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vova2212 [387]
3 years ago
10

Which are perfect square trinomials? Select two options. x2 − 9 x2 −100 x2 − 4x + 4 x2 + 10x + 25 x2 + 15x + 36

Mathematics
2 answers:
s344n2d4d5 [400]3 years ago
5 0

Answer:

C and D

Step-by-step explanation:

i just took the quiz

vivado [14]3 years ago
5 0

Answer:

C and D

Step-by-step explanation:

I took the quiz and got 100

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Which of the following expressions is equivalent to -5.4 - (2.1 + 1.8)?
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Answer:

its the first one

Step-by-step explanation:

7 0
3 years ago
If 2x squared + 8y = 121. 5, what is x
mezya [45]
X=30.4-2y
(step by step) :
5 0
3 years ago
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
Read 2 more answers
95 divided by 4 and 34 divided by 2
MaRussiya [10]
95/4=\frac{95}{4}=23\frac{3}{4} or 23r3

and

34/2=\frac{34}{2}=17
5 0
3 years ago
Read 2 more answers
Match each circumference to the corresponding radius or diameter of the circle. (All circumferences are approximate.) Tiles circ
Pavlova-9 [17]
We solve for the diameters of radius of the given circles by using the equation,
                                      C = πD
                    or               C = 2πr
where D and r are diameter and radius, respectively.

Solving,
      C = 34.54 units
                                D = C/π = 34.54 units / 3.14 = 11 units
                                r = C / 2π = 34.54 units/2(3.14) = 5.5 units

      C = 59.66 units
                                D = C/π  = 59.66 units/3.14 = 19 units
                                 r = D2 = 9.5 units
      C = 23.236 units
                                D = 23.236 units/3.14 = 7.4 units
                                 r = D/2 = 3.7 units
       C = 23.376 units
                                 D = 23.376 units / 3.14 = 7.4 units
                                    r = D/2 = 3.7 units
        C = 13.188 units
                                  D = 13.188 units/3.14 = 4.2 units
                                    r = D/2 = 2.1 units


8 0
4 years ago
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