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2 years ago

What is the square root of 134

Mathematics

2 answers:

scoundrel [369]2 years ago

8 0

Answer:

11.5758369028

Step-by-step explanation:

OleMash [197]2 years ago

4 0

Answer:

11.5758369028

Step-by-step explanation:

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In ΔABC shown below, ∠BAC is congruent to ∠BCA: Triangle ABC, where angles A and C are congruent Given: Base ∠BAC and ∠ACB are c

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Write the equation of the line that passes through the points (2, 1) and (5, -8)

My name is Ann [436]

Answer:

The answer is y = -3x + 7.

Step-by-step explanation:

❃Incase you forgot what the linear equation formula is ☟

$y = mx + b$

❃Incase you also forgot, what the slope formula is ☟

$m = \frac{y_2 - y_1}{x_2 - x_1}$

➊ First: We are going to be solving for the slope.

$m = \frac{ - 8 - 1}{5 - 2} = \frac{ - 9}{ \: \: \: 3} = - 3$

➋Second: We find the y-intercept.

$y = mx + b \\ 1 = - 3(2) + b \\ 1 = - 6 + b \\ \frac{ + 6 = + 6 \: \: \: \: \: \: \: \: \: }{7 = b}$

➌Third: Plug in.

$y = - 3x + 7$

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1 year ago

(1+tan^2 A)cot A/cosec^2 A= tanA

Rufina [12.5K]

Answer:

see explanation

Step-by-step explanation:

Using the trigonometric identities

1 + tan²A = sec²A , secA = $\frac{1}{cosA}$ , cosecA = $\frac{1}{sinA}$ , cotA = $\frac{cosA}{sinA}$ , tanA = $\frac{sinA}{cosA}$

Consider the left side

$\frac{(1+tan^2A)cotA}{cosec^2A}$

= $\frac{sec^2AcotA}{cosec^2A}$

= $\frac{\frac{1}{cos^2A}(\frac{cosA}{sinA}) }{cosec^2A}$

= $\frac{\frac{1}{cosAsinA} }{\frac{1}{sin^2A} }$

= $\frac{1}{cosAsinA}$ × sin²A

= $\frac{sinA}{cosA}$

= tanA = right side , thus proven

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2 years ago