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svetlana [45]
2 years ago
12

Apply the distributive property to factor out the greatest common factor of all three terms. 24c+36d+18

Mathematics
1 answer:
mihalych1998 [28]2 years ago
7 0

Answer:

GCF = 6

Step-by-step explanation:

24c+36d+18

6(4c)+6(6d)+6(3)

6(4c+6d+3)

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Given that a function, g, has a domain of -1 ≤ x ≤ 4 and a range of 0 ≤ g(x) ≤ 18 and that g(-1) = 2 and g(2) = 8, select the st
notka56 [123]

Options

  • (A)g(5) = 12
  • (B)g(1) = -2
  • (C)g(2) = 4
  • (D)g(3) = 18

Answer:

(D)g(3) = 18

Step-by-step explanation:

Given that the function, g, has a domain of -1 ≤ x ≤ 4 and a range of                       0 ≤ g(x) ≤ 18 and that g(-1) = 2 and g(2) = 8

Then the following properties must hold

  1. The value(s) of x must be between -1 and 4
  2. The values of g(x) must be between 0 and 18.
  3. g(-1)=2
  4. g(2)=9

We consider the options and state why they are true or otherwise.

<u>Option A: g(5)=12</u>

The value of x=5. This contradicts property 1 stated above. Therefore, it is not true.

<u>Option B: g(1) = -2 </u>

The value of g(x)=-2. This contradicts property 2 stated above. Therefore, it is not true.

<u>Option C: g(2) = 4 </u>

The value of g(2)=4. However by property 4 stated above, g(2)=9. Therefore, it is not true.

<u>Option D: g(3) = 18</u>

This statement can be true as its domain is in between -1 and 4 and its range is in between 0 and 18.

Therefore, Option D could be true.

3 0
2 years ago
A diagram of a deodorant container is shown It is made up of a cylinder and half of a sphere. (Radius = 1.6) (Height=6.2) a.) wh
viva [34]
A) 8.58 units
B)36.984 units
C)45.564 units
6 0
2 years ago
Deal with these relations on the set of real numbers: R₁ = {(a, b) ∈ R² | a &gt; b}, the "greater than" relation, R₂ = {(a, b) ∈
uranmaximum [27]

Answer:

a) R1ºR1 = R1

b) R1ºR2 = R1

c) R1ºR3 = \{ (a,b) \in R^2 \}

d) R1ºR4 = \{ (a,b) \in R^2 \}

e) R1ºR5 = R1

f) R1ºR6 = \{ (a,b) \in R^2 \}

g) R2ºR3 = \{ (a,b) \in R^2 \}

h) R3ºR3 = R3

Step-by-step explanation:

R1ºR1

(<em>a,c</em>) is in R1ºR1 if there exists <em>b</em> such that (<em>a,b</em>) is in R1 and (<em>b,c</em>) is in R1. This means that a > b, and b > c. That can only happen if a > c. Therefore R1ºR1 = R1

R1ºR2  

This case is similar to the previous one. (<em>a,c</em>) is in R1ºR2 if there exists <em>b</em> such that (<em>a,b</em>) is in R2 and (<em>b,c</em>) is in R1. This means that a ≥ b, and b > c. That can only happen if a > c. Hence R1ºR2 = R1

R1ºR3

(a,c) is in R1ºR3 if there exists b such that a < b and b > c. Independently of which values we use for a and c, there always exist a value of b big enough so that b is bigger than both a and c, fulfilling the conditions. We conclude that any pair of real numbers are related.

R1ºR4

This is similar to the previous one. Independently of the values (a,c) we choose, there is always going to be a value b big enough such that a ≤ b and b > c. As a result any pair of real numbers are related.

R1ºR5

If a and c are related, then there exists b such that (a,b) is in R5 and (b,c) is in R1. Because of how R5 is defined, b must be equal to a. Therefore, (a,c) is in R1. This proves that R1ºR5 = R1

R1ºR6

The relation R6 is less restrictive than the relation R3, if we find 2 numbers, one smaller than the other, in particular we find 2 different numbers. If we had 2 numbers a and c, we can find a number b big enough such that a<b and b >c. In particular, b is different from a, so (a,b) is in R6 and (b,c) is in R1, which implies that (a,c) is in R1ºR6. Since we took 2 arbitrary numbers, then any pair of real numbers are related.

R2ºR3

This is similar to the case R1ºR3, only with the difference that we can take b to be equal to a as long as it is bigger than c. We conclude that any pair of real numbers are related.

R3ºR3

If a and c are real numbers such that there exist b fulfilling the relations a < b and b < c, then necessarily a < c. If a < c, then we can use any number in between as our b. Therefore R3ºR3 = R3

I hope you find this answer useful!

5 0
2 years ago
A football filters that is 60 m wide and 110 m long is being paved over to make a parking lot. The builder ordered 660,000,000 c
ad-work [718]
The answer is 10 cm

Imagine cement cover as a rectangle which volume is 660,000,000 cm3. So this rectangle has width (w = 60m), length (l = 110m), and height (h = ?). The height of the rectangle is actually a thickness of the cement layer. So, we will use the formula for the volume (V) of the rectangle to calculate the thickness:
V = w · l · h

It is given:
V = 660,000,000 cm³
w = 60 m = 60 · 100 cm = 6,000 cm
l = 110 m = 110 · 100 cm = 11,000 cm
h = ?

Using the formula: V = w · l · h
660,000,000 = 6,000 · 11,000 · h
660,000,000 = 66,000,000 · h
⇒ h = 660,000,000 ÷ 66,000,000 = 10 cm
4 0
2 years ago
An ore ship is traveling west toward Duluth on Lake Superior at 18 miles per hour. The bearing of Split Rock Lighthouse is 285de
goblinko [34]
We are given
v = 18 m/hr west
θ1 = 285°
θ2 = 340°

After 1 hours, the distance traveled by the ship is
dv = 18 mi

The distance between the ship and the lighthouse is
d = 18 / cos 340
Solve for d<span />
5 0
2 years ago
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