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Nitella [24]
1 year ago
6

Consider a function that goes through the two points (0, 5) and (1, 20). Find the formula for the function if(a) the function is

linear (of the formf(x) =mx b)
Mathematics
1 answer:
EastWind [94]1 year ago
7 0

The formula for the linear function which is passing the points (0, 5) and (1, 20) is f(x) = 15x + 5.

According to the given question.

The linear form of the function is

f(x) = mx + b

Also, the function is passing through the points (0, 5) and (1, 20).

So, the given points (0, 5) and (1, 20) must satisfy f(x) = mx + b.

Now,

At (0, 5)

f(0) = m(0) + b

⇒ 5 = 0 + b

⇒ b = 5 ..(i)

Also,

at (1, 20)

f(1) = m(1) + b

⇒ 20 = m + 5     (from i)

⇒ m = 20 - 5

⇒ m = 15

Therefore, the formula for the linear function which is passing the points (0, 5) and (1, 20) is given by

f(x) = 15(x) + 5   (on substituting the vale of m and b in f(x) = mx + b).

⇒ f(x) = 15x + 5

Hemce, the formula of the function is f(x) = 15x + 5.

Find out more information about function here:

brainly.com/question/21107621

#SPJ4

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at a sale a desk is being sold for 24% of the regular price. the sale price is $182.40 what is the regular price
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x=$760

therefore

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notka56 [123]

Answer:

AC = { 4, 5, 6, 7 }

Step-by-step explanation:

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In both triangles the triangle inequality can help find the possible value( s ) of AD, as this inequality only restricts some of the possible values with which AC can take. At the same time AC is shared among the two triangles, so if we can apply the Triangle Inequality to both of these triangles, the value of AC can be " further restricted. "

Triangle 1. BC - AB < AC < BC + AB,\\6 - 3 < AC < 6 + 3,\\3 < AC < 9\\\\Triangle2. AD - CD < AC < AD + CD,\\4 - 4 < AC < 4 + 4,\\0 < AC < 8

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3 years ago
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An arithmetic sequence has a common difference of 5 and the 3rd term is 15. What is the 5th term?
trapecia [35]

Answer:

an=8n−14

Step-by-step explanation:

for an AP

where

a

1

=

the first term

d

=

the common difference

then we have

a

1

,

(

a

1

+

d

)

,

(

a

1

+

2

d

)

,

...

,

a

1

+

(

(

n

−

1

)

d

)

,

.

.

we are given the third term

10

=

a

1

+

2

d

−

−

(

1

)

and the fifth term

26

=

a

1

+

4

d

−

−

(

2

)

subtract

(

2

)

−

(

1

)

16

=

2

d

⇒

d

=

8

sub into

(

1

)

10

=

a

1

+

2

×

8

⇒

a

1

=

−

6

so the nth term

a

n

=

a

+

(

n

−

1

)

will be

a

n

=

−

6

+

8

(

n

−

1

)

a

n

=

8

n

−

14

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