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3 years ago

4. Square ABCD is shown. What is the length of diagonal AC? Round to the nearest tenth, if necessary.

Mathematics

1 answer:

zhuklara [117]3 years ago

8 0

Answer:

hey! You have to use the trigonometric formula because it's a rectangle triangle:

(a^2)+(b^2)=(c^2)

(11^2)+(11^2)=(c^2)

√(242)=√(c^2)

15.56=c

So basically, the answer is D!

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3 years ago

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3 years ago

Prove : sin²θ + cos²θ = 1 thankyou ~

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Answer:

See below

Step-by-step explanation:

Here we need to prove that ,

$\sf\longrightarrow sin^2\theta + cos^2\theta = 1$

Imagine a right angled triangle with one of its acute angle as $\theta$ .

The side opposite to this angle will be perpendicular .
Also we know that ,

$\sf\longrightarrow sin\theta =\dfrac{p}{h} \\$

$\sf\longrightarrow cos\theta =\dfrac{b}{h}$

And by Pythagoras theorem ,

$\sf\longrightarrow h^2 = p^2+b^2 \dots (i)$

Where the symbols have their usual meaning.

Now , taking LHS ,

$\sf\longrightarrow sin^2\theta +cos^2\theta$

Substituting the respective values,

$\sf\longrightarrow \bigg(\dfrac{p}{h}\bigg)^2+\bigg(\dfrac{b}{h}\bigg)^2\\$

$\sf\longrightarrow \dfrac{p^2}{h^2}+\dfrac{b^2}{h^2}\\$

$\sf\longrightarrow \dfrac{p^2+b^2}{h^2}$

From equation (i) ,

$\sf\longrightarrow\cancel{ \dfrac{h^2}{h^2}}\\$

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Since LHS = RHS ,

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2 years ago

Suppose that an insect population’s density, in thousands per acre, during year n, can be modeled by the recursive formula: a1 =

vlabodo [156]

Answer:

The population alternates between increasing and decreasing

Step-by-step explanation:

The options of the question are

A) The population density decreases each year.

B) The population density increases each year.

C) The population density remains constant.

D) The population alternates between increasing and decreasing

we have

$a_n=2.9(a_n_-_1)-0.2(a_n_-_1)^2$

$a_1=8$

Find the value of $a_2$

For n=2

$a_2=2.9(a_1)-0.2(a_1)^2$

$a_2=2.9(8)-0.2(8)^2=10.4$

Find the value of $a_3$

For n=3

$a_3=2.9(a_2)-0.2(a_2)^2$

$a_3=2.9(10.4)-0.2(10.4)^2=8.528$

For n=4

$a_4=2.9(a_3)-0.2(a_3)^2$

$a_4=2.9(8.528)-0.2(8.528)^2=10.1858$

For n=5

$a_5=2.9(a_4)-0.2(a_4)^2$

$a_5=2.9(10.1858)-0.2(10.1858)^2=8.7887$

therefore

The population alternates between increasing and decreasing

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