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babymother [125]
3 years ago
9

If a = 3 and b = 2, what is the value of the expression 11 + a - b ?

Mathematics
1 answer:
Vadim26 [7]3 years ago
4 0
Substitute the values of a and b into the expression:
11 + 3 - 2
Add first because of order of operations (perform addition and subtraction in the order of occurrence from left to right):
14 - 2
Subtract (same justification as above just with subtraction instead):
12
Hope this helps!
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Five more then six times a number is 17.what is the number
postnew [5]

Answer:

2

Step-by-step explanation:

You set up and equation:

5+6x = 17

You solve it and get 2.

7 0
3 years ago
Read 2 more answers
Help please! Calculate the exact value of cos (a-b) given that sin a= 12/13 with pi/2
zloy xaker [14]

Answer:

56/65

Step-by-step explanation:

First, we know that cos(a-b) = cos(a)cos(b) + sin(a)sin(b)

We know what sin(a) and sin(b) are, and to get cos(a), we can take the equation sin²a + cos²a = 1

Thus,

(12/13)² + cos²a = 1

1 - (12/13)² = cos²a

1- 144/169 = cos²a

cos²a = 25/169

cos(a) = 5/13

Similarly,

(3/5)² + cos²b = 1

1 - (3/5)² = cos²b

1 - 9/25 = cos²b

cos²b = 16/25

cos(b) = 4/5

Our answer is

cos(a-b) = cos(a)cos(b) + sin(a)sin(b)

cos(a-b)  = (5/13)(4/5) + (12/13)(3/5)

cos(a-b)  = 20/65 + 36/65

cos(a-b)  = 56/65

4 0
2 years ago
F(x)=5x+6;g(x)=7x-2 qfjnd (f+g)(x)
Likurg_2 [28]

Answer:

12x+4

Step-by-step explanation:

Use the given functions to set up and simplify  ( f + g ) ( x )

7 0
2 years ago
Oman said that it is impossible to raise a number to the power of 2 and get a value less than the original number.Do you agree w
Yuri [45]

Answer:

Yes.

Step-by-step explanation:

Any number raised to the power of 2 will be greater than the original number.

A positive number:

4² = 16

A negative number:

-3² = 9

Or:

0² = 0

6 0
3 years ago
Find a particular solution to the differential equation y′′+2y′+y=2t2−t+3e−2t. y′′+2y′+y=2t2−t+3e−2t.
Jlenok [28]
y''+2y'+y=2t^2-t+3e^{-2t}

The characteristic equation is

r^2+2r+1=(r+1)^2=0\implies r=-1

so the characteristic solution is

y_c=C_1e^{-t}+C_2te^{-t}

For the particular solution, we can try looking for a solution of the form

y_p=at^2+bt+c+de^{-2t}
\implies{y_p}'=2at+b-2de^{-2t}
\implies{y_p}''=2a+4de^{-2t}

Substituting into the ODE, we have

(2a+4de^{-2t})+2(2at+b-2de^{-2t})+(at^2+bt+c+de^{-2t})=2t^2-t+3e^{-2t}
at^2+(4a+b)t+(2a+2b+c)+de^{-2t}=2t^2-t+3e^{-2t}
\implies\begin{cases}a=2\\4a+b=-1\\2a+2b+c=0\\d=3\end{cases}\implies a=2,b=-9,c=14,d=3

So the general solution to the ODE is

y=y_c+y_p
y=C_1e^{-t}+C_2te^{-t}+2t^2-9t+14+3e^{-2t}
4 0
3 years ago
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