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lidiya [134]
3 years ago
9

For a given function ƒ(x), the operation ƒ(3x) will

Mathematics
2 answers:
kherson [118]3 years ago
8 0

Answer:

ill do it if you answer this

In a math class with 26 students, a test was given the same day that an assignment was due. There were 17 students who passed the test and 18 students who completed the assignment. There were 15 students who passed the test and also completed the assignment. What is the probability that a student who failed the test completed the homework?

Step-by-step explanation:

const2013 [10]3 years ago
3 0
The answer is C I think I’m completely guessing here
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For the inequality, -4x + 5 >-5
vodka [1.7K]

Answer:

x<2.5

Step-by-step explanation:

-4x +5>-5

-4x>-5-5

-4x>-10

x<10/4(on multiplying (-)ve sign both the sides inequality changes)

x < 2.5

8 0
3 years ago
Could anyone PLEASE explain how to do 56?
egoroff_w [7]
Can i get brainliest if my answer is true hehehehe :p

8 0
3 years ago
Evaluate using the values given :<br> 3 pm; use m = 4, and p = 3
Goryan [66]

Answer:

36

General Formulas and Concepts:

<u>Pre-Algebra</u>

Order of Operations: BPEMDAS

  1. Brackets
  2. Parenthesis
  3. Exponents
  4. Multiplication
  5. Division
  6. Addition
  7. Subtraction
  • Left to Right

Step-by-step explanation:

<u>Step 1: Define</u>

3pm

m = 4

p = 3

<u>Step 2: Evaluate</u>

  1. Substitute:                   3(3)(4)
  2. Multiply:                       9(4)
  3. Multiply:                       36
5 0
3 years ago
In evaluating a double integral over a region D, a sum of iterated integrals was obtained as follows:
BabaBlast [244]

Answer

a=0, b=2

g_1(x)=\frac{5x}{2},  g_2(x)=7-x

Step-by-step explanation:

Given that

\int \int   Df(x,y)dA=\int_0 ^5\int _0 ^ {\frac {2y}{5}} f(x,y)dxdy+\int_5^7\int_0^{7-y} f(x,y)dxdy\; \cdots (i)

For the term  \int_0 ^5\int _0 ^ {\frac {2y}{5}} f(x,y)dxdy.

Limits for x is from x=0 to x=\frac {2y}{5} and for y is from y=0 to y=5  and the region D, for this double integration is the shaded region as shown in graph 1.

Now, reverse the order of integration, first integrate with respect to y then with respect to x . So, the limits of y become from y=\frac{5x}{2} to y=5 and limits of x become from x=0 to x=2 as shown in graph 2.

So, on reversing the order of integration, this double integration can be written as

\int_0 ^5\int _0 ^ {\frac {2y}{5}} f(x,y)dxdy=\int_0 ^2\int _ {\frac {5x}{2}}^5 f(x,y)dydx\; \cdots (ii)

Similarly, for the other term  \int_5 ^7\int _0 ^ {7-y} f(x,y)dxdy.

Limits for x is from x=0 to x=7-y and limits for y is from y=5 to y=7  and the region D, for this double integration is the shaded region as shown in graph 3.

Now, reverse the order of integration, first integrate with respect to y then with respect to x . So, the limits of y become from y=5 to y=7-x and limits of x become from x=0 to x=2 as shown in graph 4.

So, on reversing the order of integration, this double integration can be written as

\int_5 ^7\int _0 ^ {7-y} f(x,y)dxdy=\int_0 ^2\int _5 ^ {7-x} f(x,y)dydx\;\cdots (iii)

Hence, from equations (i), (ii) and (iii) , on reversing the order of integration, the required expression is

\int \int   Df(x,y)dA=\int_0 ^2\int _ {\frac {5x}{2}}^5 f(x,y)dydx+\int_0 ^2\int _5 ^ {7-x} f(x,y)dydx

\Rightarrow \int \int   Df(x,y)dA=\int_0 ^2\left(\int _ {\frac {5x}{2}}^5 f(x,y)+\int _5 ^ {7-x} f(x,y)\right)dydx

\Rightarrow \int \int   Df(x,y)dA=\int_0 ^2\int _ {\frac {5x}{2}}^{7-x} f(x,y)dydx\; \cdots (iv)

Now, compare the RHS of the equation (iv) with

\int_a^b\int_{g_1(x)}^{g_2(x)} f(x,y)dydx

We have,

a=0, b=2, g_1(x)=\frac{5x}{2} and g_2(x)=7-x.

3 0
3 years ago
Is 1.68 greater than 1.680
Delicious77 [7]
No they are both equal it would be like 5 being greater than 05
3 0
3 years ago
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