The area of the green square is 9ft. The area of the yellow square is 25 ft.

Mathematics

1 answer:

Alex_Xolod [135]3 years ago

3 0

Answer:

Step-by-step explanation:

a²+b²=c²

3²+b²=5²

b²=5²-3²

b²=25-9=16 answer is 16

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HELP WITH 20 PLEASE!

WINSTONCH [101]

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Question from Amanda, a parent:

what is the area of a circle with a circumference of 3000 metres?

Amanda,

The area of a circle is given by the formula A = π r2, where A is the area and r is the radius. The circumference of a circle is C = 2 π r.

If we "solve for r" in the second equation, we have r = C / (2 π ). Now we use this to replace r in the first equation: A = π [ C / (2 π ) ]2.

When we simplify this, we get A = C2 / (4 π).

Now you can put your value of C into this equation and find A. Remember, π is about 3.1416.

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3 years ago

5+1+(-3)+...+(-395)=

sleet_krkn [62]

$= \frac{(-395+5)(-395-5-4)}{2.(-4)} \\ \\ \\=-19695 \\ \\ \\ Good Luck:))$

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4 years ago

Solve irrational equation pls

rusak2 [61]

$\hbox{Domain:}\\
x^2+x-2\geq0 \wedge x^2-4x+3\geq0 \wedge x^2-1\geq0\\
x^2-x+2x-2\geq0 \wedge x^2-x-3x+3\geq0 \wedge x^2\geq1\\
x(x-1)+2(x-1)\geq 0 \wedge x(x-1)-3(x-1)\geq0 \wedge (x\geq 1 \vee x\leq-1)\\
(x+2)(x-1)\geq0 \wedge (x-3)(x-1)\geq0\wedge x\in(-\infty,-1\rangle\cup\langle1,\infty)\\
x\in(-\infty,-2\rangle\cup\langle1,\infty) \wedge x\in(-\infty,1\rangle \cup\langle3,\infty) \wedge x\in(-\infty,-1\rangle\cup\langle1,\infty)\\
x\in(-\infty,-2\rangle\cup\langle3,\infty)$

$
\sqrt{x^2+x-2}+\sqrt{x^2-4x+3}=\sqrt{x^2-1}\\
x^2-1=x^2+x-2+2\sqrt{(x^2+x-2)(x^2-4x+3)}+x^2-4x+3\\
2\sqrt{(x^2+x-2)(x^2-4x+3)}=-x^2+3x-2\\
\sqrt{(x^2+x-2)(x^2-4x+3)}=\dfrac{-x^2+3x-2}{2}\\
(x^2+x-2)(x^2-4x+3)=\left(\dfrac{-x^2+3x-2}{2}\right)^2\\
(x+2)(x-1)(x-3)(x-1)=\left(\dfrac{-x^2+x+2x-2}{2}\right)^2\\
(x+2)(x-3)(x-1)^2=\left(\dfrac{-x(x-1)+2(x-1)}{2}\right)^2\\
(x+2)(x-3)(x-1)^2=\left(\dfrac{-(x-2)(x-1)}{2}\right)^2\\
(x+2)(x-3)(x-1)^2=\dfrac{(x-2)^2(x-1)^2}{4}\\
4(x+2)(x-3)(x-1)^2=(x-2)^2(x-1)^2\\$
$
4(x+2)(x-3)(x-1)^2-(x-2)^2(x-1)^2=0\\
(x-1)^2(4(x+2)(x-3)-(x-2)^2)=0\\
(x-1)^2(4(x^2-3x+2x-6)-(x^2-4x+4))=0\\
(x-1)^2(4x^2-4x-24-x^2+4x-4)=0\\
(x-1)^2(3x^2-28)=0\\
x-1=0 \vee 3x^2-28=0\\
x=1 \vee 3x^2=28\\
x=1 \vee x^2=\dfrac{28}{3}\\
x=1 \vee x=\sqrt{\dfrac{28}{3}} \vee x=-\sqrt{\dfrac{28}{3}}\\$

There's one more condition I forgot about
$-(x-2)(x-1)\geq0\\
x\in\langle1,2\rangle\\$

Finally
$x\in(-\infty,-2\rangle\cup\langle3,\infty) \wedge x\in\langle1,2\rangle \wedge x=\{1,\sqrt{\dfrac{28}{3}}, -\sqrt{\dfrac{28}{3}}\}\\
\boxed{\boxed{x=1}}$

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3 years ago

Given solve triangle ABC. Round the answer to the nearest hundredth. A. A- 9.590, B8841°, c-11.89 B. A 9.590, B-8841, c 12.17 C.

nikitadnepr [17]

I think the answer is b

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4 years ago

Can somebody help me with this

Lera25 [3.4K]

Answer:

Step-by-step explanation:

It’s just like adding except with a variable so 5x+3x is like 5+3 except with x so 8x+3=16