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Alla [95]
2 years ago
9

Evaluate.

Mathematics
2 answers:
mafiozo [28]2 years ago
8 0

Answer:

-539.25

Step-by-step explanation:

(w^2x−3)÷10⋅z

w=−9, x = 2.7, and z=−25

((-9)^2*2.7−3)÷10⋅(-25)

Parentheses first

The exponent in the parentheses

(81*2.7−3)÷10⋅(-25)

Then multiply

(218.7−3)÷10⋅(-25)

Then subtract

(215.7)÷10⋅(-25)

Now multiply and divide from left to right

21.57*(-25)

-539.25

RoseWind [281]2 years ago
4 0

Same Question here answered by me correctly with explanation.

Visit the link below for your answer

brainly.com/question/24948989?

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five times the difference of twice a number and three, decreased by the sum of the number and eight, equals thirteen
trapecia [35]
5x(2y-3)-(y+8)=13 I believe this is correct
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Jay bought 6 bags of maize at shs 2700 each.He sold the bags at shs.3300 each. what was his profit?​
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Answer:

3300-2700= 600 x 6 = 3600

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A 100-meter-long christmas train needs 30 seconds to cross a 400-meter-long bridge. assuming the train goes at a steady speed, h
san4es73 [151]
Length of train : 100
length of bridge : 400
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3 years ago
Calculate the area of a rectangle with an altitude of 30cm and a diagnol of 50cm
romanna [79]

Answer:

2000cm²

Step-by-step explanation:

find the other side using pythagorean theorem

50²=30²+x²

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7 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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