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Mathematics

1 answer:

Yuki888 [10]3 years ago

7 0

Dilation about the origin multiplies every coordinate by the scale factor.

C' = 4(-5, 2) = (-20, 8)

A' = 4(-4, 4) = (-16, 16)

T' = 4(-1, 2) = (-4, 8)

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rjkz [21]

Answer:

Step-by-step explanation:

We know that angles BOC and AOB are complementary angles, that is, the sum 90°. Additionally, we know that

$\angle BOC=3x+30$ and $\angle AOB = 6x-12$ , which can be expressed as

$3x+30+6x-12=90$

Let's solve for $x$

$9x=72\\x=\frac{72}{9}\\ x=8$

Using this vale, we find the measure of angle AOB

$\angle AOB = 6x-12=6(8)-12 =48-12=36\°$

Therefore, the answer is 36°.

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4 years ago

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Tems11 [23]

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3 years ago

Integrate this for me please

mel-nik [20]

$\int\limits_{0}^{\frac{\pi }{6}}tan^2(x)sec^2(x)\cdot dx
\\------------------\\
u=tan(x)\implies \frac{du}{dx}=sec^2(x)\implies \frac{du}{sec^2(x)}=dx
\\------------------\\
\int\limits_{0}^{\frac{\pi }{6}}u^2sec^2(x)\cdot \cfrac{du}{sec^2(x)}\implies 
\int\limits_{0}^{\frac{\pi }{6}}u^2\cdot du
\\ \quad \\
$

$\textit{now, we need to change the bounds as well, so}
\\------------------\\
u(0)=tan(0)\implies 0
\\ \quad \\

u\left( \frac{\pi }{6} \right)=tan\left( \frac{\pi }{6} \right)\implies \frac{1}{\sqrt{3}}
\\------------------\\
thus\implies \int\limits_{0}^{\frac{1 }{\sqrt{3}}}u^2\cdot du$

and surely you can take it from there,
recall, that, since we changed the bounds, with the u(x),
you don't need to change the variable "u", and simply,
get the integral of it, simple enough, and apply those bounds

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3 years ago

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Artist 52 [7]

since she already has 18 cups flour she needs 6 cups of sugar

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3 years ago

A circle has a radius of 3 cm. What is the circles circumference expressed as a decimal

Aloiza [94]

Answer:

18.85 cm