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3 years ago

Pllzzz help I’ll mark brainliest

Mathematics

2 answers:

Viefleur [7K]3 years ago

7 0

ACB=26.............. :)

icang [17]3 years ago

3 0

Answer:

26°

Step-by-step explanation:

AC=tan 64

BC=x=sec 64

Using sine rule

(Sin 64)/tan 64 = (sin BAC)/sec 64

Cos 64 = (sin BAC)(cos 64)

Sin BAC= 1

BAC=90

ACB+ 64+90=180

ACB= 26°

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State the domain, the range, and the intervals on which function is increasing, decreasing, or constant in interval notation

aleksandr82 [10.1K]

Answer:

Domain: $[-1,\infty)$

Range: $(-\infty, 3]$

Decreases over: $(-1,-\infty)$

Step-by-step explanation:

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$f(x) = -\sqrt{x + 1} + 3$ --- Missing from the question

Solving (a): The domain

To get the domain, the expression under the square root must not be negative.

In other words:

$x + 1 \ge 0$

Solve for x

$x + 1 -1\ge 0-1$

$x \ge -1$

Hence, the domain is:

$[-1,\infty)$

To get the range, we plot the values of the domain in the expression.

$x = -1$

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$f(-1) = -\sqrt{-1+1} +3$

$f(-1) = -\sqrt{0} +3$

$f(-1) = -0 +3$

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$f(\infty) = -\sqrt{\infty}+3$

$f(\infty) = -\infty+3$

$f(\infty) = -\infty$

So, the range is: $(-\infty, 3]$

To get the interval where the function increases or not, we simply plot the graph of f(x).

See attachment for graph.

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