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3 years ago

Pls helllllllllllllpppppppppp mmmmmmmmeeeeee

Mathematics

2 answers:

Masja [62]3 years ago

7 0

Answer:

Hello! answer: 36

Step-by-step explanation:

6 × 7 = 42 3 × 2 = 6 42 - 6 = 36 therefore the area is 36 hope that helps!

ElenaW [278]3 years ago

7 0

Answer:

40cm^2

perimeter:24cm

Step-by-step explanation:

Area: First of all, you can see that the missing height of the shape is 3cm. By adding 4 and 3, you get 7. Then, by multiplying 7 and the length of the shape (6cm), you get 42cm^2. After that, you have to multiply 2cm and 3cm to get the blank area in the bottom right corner to get 6cm^2. Finally, you can subtract 42cm^2(the whole area) by 6cm^2 (the blank area) to get 40cm^2.

Perimeter: Okay! To get the first missing measurement (the one on the left), you have to imagine 4cm being moved to that side. 3cm is parallel to the missing blank space under the 4cm. This means you have to get 7cm (you add 4 and 3). To get the other missing measurement (bottom of the shape) you must see that 6cm would be parallel to the bottom. Since 2cm is also parallel to the bottom, you would have to subtract 2 from 6 to get the bottom of the shape's length, which is 4. Finally, you would have to add 6, 4, 3, 2, 4 and 7 together to get 24cm.

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mr Goodwill [35]

we'll start off by grouping some

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so we have a missing guy at the end in order to get the a perfect square trinomial from that group, hmmm, what is it anyway?

well, let's recall that a perfect square trinomial is

$\bf \qquad \textit{perfect square trinomial} \\\\ (a\pm b)^2\implies a^2\pm \stackrel{\stackrel{\text{\small 2}\cdot \sqrt{\textit{\small a}^2}\cdot \sqrt{\textit{\small b}^2}}{\downarrow }}{2ab} + b^2$

so we know that the middle term in the trinomial, is really 2 times the other two without the exponent, well, in our case, the middle term is just "x", well is really -x, but we'll add the minus later, we only use the positive coefficient and variable, so we'll use "x" to find the last term.

$\bf \stackrel{\textit{middle term}}{2(x)(?)}=\stackrel{\textit{middle term}}{x}\implies ?=\cfrac{x}{2x}\implies ?=\cfrac{1}{2}$

so, there's our fellow, however, let's recall that all we're doing is borrowing from our very good friend Mr Zero, 0, so if we add (1/2)², we also have to subtract (1/2)²

$\bf \left( x^2 -x +\left[ \cfrac{1}{2} \right]^2-\left[ \cfrac{1}{2} \right]^2 \right)=6\implies \left( x^2 -x +\left[ \cfrac{1}{2} \right]^2 \right)-\left[ \cfrac{1}{2} \right]^2=6 \\\\\\ \left(x-\cfrac{1}{2} \right)^2=6+\cfrac{1}{4}\implies \left(x-\cfrac{1}{2} \right)^2=\cfrac{25}{4}\implies x-\cfrac{1}{2}=\sqrt{\cfrac{25}{4}} \\\\\\ x-\cfrac{1}{2}=\cfrac{\sqrt{25}}{\sqrt{4}}\implies x-\cfrac{1}{2}=\cfrac{5}{2}\implies x=\cfrac{5}{2}+\cfrac{1}{2}\implies x=\cfrac{6}{2}\implies \boxed{x=3}$

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