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Maru [420]
2 years ago
7

I can walk at 6km/hr and run at 15km/hr. On a journey I spend as long walking as I do running. If I had walked twice as long on

the journey it would have taken me six minutes longer find how far I ran​
Mathematics
1 answer:
const2013 [10]2 years ago
5 0

He ran for 12 km in his journey.

<h3>What is an Equation ?</h3>

An equation is formed when two algebraic expressions are equated by an equal sign.

It is given that

Speed of Walking =6km/hr  

Speed of Running = 15km/hr

Let he walks and run x hours ( as given)

he covered distance = 21 x km

The total time taken in the journey is 2x

for the same distance

It is also given that

if for the total journey

if he walks walked twice as long on the journey

He walked 6x in the journey so new distance that he walked is  12x  , then the total journey time = 6 minutes longer

Converting speed from km / hr to km/ min

1 km / hr = 60 km/ min

and forming an equation with the given data

12x * 60 /6 + 9x *60/15 = 2x * 60 + 6

120x + 36x = 120x +6

150x = 120

x = 120/150

x = 0.8 hours.

Therefore he spent 0.8 hours walking and 0.8 hours running and He ran 15 * 0.8 = 12 km

To know more about Equation

brainly.com/question/10413253

#SPJ1

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Digiron [165]

The number of ways to choose the colors is 36

<h3>How to determine the number of ways?</h3>

The given parameters are:

Colors = 9

Colors to choose = 7

Since order does not matter, then it is combination

This is calculated using:

^nC_r = \frac{n!}{(n - r)!r!}

This gives

^nC_r = \frac{9!}{7!2!}

Evaluate

^nC_r = 36

Hence, the number of ways is 36

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6 0
2 years ago
Help with these two and explain pls and thankyouuu
Usimov [2.4K]

Answer:

3-3x and 3

Step-by-step explanation:

f(x) = 3x-3

g(x) = -x

g(f(x)) in your g(x) substitute x with f(x)

g( f(x) ) = g( 3x-3) = -(3x-3) = -3x+3

is the the second choice from top because -3x+3 is the same with 3 -3x

g(f(x)) =3-3x

g(f(0)) always work from inside first, so first find f(0) than g(f(0))

f(0) = f(x=0) = 3*0-3 = -3

Now we do the g(f(0)) = g( -3) is the same as  

g(x) except that we substitute x for -3, g(x= -3) = -(-3) =3

g(f(0))=3

*an easier way to find the answer for the second problem would have been that we knew that f(g(x) ) = 3-3x so f(g(0)) =3-3*0 = 3

8 0
3 years ago
What is the product?
myrzilka [38]

Answer6x3-33x2+45x-6

Step-by-step explanation:

8 0
3 years ago
Read 2 more answers
I need help with 6-8
Elina [12.6K]

Answer:

6. y² = 169x

7. y² = 12x

8. y^{2} = \frac{1}{2}x

Step-by-step explanation:

The equation of a quadratic function in vertex form is given by  

(y - α)² = 4a(x - β)

Where, (α,β) is the vertex of the function and a is the distance from vertex to its focus.

6. Here, (α,β) ≡ (0,0) and a point on the equation is (1,13)

So, the equation will be y² = 4ax

⇒ 13² = 4a(1)  

⇒ 4a = 169

Therefore, the original equation will be y² = 169x  (Answer)

7. Here also the vertex is (0,0) and a point on the equation is (3,-6),

So, (-6)² = 4a(3)  

⇒ 4a = 12

So, the equation is y² = 12x (Answer)

8. Here also the vertex is (0,0) and a point on the equation is (\frac{1}{2},\frac{1}{2}),

So, (\frac{1}{2})^{2} = 4a(\frac{1}{2})

⇒ 4a = \frac{1}{2}

So, the equation is y^{2} = \frac{1}{2}x (Answer)

4 0
4 years ago
Determine which of the following sets of three points constitute the vertices of a right triangle: (a) 3 + 5i,2 +2i,5i; (b)2i,3
Illusion [34]

Answer:

Option (c) is correct

Step-by-step explanation:

Case (a)

A = 3 + 5i = (3, 5)

B = 2 + 2i = (2, 2)

C = 5i = (0, 5)

Use the distance formula to find the distance between two points

AB = \sqrt{(2-3)^{2}+(2-5)^{2}}=\sqrt{10}

BC = \sqrt{(0-2)^{2}+(5-2)^{2}}=\sqrt{13}

CA = \sqrt{(0-3)^{2}+(5-5)^{2}}=\sqrt{9}

For the triangle to be right angles triangle

BC^{2}=AB^{2}+CA^{2}

Here, it is not valid, so these are not the points of a right angled triangle.

Case (b)

A = 2i = (0, 2)

B = 3 + 5i = (3, 5)

C = 4 + i = (4, 1)

Use the distance formula to find the distance between two points

AB = \sqrt{(3-0)^{2}+(5-2)^{2}}=\sqrt{18}

BC = \sqrt{(4-3)^{2}+(1-5)^{2}}=\sqrt{17}

CA = \sqrt{(4-0)^{2}+(1-2)^{2}}=\sqrt{17}

For the triangle to be right angles triangle

AB^{2}=BC^{2}+CA^{2}

Here, it is not valid, so these are not the points of a right angled triangle.

Case (c)

A = 6 + 4i = (6, 4)

B = 7 + 5i = (7, 5)

C = 8 + 4i = (8, 4)

Use the distance formula to find the distance between two points

AB = \sqrt{(7-6)^{2}+(5-4)^{2}}=\sqrt{2}

BC = \sqrt{(8-7)^{2}+(4-5)^{2}}=\sqrt{2}

CA = \sqrt{(8-6)^{2}+(4-4)^{2}}=\sqrt{4}

For the triangle to be right angles triangle

CA^{2}=BC^{2}+AB^{2}

Here, it is valid, so these are the points of a right angled triangle.

7 0
3 years ago
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