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LUCKY_DIMON [66]
3 years ago
13

The function f(x)=x^2+4.2x+3.6 has an axis of symmetof x=-2-. What is the minimum value of the function

Mathematics
1 answer:
Pepsi [2]3 years ago
8 0

ANSWER

The minimum value is -0.8

EXPLANATION

The given

f(x)=x^2+4.2x+3.6

The minimum value is the y-value of the vertex (minimum point).

To find the minimum value of the function, we substitute x=-2 into the function to get;

f( - 2)=( - 2)^2+4.2( - 2)+3.6

f( - 2)=4 - 8.4+3.6 =  - 0.8

The minimum value is -0.8

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AM is a median in △ABC (M∈ BC ). A line drawn through point M intersects AB at its midpoint P. Find areas of △APC and △PMC, if A
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Answer:

The area of APC is 70m². The area of triangle PMC is 35m².

Step-by-step explanation:

Let the area of triangle ABC be x.

It is given that AM is median, it means AM divides the area of triangle in two equal parts.

\text{Area of }\triangle ACM=\text{Area of }\triangle ABM=\frac{x}{2}    .....(1)

The point P is the midpoint of AB, therefore the area of APC and BPC are equal.

\text{Area of }\triangle APC=\text{Area of }\triangle BPC=\frac{x}{2}          ......(2)

The point P is midpoint of AB therefore the line PM divide the area of triangle ABM in two equal parts. The area of triangle APM and BPM are equal.

\text{Area of }\triangle APM=\text{Area of }\triangle BPM=\frac{x}{4}        .....(3)

The area of triangle APM is 35m².

\text{Area of }\triangle APM=\frac{x}{4}

35=\frac{x}{4}

x=140

Therefore the area of triangle ABC is 140m².

Using equation (2).

\text{Area of }\triangle APC=\frac{x}{2}

\text{Area of }\triangle APC=\frac{140}{2}

\text{Area of }\triangle APC=70

Therefore the area of triangle APC is 70m².

Using equation (3), we can say that the area of triangle BPM is 35m² and by using equation (2), we can say that the area of triangle BPC is 70m².

\triangle BPC=\triangle BPM+\triangle PMC

70=35+\triangle PMC

35=\triangle PMC

Therefore the area of triangle PMC is 35m².

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