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liq [111]
3 years ago
12

Complete 4,5,6 for 25 points.

Mathematics
1 answer:
NemiM [27]3 years ago
7 0

Answer:

Step-by-step explanation:

4. A = P + Prt

   a. Solve for r

    Prt = P - A

    r = (P - A)/Pt

    b. Solve for t

         Prt = P - A

           t = (P - A)/Pr

5. Ax + By = C

   a. By = C - Ax

        y = (C - Ax)/B

   b. Ax = C - By

       x = (C - By)/A

6. y = mx + b

   a. mx + b = y

       mx = y - b

       x = (y - b)/m

    b. mx + b = y

        mx = y - b

        m= (y - b)/x

   c. mx + b = y

       b = y - mx

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Given the function f(x) = x^4 + 3x^3 - 2x^2 - 6x - 1, use intermediate theorem to decide which of the following intervals contai
marta [7]

f(x) = x^4 + 3x^3 - 2x^2 - 6x - 1

Lets check with every option

(a) [-4,-3]

We plug in -4  for x  and -3 for x

f(-4) = (-4)^4 + 3(-4)^3 - 2(-4)^2 - 6(-4) - 1= 55

f(-3) = (-3)^4 + 3(-3)^3 - 2(-3)^2 - 6(-3) - 1= -1

f(-4) is positive and f(-3) is negative. there is some value at x=c on the interval [-4,-3] where f(c)=0. so there exists atleast one zero on this interval.

(b) [-3,-2]

We plug in -3  for x  and -2 for x

f(-3) = (-3)^4 + 3(-3)^3 - 2(-3)^2 - 6(-3) - 1= -1

f(-2) = (-2)^4 + 3(-2)^3 - 2(-2)^2 - 6(-2) - 1= -5

f(-2) is negative and f(-3) is negative. there is no value at x=c on the interval [-3,-2] where f(c)=0.  

(c) [-2,-1]

We plug in -2  for x  and -1 for x

f(-2) = (-2)^4 + 3(-2)^3 - 2(-2)^2 - 6(-2) - 1= -5

f(-1) = (-1)^4 + 3(-1)^3 - 2(-1)^2 - 6(-1) - 1= 1

f(-2) is negative and f(-1) is positive. there is some value at x=c on the interval [-2,-1] where f(c)=0. so there exists atleast one zero on this interval.

(d) [-1,0]

We plug in -1  for x  and 0 for x

f(-1) = (-1)^4 + 3(-1)^3 - 2(-1)^2 - 6(-1) - 1= 1

f(0) = (0)^4 + 3(0)^3 - 2(0)^2 - 6(0) - 1= -1

f(-1) is positive and f(0) is negative. there is some value at x=c on the interval [-1,0] where f(c)=0. so there exists atleast one zero on this interval.

(e) [0,1]

We plug in 0  for x  and 1 for x

f(0) = (0)^4 + 3(0)^3 - 2(0)^2 - 6(0) - 1= -1

f(1) = (1)^4 + 3(1)^3 - 2(1)^2 - 6(1) - 1= -5

f(0) is negative and f(1) is negative. there is no value at x=c on the interval [0,1] where f(c)=0.  

(f) [1,2]

We plug in 1  for x  and 2 for x

f(1) = (1)^4 + 3(1)^3 - 2(1)^2 - 6(1) - 1= -5

f(2) = (2)^4 + 3(2)^3 - 2(2)^2 - 6(2) - 1= 19

f(-4) is positive and f(-3) is negative. there is some value at x=c on the interval [-4,-3] where f(c)=0. so there exists atleast one zero on this interval.

so answers are (a) [-4,-3], (c) [-2,-1],  (d) [-1,0], (f) [1,2]

8 0
3 years ago
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Verdich [7]

Answer:

A. 35

Step-by-step explanation:

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[100 ÷ (27 – 22)] – 3 • 5 + 25
oksian1 [2.3K]
Answer is 30
you can use PEMDAS to help you on further questions like this
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What is the value of x in this proportion? 5/6=−4x/+2 x=−2 4/5 x=−4 2/5 x=−5 1/5 x=−6 4/5
insens350 [35]

Answer:

x= -4 2/5

Step-by-step explanation:

firstly cross multiply 5/6=-4x/+2

5×+2=6×-4x

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make x subject of formula

10/-24= ans

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Will mark brainless for first response!!!
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