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Natasha2012 [34]
3 years ago
15

HELP PLEASE! I need the answer please! ​

Mathematics
1 answer:
cestrela7 [59]3 years ago
7 0

Answer:

The triangles are similar in that the smaller triangle was dilated by a scale factor of 1/4 which means that the length of the missing side is 20m

Step-by-step explanation:

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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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How do find the answer in subtractiion problem, when the top number is smaller than the bottom
Travka [436]
Use a number line if you need 4/10 - 6/10 = -2/10a simpler way is to do the opposite 6/10 - 4/10 = 2/10  then the answer would be the opposite as well. so, -2/10
8 0
4 years ago
Read 2 more answers
What is the area enclosed by the curves???
slava [35]
The x and y axis
i think
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3 years ago
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A recipe states that 5 servings of peanuts have 1,250 calories. Write an equation to represent the relationship between the numb
Nataly_w [17]

dont care idiot hahahahahahahahahah

4 0
3 years ago
Paul has three cube-shaped boxes. Each box is a different size
qwelly [4]

Answer:

See explanation

Step-by-step explanation:

Paul has three cube-shaped boxes. Each box is a different size and they are stacked from the largest to the smallest. Some information about the boxes is given below.

  • The combined volume of the three boxes is 1,197 cubic inches.
  • The area of one face of the medium box is 49 square inches.
  • The volume of the smallest box is 218 cubic inches less than the volume of the medium box.

1. The medium box has the area of one face of 49 square inches, then

a^2=49\\ \\a=7\ inches

is the side length.

The volume of the medium box is

a^3=7^3=343\ in^3.

2. The volume of the smallest box is 218 cubic inches less than the volume of the medium box, then the volume of the smallest box is

343-218=125\ in^3.

Ib is the side length, then

b^3=125\\ \\b=5\ inches

The area of one face is

b^2=5^2=25\ in^2.

3. The volume of the largest box is

1,197-343-125=729\ in^3,

then if c is the side length,

c^3=729\\ \\c=9\ inches.

4. The total height of the stack is the sum of all sides lengths:

a+b+c=7+5+9=21\ inches

5. Find the surface area of each box:

  • small 6b^2=6\cdot 5^2=6\cdot 25=150\ in^2;
  • medium 6a^2=6\cdot 7^2=6\cdot 49=294\ in^2;
  • large 6c^2=6\cdot 9^2=6\cdot 81=486\ in^2.

In total, Paul needs

150+294+486=930\ in^2

of wrapping paper. He has 1,000 square inches, so Paul has enough paper to wrap all 3 boxes.

8 0
4 years ago
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