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Inga [223]
3 years ago
15

Solve the equation 5.5(x-3)=17+10.5 in systems of equations. How can you find your solution?

Mathematics
1 answer:
Readme [11.4K]3 years ago
3 0

Answer:

x= 8

Step-by-step explanation:

5.5(x-3)=17+10.5

5.5x - 16.5= 27.5

5.5x= 44

x= 8

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Which is a simplified form of the expression -4(3r – 2) – 2(r + 1)?
Delicious77 [7]

Answer:

2 (3 - 7 r)

Step-by-step explanation:

Simplify the following:

-4 (3 r - 2) - 2 (r + 1)

-4 (3 r - 2) = 8 - 12 r:

8 - 12 r - 2 (r + 1)

-2 (r + 1) = -2 r - 2:

-12 r + -2 r - 2 + 8

Grouping like terms, -2 r - 12 r - 2 + 8 = (8 - 2) + (-12 r - 2 r):

(8 - 2) + (-12 r - 2 r)

-12 r - 2 r = -14 r:

-14 r + (8 - 2)

8 - 2 = 6:

6 - 14 r

Factor 2 out of 6 - 14 r:

Answer: 2 (3 - 7 r)

6 0
3 years ago
At an animal shelter, the ratio of cats to dogs is 5:6. If there are 54 cats, how many dogs are there at the shelter?
Neko [114]

There would be 45 dogs!

8 0
3 years ago
Geometry please help. What is the perimeter of XYZ
Morgarella [4.7K]

Answer:

39

Step-by-step explanation:

Angle AQC, AQB and BQC are 120deg. (360/3=120). Therefore, Angle BZC, CXA and AYB are 60deg. Because, angle BZC is half of angle BQC. same for the other three angles. As, all 3 angles are 60deg, it is an equilateral triangle. So, YZ is 13 therefore as all 3 sides are equal its 13*3=39.

6 0
3 years ago
On a coordinate plane, triangle A B C and parallelogram G H J K are shown. Triangle A B C has points (2, 0), (1, negative 6), (n
valentina_108 [34]

Answer:

The area of △ABC is 2 square units greater than the area of parallelogram GHJK.

Step-by-step explanation:

4 0
3 years ago
Calculus 2: I need help with u substitution and intergration by parts. Can someone help me understand the concept?​
snow_lady [41]

Computing an integral by substitution is the reverse of the chain rule for computing the derivative. Substitution is intended to rewrite a complicated-looking integral involving the derivative of some component expression as another much simpler integral. For example, if u=f(x), then \mathrm du=f'(x)\,\mathrm dx and

\displaystyle\int f(x)f'(x)\,\mathrm dx=\int u\,\mathrm du

###

Integration by parts is the reverse of the product rule for derivatives:

(f(x)g(x))'=f'(x)g(x)+f(x)g'(x)\implies f(x)g'(x)=(f(x)g(x))'-f'(x)g(x)

Integrating both sides with respect to x gives

\displaystyle\int f(x)g'(x)\,\mathrm dx=\int(f(x)g(x))'\,\mathrm dx-\int f'(x)g(x)\,\mathrm dx

\displaystyle\int f(x)g'(x)\,\mathrm dx=f(x)g(x)-\int f'(x)g(x)\,\mathrm dx

###

Personally, I think the best way to grasp the idea behind the two methods is to practice. You start to notice patterns to the point where knowing which is the "right" method to use becomes second nature.

4 0
3 years ago
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