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Katarina [22]
2 years ago
10

What is important when learning about trig?

Mathematics
2 answers:
vlabodo [156]2 years ago
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

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For f(x) = square root of x, estimate f(17) to the nearest integer.
Tomtit [17]

f(x) = sqrt(x)

f(17) = sqrt(17)

it is close to sqrt(16)=4

so to the nearest integer

f(17) = 4

4 0
3 years ago
Ravi needs to buy some pencils. Brand A has 36 pencils for $8.52. Brand B has a pack of 48 pencils for $9.98. Find the unit pric
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What fraction of a day is 14 hours
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7 0
3 years ago
Read 2 more answers
2v^2-12 =-12v <br> Would really appreciate it loves❤️
Black_prince [1.1K]

For this case we have the following equation:

2v ^ 2-12 = -12v

Rewriting we have:

2v ^ 2 + 12v-12 = 0

Dividing by 2 to both sides of the equation:

v ^ 2 + 6v-6 = 0

We apply the quadratic formula:

x = \frac {-b \pm \sqrt {b ^ 2-4 (a) (c)}} {2a}

We have to:

a = 1\\b = 6\\c = -6

Substituting:

x = \frac {-6 \pm \sqrt {6 ^ 2-4 (1) (- 6)}} {2 (1)}\\x = \frac {-6 \pm \sqrt {36 + 24}} {2}\\x = \frac {-6 \pm \sqrt {60}} {2}\\x = \frac {-6 \pm \sqrt {4 * 15}} {2}\\x = \frac {-6 \pm2 \sqrt {15}} {2}

Thus, we have two roots:

x_ {1} = - 3+ \sqrt {15}\\x_ {2} = - 3- \sqrt {15}

ANswer:

x_ {1} = - 3+ \sqrt {15}\\x_ {2} = - 3- \sqrt {15}

7 0
3 years ago
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