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avanturin [10]
3 years ago
15

A car dealer offers 6 different models, 4 colors, and 3 kinds of transmissions. How many unique cars could you choose from at th

is dealership?
Mathematics
2 answers:
Mkey [24]3 years ago
8 0
6 × 4 × 3 = 72 cars.

72 different type of cars.
Naily [24]3 years ago
7 0

Answer: 72

Step-by-step explanation:

Given: The number of difference models of cars = 6

The number of different colors of cars = 4

The number of different kinds of transmissions= 3

Then, the number of unique cars will be given by :-

N=6\times4\times3=72

Therefore, the number of unique cars dealer can offer= 72.

Hence we can choose a unique car from 72 cars.

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4/3(6y+21)-17>43 help pls
dezoksy [38]

Answer:

Step-by-step explanation:

4/3(6y+21)-17=43

move all terms to the left:

4/3(6y+21)-17-(43)=0

Domain of the equation: 3(6y+21)!=0

y∈R

We add all the numbers together, and all the variables

4/3(6y+21)-60=0

You multiply all the terms by the denominator

-60*3(6y+21)+4=0

multiply elements

8 0
3 years ago
X(21)+x(4)=300 I need solving
Genrish500 [490]

Answer:

12

Step-by-step explanation:

21x + 4x = 300         Put brackets around the first two terms.

(21x + 4x) = 300        Pull the x outside the brackets. Use the distributive property.  Combine the left

25x = 300                 Divide by 25

25x/25=300/25        Do the division

x = 12

                 

5 0
4 years ago
(tan145 + tan 85)/(1-tan145 tan85)
pshichka [43]

Answer:

Step-by-step explanation:

3 0
4 years ago
Can someone help me plz
andrew-mc [135]
Use hypotenuse leg theorem.

7 0
3 years ago
Hich function has a range of {y|y ≤ 5}?
marshall27 [118]

Answer:

f(x)=-(x-4)^{2}+5

Step-by-step explanation:

we know that

The equation of a vertical parabola into vertex form is equal to

y=a(x-h)^{2}+k

where

(h,k) is the vertex of the parabola

if a>0 -----> the parabola open upward (vertex is a minimum)

if a<0 -----> the parabola open downward (vertex is a maximum)

<u>Verify each case</u>

case A) f(x)=(x-4)^{2}+5

The vertex is the point (4,5)

a=1

a>0 -----> the parabola open upward (vertex is a minimum)

The range is the interval--------> [5,∞)

y\geq5

case B) f(x)=-(x-4)^{2}+5

The vertex is the point (4,5)

a=-1

a<0 -----> the parabola open downward (vertex is a maximum)

The range is the interval--------> (-∞,5]

y\leq 5

case C) f(x)=(x-5)^{2}+4

The vertex is the point (5,4)

a=1

a>0 -----> the parabola open upward (vertex is a minimum)

The range is the interval--------> [4,∞)

y\geq4

case D) f(x)=-(x-5)^{2}+4

The vertex is the point (5,4)

a=-1

a<0 -----> the parabola open downward (vertex is a maximum)

The range is the interval--------> (-∞,4]

y\leq 4

4 0
3 years ago
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