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2 years ago

What is important when learning about trig?

2 answers:

7 0

Before beginning trigonometry, you need be comfortable with algebra and geometry. You should be comfortable handling algebraic expressions and solving problems after studying algebra. You should know about similar triangles, the Pythagorean theorem, and a few other things from geometry, but not much else. Students can use trigonometry to calculate the precise angle of a triangle's sides, the distance between distinct points on a triangle, and other information that is useful in a number of situations. Trigonometry is an essential component of ICSE Class 10 Mathematics since it combines memorization, conceptual knowledge, and problem-solving abilities. Because many of the earth's natural formations resemble triangles, it aids children in their comprehension of the globe.

Hope this helps

~ ROR

frutty [35]2 years ago

3 0

Answer:

-You should already be familiar with algebra and geometry before learning trigonometry. From algebra, you should be comfortable with manipulating algebraic expressions and solving equations. From geometry, you should know about similar triangles, the Pythagorean theorem, and a few other things, but not a great deal.

-Measure the lengths of the sides of sets of similar right angled triangles and find the ratio of sides.

-Investigate the relationship between these ratios and the angle size.

-Use calculators or tables to find the sine, cosine and tangent of angles.

Step-by-step explanation:

basic knowledge i guess

You might be interested in

Solving Rational Inequalities and use sign diagram to sketch the graph. Image attached for better understanding.

hjlf

Answer:

x ∈ (-∞ , -2) ∪ (1, 3)

Step-by-step explanation:

The expression is already factored. Note that for the polynomial that appears in the numerator $(x-3)(x + 2)$ there are 2 roots:

$x = 3\\x = -2$

For the polynomial that appears in the denominator there is 1 root:

$x = 1$

Note that $x = 1$ does not belong to the domain of f(x) because it zeroes the denominator of the function and the division between zero is not defined.

With these three roots we do the study of signs to find out when $f(x)$

Observe the attached image

Note that:

$(x-3)$ when $x$

$(x + 2)$ when $x$

$(x-1)$ when $x$

Finally, we have the solution:

x ∈ (-∞ , -2) ∪ (1, 3)

5 0

3 years ago

How to solve this problem

Agata [3.3K]

So first, you have to get two of the same variables to cancel out. Let's do this for x. In order for the x's to cancel out, we could multiply the bottom problem by 2.
(2) 3x-6y=24
After multiplying all the numbers by 2, you get the equation 6x-12y=48
The set of equations is now
-6x+2y=12
6x-12y=48
Now you can add them. The x variables cancel out, so you are left with the y variable.
2y+-12y=-10y and 12+48=60
Then you would divide 60 by -10 to get y=-6.
You would plug the answer for y into one of the original equations, lets do the top one. -6x+2y=12 becomes -6x+2(-6)=12
You'd multiply the 2 and -6 to get -12 so the equation is
-6x-12=12
The negative 12 turn positive and you add to both sides to get the -6x alone.
-6x-12=12
+12=12
-6x=24
Then divide 24 by -6
X=4

(-4,-6) is your final answer.

7 0

2 years ago

What value will make the equation true? <br> - 0.8 - ? = - 1 ½

alekssr [168]

Answer:

$the \: value \: of \: x = 0.7$

8 0

3 years ago

Help me ahhhhhhhhhhhhhhhhhh

shepuryov [24]

I got m 7/3 I think that's the answer

7 0

3 years ago

(gh)^0 I'm very confused can you show me how to solve this

lana [24]

The expression is:

$(9d^{10})^{-2}$

Simplify the expression:

1. Multiply the exponent outside the parenthesis with every exponent inside the parenthesis,

$\begin{gathered} (9d^{10})^{-2}=9^2d^{(-10)\times2} \\ (9d^{10})^{-2}=9^{-2}d^{-20} \end{gathered}$

Since, exponents in negative form are express as fraction with 1 as numerator.

$\begin{gathered} \mleft(9d^{\mleft\{10\mright\}}\mright)^{\mleft\{-2\mright\}}=9^{-2}d^{\mleft\{-20\mright\}} \\ \mleft(9d^{\mleft\{10\mright\}}\mright)^{\mleft\{-2\mright\}}=\frac{1}{9^2}\times d^{\mleft\{-20\mright\}} \\ \mleft(9d^{\mleft\{10\mright\}}\mright)^{\mleft\{-2\mright\}}=\frac{1}{81}^{}d^{\mleft\{-20\mright\}} \end{gathered}$

Answer: 1/81d^(-20)

3 0

1 year ago