All categories

3 years ago

So i need help with this pls i suck at algebra

Mathematics

1 answer:

Ronch [10]3 years ago

5 0

Answer:

The 5x^2 vs -5x^2 will reflect over "X" axis

the +1 vs -2 will shift the graph down three units

the first answer is the correct answer

Step-by-step explanation:

You might be interested in

NEED HELP ASAP!!!!!!!!!!!

zhenek [66]

The equation $x=tan^{-1}(\frac{12}{5})$ can be used to find the measure of ∠BAC ⇒ 2nd answer

Step-by-step explanation:

Let us revise the trigonometry ratios in the right triangle ABC, where B is the right angle, AC is the hypotenuse, AB and BC are the legs of the triangle

The trigonometry ratios of the ∠BAC, the opposite side to this angle is BC and the adjacent side to it is AB are

$sin(BAC)=\frac{opposite}{hypotenuse}=\frac{BC}{AC}$
$cos(BAC)=\frac{adjacent}{hypotenuse}=\frac{AB}{AC}$
$tan(BAC)=\frac{opposite}{adjacent}=\frac{BC}{AB}$

In Δ ABC

∵ ∠ BCA is a right angle

∴ The hypotenuse is AB

∵ The adjacent side to ∠CAB is AC

∵ The opposite side to ∠CAB is BC

∵ AB = 13 units ⇒ hypotenuse

∵ CB = 12 units ⇒ opposite

∵ AC = 5 units ⇒ adjacent

- Let us find the trigonometry ratios of angle BAC

∵ m∠CAB is x

∵ $sin(x)=\frac{BC}{AB}$

∴ $sin(x)=\frac{12}{13}$

∴ $x=sin^{-1}(\frac{12}{13})$

∵ $cos(x)=\frac{AC}{AB}$

∴ $cos(x)=\frac{5}{13}$

∴ $x=cos^{-1}(\frac{5}{13})$

∵ $tan(x)=\frac{BC}{AC}$

∴ $tan(x)=\frac{12}{5}$

∴ $x=tan^{-1}(\frac{12}{5})$

The equation $x=tan^{-1}(\frac{12}{5})$ can be used to find the measure of ∠BAC

Learn more:

You can learn more about the trigonometry ratios in brainly.com/question/4924817

#LearnwithBrainly

4 0

3 years ago

Read 2 more answers

What is the equation of the line with a slope of 3/4 that goes through the point (4, 1) in slope-intercept form?

Svetach [21]

Answer:

$y=\frac{3}{4}x+-2$

Step-by-step explanation:

We do not have enough information for slope intercept form. But we can use the point-slope formula to find the information. The formula is $y -y_{1} =m(x -x_{1})$ where we substitute a point (x,y) for $(x_{1},y_{1})$ .