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3 years ago

12

CHECK MY ANSWER??

1 answer:

Tatiana [17]3 years ago

4 0

You are partially correct :) 2^SQR(10) x SQR(10) = 20
The answer is just 20

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-3|x-4|-2>-11<br> please explain

Anastaziya [24]

Answer:

Ang answer kay jdlfjehweorj2jeihdjeiegfiwh11111111

7 0

3 years ago

What is the equation of the line shown in this graph? A function graph of a line with two points (-4,1) and (2,1) with an x axis

alisha [4.7K]

Answer:

y = 1

Step-by-step explanation:

You should be able to tell by inspection that the line is horizontal at y = 1.

7 0

3 years ago

How is this worked out ? im confused.

Rudik [331]

Answer:

From the given ratio for $tan\theta=\frac{4}{5}$ we found the values are

$sin\theta=4$

$cos\theta=5$

$csc\theta=\frac{1}{4}$

$sec\theta=\frac{1}{5}$

$cot\theta=\frac{5}{4}}$

Step-by-step explanation:

Given the ratio for $tan\theta=\frac{4}{5}$

We have to find the rest of trignometric functions:

$tan\theta=\frac{4}{5}\hfill (1)$

$tan\theta$ can be written as

$tan\theta=\frac{sin\theta}{cos\theta}\hfill (2)$

Comparing equations (1) and (2) we get

$tan\theta=\frac{sin\theta}{cos\theta}=\frac{4}{5}$

$\frac{sin\theta}{cos\theta}=\frac{4}{5}$

Equating the coressponding numerator and denominator respectively

Therefore $sin\theta=4$ and $cos\theta=5$

We can find $csc\theta$

$csc\theta=\frac{1}{sin\theta}$

$=\frac{1}{4}$ (since $sin\theta=4$ )

$csc\theta=\frac{1}{4}$

We can find $sec\theta$

$sec\theta=\frac{1}{cos\theta}$

$=\frac{1}{5}$ (since $cos\theta=5$ )

$sec\theta=\frac{1}{5}$

To find $cot\theta$ :

$cot\theta=\frac{1}{tan\theta}$

$=\frac{1}{\frac{4}{5}}$

$=1\times \frac{5}{4}$

$=\frac{5}{4}}$

$cot\theta=\frac{5}{4}}$

7 0

4 years ago

Ad libitum [116K]

D I’m pretty sure because 8 times itself = 64

5 0

3 years ago

Read 2 more answers

Given: Which additional congruence statement could you use to prove that ∆BJK ≅ ∆CFH by the HL Theorem? ∠BJK ≅ ∠CFH ∠B ≅ ∠C

earnstyle [38]

Answer:

∠BJK ≅ ∠CFH

Step-by-step explanation:

∆BJK ≅ ∆CFH then we say ∠BJK ≅ ∠CFH

Only when we say ∠BJK ≅ ∠CFH we can say (expand) and speak of lengths by comparing two of each ∠BJA ≅ ∠AFC whilst ∠BK≅ ∠ HC then repeat with∠BKJ ≅ ∠CHF etc.

3 0

3 years ago

Read 2 more answers