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guajiro [1.7K]
2 years ago
9

On a coordinate plane, a straight line with a negative slope, labeled f of x, crosses the y-axis at (0, 4), and the x-axis at (4

, 0). Which is true regarding the graphed function f(x)?
Mathematics
1 answer:
Katyanochek1 [597]2 years ago
3 0

Answer:

A) f(-3) = g(-4)

Step-by-step explanation: hope it helps

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1+7 and 7+1 are the same equations. The numbers are just switched around .
Example:
1+2=3
2+1+3

<span>They add up to the same answer no matter where they are placed, therefore knowing 1+7 helps you find the sum of 7+1 (again, because they are the same)  </span>
8 0
2 years ago
Sandy evaluated the expression below. (negative 2) cubed (6 minus 3) minus 5 (2 + 3) = (negative 2) cubed (3) minus 5 (5) = 8 (3
nasty-shy [4]

Answer:

should be - 8

Step-by-step explanation:

-2*-2=4 4*-2=-8

4 0
2 years ago
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Two production lines are used to pack sugar into 5 kg bags. Line 1 produces twice as many bags as does line 2. One percent of ba
enyata [817]

Answer:

P(Bag is Defective) = 0.0167

Step-by-step explanation:

Line 1 produces twice as many bags as line 2. Let x be the number of bags produced by line 2.

No. of bags produced by line 2 = x

No. of bags produced by line 1 = 2x

Probability that the bag has been produced by line 1 can be written as:

P(Line 1) = No. of bags produced by line 1/Total no. of bags

             = 2x/(x+2x)

             = 2x/3x

P(Line 1) = 2/3. Similarly,

P(Line 2) = x/3x

P(Line 2) = 1/3

1% bags produced by line 1 are defective so the probability of line 1 producing a defective bag is:

P(Defective|Line 1) = 0.01

3% of bags from line 2 are defective, so:

P(Defective|Line 2) = 0.03

b. The probability that the chosen bag is defective can be calculated through the conditional probability formula:

P(A|B) = P(A∩B)/P(B)

<u>P(A∩B) = P(A|B)*P(B)</u>

The chosen defective bag can be either from line 1 or from line 2. So, the probability that the chosen bag is defective is:

P(Bag is Defective) = P(Defective and from Line 1) + P(Defective and from Line 2)

                                = P(D∩Line 1) + P(D∩Line 2)

                                = P(Defective|Line 1)*P(Line 1) + P(Defective|Line 2)*P(Line 2)

                                = (0.01)*(2/3) + (0.03)(1/3)

P(Bag is Defective) = 0.0167

7 0
2 years ago
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Hey ^-^ can someone please help me with this problem:
soldi70 [24.7K]

Answer:   8 (Pi - sqrt(3))

Discussion:

The area of the shaded region is that of the semicircle minus the area of the triangle..

Area of semicircle = 1/2 * Pi * R^2        

   Where R^2 is the square of the radius of the circle. In our case, R ( = OC)

    = 4 so the semicircle area is

   (1/2) * Pi * (4^2) = (1/2) * Pi * 16 = 8 Pi

Area of triangle.

  First of all, angle ACB is a right angle ( i.e. 90 degrees).

    * This is the Theorem of Thales from elementary Plane Geometry. *

 so by Pythagoras

   AC^2 + BC^2 = AB^2

But CB = 4 (given) and AB = 4*2 = 8 ( the diameter is twice the radius).

Substituting these in Pythagoras gives

   AC^2 + 4^2 = 8^2 or

   AC^2 = 8^2 - 4^2- = 64 - 16 = 48

   Hence AC = sqrt(48) = sqrt (16*3) = 4 * sqrt(3)

We are almost done! The area of the triangle is given by

  (1/2) b * h = (1/2)  BC * AC = (1/2) 4 * (4 * sqrt(3)) =  8 sqrt(3)

We conclude the area area of the shaded part is

 8 PI - 8 sqrt(3)   = 8 (Pi - sqrt(3))

Note that sqrt(3) is approx  1.7 so (PI - sqrt(3)) is a positive number, as it better well be!

6 0
2 years ago
Frank left his house at 7 a.m. and drove to the airport at a speed of 50 mph. Lance left his house at 6 a.m. and drove to the sa
frosja888 [35]

Answer:

10 AM

Step-by-step explanation:

Frank drove at a speed of 50 mph, and traveled a distance of 150 miles.  This means he traveled for

150/50 = 3 hours.

7 AM + 3 hours = 10 AM

3 0
2 years ago
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