]

Answer:

option c

Step-by-step explanation:

(4 - 2i) + (12 + 7i)

multiply the brackets first

4 - 2i + 12 + 7i

combine like terms

16 + 5i

3 0
3 years ago
Read 2 more answers
What is the answer for this problem
neonofarm [45]
Common denominator- 70
24/35 =  48/70
7/10=  49/70
7/10 is larger.
5 0
3 years ago
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