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julia-pushkina [17]
3 years ago
5

Write the function a a product of linear factor by grouping or using the x method or a combination of both

Mathematics
1 answer:
Sladkaya [172]3 years ago
4 0
<h3><u>Answer:</u></h3>

\boxed{\boxed{\pink{\sf Option \ A \ is \ correct .}}}

<h3><u>Step-by-step explanation:</u></h3>

Given function to us is :-

\bf \implies g(x) = x^2 - 9

And we , need to write the function a a product of linear factor by grouping or using the x method or a combination of both . So let's factorise this ,

\bf \implies g(x) = x^2 - 9 \\\\\bf\implies g(x) = x^2-3^2\\\\\bf\implies \boxed{\red{\bf g(x) = (x+3)(x-3) }}\:\:\bigg\lgroup \blue{\tt Using \ (a+b)(a-b) \ = a^2-b^2 }\bigg\rgroup

I have also attached the graph of x²-9.

<h3><u>Hence </u><u>option</u><u> </u><u>A</u><u> </u><u>is</u><u> </u><u>corr</u><u>ect</u><u> </u><u>.</u></h3>

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Answer:

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Step-by-step explanation:

Convert the problem to an equation using the percentage formula: P% * X = Y.

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2 years ago
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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
Snowcat [4.5K]

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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