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LenaWriter [7]
2 years ago
5

The sum of the first and third of three consecutive even integers is 154. Find the three even integers.

Mathematics
1 answer:
topjm [15]2 years ago
5 0
X + (x+2) = 154
combine like terms
2x + 2 = 154
subtract 2
2x = 154
divide by 2
x = 76

so the two numbers are 76 and 78
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Evaluate f(t)=-2t2+ 1 for f(-3).
podryga [215]

Answer:

7

Step-by-step explanation:

First, plug-in f(-3) into t in the function:

f(-3)=-2(-3) +1

Second, multiply -2 and -3 to get 6:

f(-3)=-2(-3)+1⇒f(-3)=6+1

Third, add 6 and 1 to get your answer:

f(-3)= 6+1⇒f(-3)=7

3 0
3 years ago
3. A large fruit dish with a 36 in. diameter is placed in the center of a banquet across and is completely filled with apples. T
Serga [27]

Answer:

a) 11%

b) 56%

Step-by-step explanation:

a) A= πr^2    6^2π= 36π    A= 18^2π = 324π

36π/324π = .111111111111111111111111 or 11%

remember to cancel out the pi symbols when you are showing your work

(put a line across the pi symbols when dividing)

b)   A= πr^2         18^2 π = 324π

     A= πr^2          12^2π = 144π  

              324 – 144  = 180

180π/324π = .5555555555555555555555556 or 56%

remember to cancel out the pi symbols when you are showing your work

(put a line across the pi symbols when dividing)

3 0
3 years ago
Let f(x)=5x3−60x+5 input the interval(s) on which f is increasing. (-inf,-2)u(2,inf) input the interval(s) on which f is decreas
o-na [289]
Answers:

(a) f is increasing at (-\infty,-2) \cup (2,\infty).

(b) f is decreasing at (-2,2).

(c) f is concave up at (2, \infty)

(d) f is concave down at (-\infty, 2)

Explanations:

(a) f is increasing when the derivative is positive. So, we find values of x such that the derivative is positive. Note that

f'(x) = 15x^2 - 60


So,


f'(x) \ \textgreater \  0
\\
\\ \Leftrightarrow 15x^2 - 60 \ \textgreater \  0
\\
\\ \Leftrightarrow 15(x - 2)(x + 2) \ \textgreater \  0
\\
\\ \Leftrightarrow \boxed{(x - 2)(x + 2) \ \textgreater \  0} \text{   (1)}

The zeroes of (x - 2)(x + 2) are 2 and -2. So we can obtain sign of (x - 2)(x + 2) by considering the following possible values of x:

-->> x < -2
-->> -2 < x < 2
--->> x > 2

If x < -2, then (x - 2) and (x + 2) are both negative. Thus, (x - 2)(x + 2) > 0.

If -2 < x < 2, then x + 2 is positive but x - 2 is negative. So, (x - 2)(x + 2) < 0.
 If x > 2, then (x - 2) and (x + 2) are both positive. Thus, (x - 2)(x + 2) > 0.

So, (x - 2)(x + 2) is positive when x < -2 or x > 2. Since

f'(x) \ \textgreater \  0 \Leftrightarrow (x - 2)(x + 2)  \ \textgreater \  0

Thus, f'(x) > 0 only when x < -2 or x > 2. Hence f is increasing at (-\infty,-2) \cup (2,\infty).

(b) f is decreasing only when the derivative of f is negative. Since

f'(x) = 15x^2 - 60

Using the similar computation in (a), 

f'(x) \ \textless \  \ 0 \\ \\ \Leftrightarrow 15x^2 - 60 \ \textless \  0 \\ \\ \Leftrightarrow 15(x - 2)(x + 2) \ \ \textless \  0 \\ \\ \Leftrightarrow \boxed{(x - 2)(x + 2) \ \textless \  0} \text{ (2)}

Based on the computation in (a), (x - 2)(x + 2) < 0 only when -2 < x < 2.

Thus, f'(x) < 0 if and only if -2 < x < 2. Hence f is decreasing at (-2, 2)

(c) f is concave up if and only if the second derivative of f is positive. Note that

f''(x) = 30x - 60

Since,

f''(x) \ \textgreater \  0&#10;\\&#10;\\ \Leftrightarrow 30x - 60 \ \textgreater \  0&#10;\\&#10;\\ \Leftrightarrow 30(x - 2) \ \textgreater \  0&#10;\\&#10;\\ \Leftrightarrow x - 2 \ \textgreater \  0&#10;\\&#10;\\ \Leftrightarrow \boxed{x \ \textgreater \  2}

Therefore, f is concave up at (2, \infty).

(d) Note that f is concave down if and only if the second derivative of f is negative. Since,

f''(x) = 30x - 60

Using the similar computation in (c), 

f''(x) \ \textless \  0 &#10;\\ \\ \Leftrightarrow 30x - 60 \ \textless \  0 &#10;\\ \\ \Leftrightarrow 30(x - 2) \ \textless \  0 &#10;\\ \\ \Leftrightarrow x - 2 \ \textless \  0 &#10;\\ \\ \Leftrightarrow \boxed{x \ \textless \  2}

Therefore, f is concave down at (-\infty, 2).
3 0
3 years ago
Solve the following please
Dima020 [189]

Answer:

x=4

Step-by-step explanation:

5 0
3 years ago
Read 2 more answers
2x + 3y = 26 3x + 5y = 40<br>find x and y I will give brainliest
Yuki888 [10]

Answer:

x = 10 and y = 2

Step-by-step explanation:

multiply first equation by 5 and 2nd equation by 3 then subtract equation 1 and 2, we get,

10x+15y= 130

9x+15y=120

( -) (-) (-)

<u> </u><u> </u><u> </u><u> </u><u> </u><u> </u><u> </u><u> </u><u> </u><u> </u><u> </u><u> </u><u> </u><u> </u><u> </u><u> </u><u> </u><u> </u><u> </u><u> </u><u> </u><u> </u><u> </u><u> </u><u> </u><u> </u>

x= 10

substituting the value of x in equation 1 we get

2x + 3y = 26

2*10 + 3y = 26

3y = 26-20

y =6/3

y= 2.i hope this helps

7 0
3 years ago
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