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slava [35]
3 years ago
13

If the ordinate is three more than twice the abscissa and x {-1, 0, -1/3}, which of the following sets of ordered pairs satisfie

s the given conditions?
A. {(-1, 1), (0, 3), (-1/3, 7/3)}
B. {(-1, 1), (0, 3), (-1/3, 11/3)}
C. {(-1, -1), (0, 3), (-1/3, 1)}
Mathematics
2 answers:
Molodets [167]3 years ago
8 0

Answer:

c

Step-by-step explanation:

lions [1.4K]3 years ago
7 0

Answer:

I think it is A not sure though

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Find the slope that passes through (4,9) and (6,8) <br>​
Aliun [14]
Answer: Y=-1/2x +11

-1/2 is the slope and 11 is the y-intercept
3 0
3 years ago
. Chuck needs to cut a piece of cardboard for an art project at school. He has four pieces of
pickupchik [31]

Answer:

6is the answer

Step-by-step explanation:

7 0
3 years ago
A rectangle hasan area of 100 square yards and a perimeter of 58 yds. What is the length of the longer side of the rectangle?
Vinil7 [7]

Answer:

The longer side of the rectangle is 25 yards.

Step-by-step explanation:

The area of a rectangle, A = 100 square yards

The perimeter of a rectangle, P = 58 yards

We formula for area and perimeter of a rectangle is given by :

A = lb

and

P = 2(l+b)

ATQ,

lb = 100 ...(1)

2(l+b) = 58

(l+b) = 29 ....(2)

We need to solve equation (1) and (2).

l = 4, b = 25

So, the longer side of the rectangle is 25 yards.

3 0
3 years ago
A bicycle manufacturing company makes a particular type of bike. Each child bike requires 4 hours to build and 4 hours to test.
nikitadnepr [17]

Answer:

4 (20) + 6(6) = 120

4(20) + 4 (6) = 100

Step-by-step explanation:

Hi, to answer this question we have to write a system of equations with the information given:

Build time (hours) = 4c + 6a = 120

Test time (hours) = 4c + 4 a = 100

Where c represents the number of child bikes and a represents adult bikes.

So, for 20 child bikes and 6 adult bikes:

4 (20) + 6(6) = 120

4(20) + 4 (6) = 100

Feel free to ask for more if needed or if you did not understand something.

7 0
3 years ago
(−4 + 5) − (6 + 7) = x x 0
weeeeeb [17]

1

Add the numbers

(

−

4

+

5

)

−

(

6

+

7

)

=

0

({\color{#c92786}{-4}}+{\color{#c92786}{5}})-(6+7)=xx^{0}

(−4+5)−(6+7)=xx0

(

1

)

−

(

6

+

7

)

=

0

({\color{#c92786}{1}})-(6+7)=xx^{0}

(1)−(6+7)=xx0

2

Add the numbers

1

−

(

6

+

7

)

=

0

1-({\color{#c92786}{6}}+{\color{#c92786}{7}})=xx^{0}

1−(6+7)=xx0

1

−

(

1

3

)

=

0

1-({\color{#c92786}{13}})=xx^{0}

1−(13)=xx0

3

Multiply the numbers

1

−

1

⋅

1

3

=

0

1{\color{#c92786}{-1}} \cdot {\color{#c92786}{13}}=xx^{0}

1−1⋅13=xx0

1

−

1

3

=

0

1{\color{#c92786}{-13}}=xx^{0}

1−13=xx0

4

Subtract the numbers

1

−

1

3

=

0

{\color{#c92786}{1-13}}=xx^{0}

1−13=xx0

−

1

2

=

0

{\color{#c92786}{-12}}=xx^{0}

−12=xx0

5

Combine exponents

−

1

2

=

0

-12={\color{#c92786}{xx^{0}}}

−12=xx0

−

1

2

=

1

-12={\color{#c92786}{x^{1}}}

−12=x1

Show less

Solution

−

1

2

=

1

3 0
3 years ago
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