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Sati [7]
3 years ago
5

Jim and Ava are 20 miles apart on a path when they start moving toward each other.Jim walks at a constant speed of 2 mph, And Av

a runs at a constant speed of 6 mph.How long does it take until Jim and Ava meet
Mathematics
1 answer:
Katarina [22]3 years ago
6 0

Answer:

It will take 2.5 hours before they meet

Step-by-step explanation:

The initial distance between Jim and Ava was 20 miles on a path before they start moving toward each other.

Jim walks at a constant speed of 2 mph,

Let time taken by Jim to cover his distance be a

Distance = speed × time

Distance covered by Jim = 2a

And Ava runs at a constant speed of 6 mph.

Let time taken by Ava to cover her distance be b

Distance = speed × time

Distance covered by Ava = 6b

When they meet, total distance covered would be 20miles. Therefore,

2a + 6b = 20

To get the common time it takes for them to cover this distance,

a = b

2a + 6a = 20

8a = 20

a =20/8 = 2.5hours

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Which graph represent the solution set of y=x^2+4x+3 and y=2x+6
Mamont248 [21]

Answer:

solution set are

(-3,0)

(1,8)

Step-by-step explanation:

we are given system of equation as

y=x^2+4x+3

y=2x+6

For finding solution set , we can draw graph of both equations

and then we can find intersection point

the intersection point will be our solution set

we get graph  as

So, solution set are

(-3,0)

(1,8)


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3 years ago
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108 hope this helps.
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What number replaces the Δ?
OleMash [197]
1 + 3 = Δ + 3

By comparing

Δ = 1.

Option A.


3 0
3 years ago
Original amount:625 end amount:550
Zolol [24]

Answer:

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Step-by-step explanation:

3 0
3 years ago
Read 2 more answers
(−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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