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2 years ago

Please can i have some help, just on this question, thank you so much!

Mathematics

1 answer:

Andrew [12]2 years ago

5 0

Answer:

a) 1/2 of 12 = 1/4 of 24

b) 1/3 of 90 = 2/3 of 45

Step-by-step explanation:

1/2 of 16 is 6, so 6 would be 1/4 of what? 6 x 4 = 24. 6 is 1/4 of 24. The other one is the same. 1/3 of 90 is 30, and 30 is 2/3 of 45.

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ivann1987 [24]

Answer:

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Step-by-step explanation:

(10x² - 10x - 6) + (x + 6)

10x² - 10x - 6 + x + 6

10x² - 9x - 6 + 6

10x² - 9x

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3 years ago

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3 years ago

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3 years ago

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kondaur [170]

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3 years ago

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Given: PRST is a square

xxTIMURxx [149]

Answer:

$(1-\sqrt{2})a^2$

Step-by-step explanation:

Consider irght triangle PRS. By the Pythagorean theorem,

$PS^2=PR^2+RS^2\\ \\PS^2=a^2+a^2\\ \\PS^2=2a^2\\ \\PS=\sqrt{2}a$

Thus,

$MS=PS-PM=\sqrt{2}a-a=(\sqrt{2}-1)a$

Consider isosceles triangle MSC. In this triangle

$MS=MC=(\sqrt{2}-1)a.$

The area of this triangle is

$A_{MSC}=\dfrac{1}{2}MS\cdot MC=\dfrac{1}{2}\cdot (\sqrt{2}-1)a\cdot (\sqrt{2}-1)a=\dfrac{(\sqrt{2}-1)^2a^2}{2}=\dfrac{(3-2\sqrt{2})a^2}{2}$

Consider right triangle PTS. The area of this triangle is

$A_{PTS}=\dfrac{1}{2}PT\cdot TS=\dfrac{1}{2}a\cdot a=\dfrac{a^2}{2}$

The area of the quadrilateral PMCT is the difference in area of triangles PTS and MSC:

$A_{PMCT}=\dfrac{(3-2\sqrt{2})a^2}{2}-\dfrac{a^2}{2}=\dfrac{(2-2\sqrt{2})a^2}{2}=(1-\sqrt{2})a^2$

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3 years ago