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Alborosie
3 years ago
11

Write a real world problem that could be modeled by a linear function whose x-intercept is 5 and y-intercept is 60.

Mathematics
1 answer:
max2010maxim [7]3 years ago
8 0
The y-intercept is 60 so say you start with 60 apples. And the x-intercept is 5 so lets say you eat them all in 5 days. So the equation would look somthing like this Y=-12x+60 and would end up having a word problem that says somthing like "Jimmy has 60 apples and eats 12 a day how many days go by before Jonny has no more apples ?"
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Step-by-step explanation:

So, it's x*x+x*-3+-4*x+3*-3

That simplifies to x^2 - 7x + 12

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What's the LCF of 12, 27, and 36?
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Find the coordinates of midpoint F.
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Answer:

The coordinates are (0,b)

Step-by-step explanation:

Here, we want to find the coordinates of the midpoint F

as we can see, F is between A and B

we proceed to use the midpoint formula

The midpoint formula is;

(x,y) = (x1 + x2)/2, (y1 + y2)/2

(x1,y1) = (-3a, b)

(x2,y2) = (3a, b)

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2 years ago
Prove that the triangle EDF is isosceles. Give reasons for your answer.
Gekata [30.6K]

Answer:

\triangle EDF is isosceles.

Step-by-step explanation:

Please have a look at the attached figure.

We are <u>given</u> the following things:

\angle EDF = y

\text{External }\angle DFG = 90 +\dfrac{y}{2}

Let us try to find out \angle E and \angle DFE. After that we will compare them.

<u>Finding </u>\angle DFE<u>:</u>

Side EG is a straight line so \angle GFE = 180

\angle GFE is sum of internal \angle DFE and external \angle DFG

\angle GFE = 180 = \angle DFE  + \angle DFG\\\Rightarrow 180 = \angle DFE + (90+\dfrac{y}{2})\\\Rightarrow \angle DFE = 180 - 90 - \dfrac{y}{2}\\\Rightarrow \angle DFE = 90 - \dfrac{y}{2} ....... (1)

<u>Finding </u>\angle E<u>:</u>

<u>Property of external angle:</u> External angle in a triangle is equal to the sum of two opposite internal angles of a triangle.

i.e. external \angle DFG = \angle E + \angle EDF

\Rightarrow 90+\dfrac{y}{2} = \angle E + y\\\Rightarrow \angle E = 90+\dfrac{y}{2}  -y\\\Rightarrow \angle E = 90-\dfrac{y}{2} ....... (2)

Comparing equations (1) and (2):

It can be clearly seen that:

\angle DFE = \angle E =90-\dfrac{y}{2}

The two angles of \triangle EDF are equal hence \triangle EDF is isosceles.

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