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3 years ago

7

Simplify negative 5 minus the square root of negative 44 negative 5 minus 4 times the square root of 11 i negative 5 minus 4 i t

imes the square root of 11 negative 5 minus 2 i times the square root of 11 negative 5 minus 2 times the square root of 11 i

1 answer:

Anika [276]3 years ago

6 0

Answer:

$-5-\sqrt{-44}=-5-2\sqrt{11}i$

Step-by-step explanation:

We want to simplify:

$-5-\sqrt{-44}$

We rewrite the expression under the radical sign to obtain:

$-5-\sqrt{4\times11\times -1}$

We split the expression under the radical sign to get:

$-5-\sqrt{4} \times \sqrt{11} \times \sqrt{-1}$

Recall that:

$\sqrt{-1}=i$

This implies that:

$-5-\sqrt{4} \times \sqrt{11} \times \sqrt{-1}=-5-2\sqrt{11}i$

Therefore $-5-\sqrt{-44}=-5-2\sqrt{11}i$

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[Will mark as brainliest]

Aleks04 [339]

The answer is: $f^{-1} = 4x^2-3,\quad\text{for } x \leq 0$

The inverse of a function $f(x)$ is another function, $f^{-1}(x)$ , with the following property:

$f(f^{-1}(x)) = f^{-1}(f(x)) = x$

In other words, the inverse of a function does exactly "the opposite" of what the original function does, and so if you compute them both in sequence you return to the starting point.

Think for example to a function that doubles the input, $f(x)=2x$ , and one that halves it: $f(x)= \frac{x}{2}$ . Their composition is clearly the identity function $f(x)=x$ , since you consider "twice the half of something", or "half the double of something".

In general, to invert a function $y=f(x)$ , you have to solve the expression for $x$ , writing an expression like $x = g(y)$ . If you manage to do so, then $g$ is the inverse of $f$ .

In your case, you have

$f(x) = y = -\frac{1}{2}\sqrt{x+3}$

Multiply both sides by $-2$ to get

$-2y = \sqrt{x+3}$

Square both sides to get

$4y^2 = x+3$

Finally, subtract 3 from both sides to get

$x = 4y^2 - 3$

Since the name of the variables doesn't really have a meaning, you can say that the inverse function is

$f^{-1}(x) = 4x^2 - 3$

As for the domain of the inverse function, remember what we said ad the beginning: if the original function goes from set A (domain) to set B (codomain), then the inverse function goes from set B (domain) to set A (codomain). This means that the inverse function is defined on an element in B if and only if that element belongs to the range of the original function, i.e. the set of the elements of the codomain $b \in B$ such that there exists $a \in A : f(a)=b$ . So, we need the range of $f(x)$ .

We know that the range of $g(x)=\sqrt{x}$ is $[0,\infty)$ . When you transform it to $g(x)=\sqrt{x+3}$ you simply translate the graph horizontally, so the range doesn't change. But when you multiply the function times $-\frac{1}{2}$ you affect both extrema of the range, turning it into $(-\infty,0]$ , which you can simply write as $x \leq 0$

4 0

3 years ago

Write an expression that can be used to multiply 6 × 198 mentally.

Mamont248 [21]

Answer:

assuming you meant 6x198 do a mental sum for 6x200 (1200) then subtract 6x2(12)to give you 1188

Step-by-step explanation:

brainly.com/question/274291

5 0

2 years ago

Read 2 more answers

Line 1 passes through the points (-4, -6) and (12,2). line 2 passes through the points (-6,-2) and (-2,0). Do the lines intersec

kifflom [539]

The lines dont intersect

7 0

3 years ago

For 50 Points!!!!!!!!!

dolphi86 [110]

Hello!

First you have to find what the temperature was at the 3 stages

It started at 87

Then went down 10

87 - 10 = 77

Then up 17

77 + 17 = 94

Then you take the difference from the current temperature from the lowest temperature

94 - 77 = 17

The answer is D

Hope this helps!

5 0

3 years ago

Read 2 more answers

An acute triangle has sides measuring 10 cm and 16 cm. The length of the third side is unknown.

saveliy_v [14]

Answer: Choice B

12.5 < x < 18.9

================================================

Explanation:

We have a triangle with these side lengths:

a = 10
b = 16
c = x = unknown

Let's assume that b = 16 is the largest side of this triangle.

By the converse of the pythagorean theorem, we need $b^2 < a^2+c^2$ to be true in order for an acute triangle to happen.

So,

$b^2 < a^2 + c^2\\\\c^2 > b^2 - a^2\\\\c > \sqrt{b^2-a^2}\\\\x > \sqrt{16^2-10^2}\\\\x > \sqrt{156}\\\\x > 12.4899959967968 \ \text{(approximate)}\\\\x > 12.5$

Now let's consider the possibility that the missing side x is actually the longest side.

Using the same theorem as before, we would say,

$c^2 < a^2 + b^2\\\\c < \sqrt{a^2 + b^2}\\\\x < \sqrt{10^2 + 16^2}\\\\x < \sqrt{356}\\\\x < 18.8679622641132 \ \text{(approximate)}\\\\x < 18.9\\\\$

We found that x > 12.5 and x < 18.9

This is the same as saying 12.5 < x and x < 18.9

Put together, they form the approximate answer of 12.5 < x < 18.9

6 0

3 years ago