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3 years ago

9

Need Answer ASAP! 30 POINTS !

2 answers:

vfiekz [6]3 years ago

5 0

Answer:

5

Step-by-step explanation:

Afina-wow [57]3 years ago

5 0

Answer:

5 -_-

Step-by-step explanation:

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If the exponential parent function,

laila [671]

Answer: $f(x) =3^{-x}-4$

Step-by-step explanation:

Some transformations for a function f(x) are shown below:

If $f(x)+k$ , the function is translated up "k" units.

If $f(x)-k$ , the function is translated down "k" units.

If $-f(x)$ , the function is reflected across the x-axis.

If $f(-x)$ , the function is reflected across the y-axis.

Therefore, knowing those transformations and given the exponential parent function:

$f(x) = 3^x$

If it is reflected across the y-axis and the it is translated down 4 units, we can determine that the resulting function is:

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

6 0

3 years ago

How to solve step by step [(7,5-1,5) x 10^-11]^2 ??

Oliga [24]

1.7,5-1,5=6
2.10-11=-1
3.6×-1=-6
4.-6+2=-4

5 0

3 years ago

Which statements about a square are always true?

blondinia [14]

All of the answers are correct except the last one

6 0

3 years ago

Read 2 more answers

A candy store sells 2 ounces of chocolate for $0.80 and 3 ounces of chocolate for $0.90. How much does the store charge per ounc

Arte-miy333 [17]

Divide the 2 ounces into the .80 which would be 0.40 per ounce for the 2 ounces chocolate and divide the 3 ounces into the .90 which would be 0.30 per ounce for the 3 ounces of chocolate

4 0

3 years ago

Given the following trigonometric ratio, enumerate the meaning ratio

Likurg_2 [28]

Answer:

The trigonometric ratios are presented below:

$\sin \theta = \frac{AC}{\sqrt{AC^{2} + BC^{2}}}$

$\cos \theta = \frac{BC}{\sqrt{AC^{2} + BC^{2}}}$

$\cot \theta = \frac{BC}{AC}$

$\sec \theta = \frac{\sqrt{AC^{2}+BC^{2}}}{BC}$

$\csc \theta = \frac{\sqrt{AC^{2}+BC^{2}}}{AC}$

Step-by-step explanation:

From Trigonometry we know the following definitions for each trigonometric ratio:

Sine

$\sin \theta = \frac{y}{h}$ (1)

Cosine

$\cos \theta = \frac{x}{h}$ (2)

Tangent

$\tan \theta = \frac{\sin \theta}{\cos \theta} = \frac{y}{x}$ (3)

Cotangent

$\cot \theta = \frac{\cos \theta}{\sin \theta} = \frac{x}{y}$ (4)

Secant

$\sec \theta = \frac{1}{\cos \theta} = \frac{h}{x}$ (5)

Cosecant

$\csc \theta = \frac{1}{\sin \theta} = \frac{h}{y}$ (6)

Where:

$x$ - Adjacent leg.

$y$ - Opposite leg.

$h$ - Hypotenuse.

The length of the hypotenuse is determined by the Pythagorean Theorem:

$h = \sqrt{x^{2}+y^{2}}$

If $y = AC$ and $x = BC$ , then the trigonometric ratios are presented below:

$\sin \theta = \frac{AC}{\sqrt{AC^{2} + BC^{2}}}$

$\cos \theta = \frac{BC}{\sqrt{AC^{2} + BC^{2}}}$

$\cot \theta = \frac{BC}{AC}$

$\sec \theta = \frac{\sqrt{AC^{2}+BC^{2}}}{BC}$

$\csc \theta = \frac{\sqrt{AC^{2}+BC^{2}}}{AC}$

6 0

3 years ago