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kumpel [21]
3 years ago
12

Solve a triangle when A=40°, B=59°, and c=14.

Mathematics
1 answer:
RideAnS [48]3 years ago
5 0

Answer:

The answer is (c) ⇒ C = 81° , a = 9.1 , b = 12.1

Step-by-step explanation:

∵ m∠A = 40° , m∠B = 59

∴ m∠C = 180 - (40 + 59) = 81

∵ c = 14

By using sin rule

∵ \frac{14}{sin81}=\frac{a}{sin40}

∴ a=\frac{14sin40}{sin81}=9.1

∵ \frac{14}{sin81}=\frac{b}{sin59}

∴ b=\frac{14sin59}{sin81}=12.1

∴ m∠C = 81° , a = 9.1 , b = 12.1

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3 years ago
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Alika [10]
This seems like a concept you’re going to learn and use so I’ll do a problem per section and explain.

Well basically this is a concept: if it’s a negative exponent it’s just 1/number^exponent.

So for example:
#1. 1/(10)^2 = 10^-2

The standard notation:
#9: You know that 1/(10)^3 you could either put it in a calculator or realize that you take the number 1, and move the decimal to the left three times. 0.001 would be the answer.

Scientific notation:
combines these topics. Let’s take #17. 0.025. Scientific notation means you write it as a number multiplied by 10^? So let’s see how many places you can move the decimal to get a number without the zeros.

0.025 => move it two places to the right. So that’s 2.5. Now 2.5 multiplied by what 10^? Would give you 0.025? It would be 10^-2. So your answer would be 2.5 x 10^-2.

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3 years ago
Consider the equation x2+4x+9=0 in standard form. Which equation shows the coefficients a, b, and c correctly substituted into t
Afina-wow [57]

Answer:

<h2>x = -2+i√5 and  -2i-√5</h2>

Step-by-step explanation:

The general form of a quadratic equation is ax²+bx+c = 0

Given the quadratic equation x²+4x+9=0 in its standard form, on comparing with the general equation we can get the value of the constant a, b and c as shown;

ax² = x²

a = 1

bx = 4x

b = 4

c = 9

The quadratic formula is given as x = -b±√(b²-4ac)/2a

Substituting the constant;

x = -4±√(4²-4(1)(9))/2(1)

x = -4 ±√(16-36)/2

x = -4±√-20/2

x = -4±(√-1*√20)/2

Note that √-1 = i

x = -4±(i√4*5)/2

x = (-4±i2√5)/2

x = -4/2±i2√5/2

x = -2±i√5

The solution to the quadratic equation are  -2+i√5 and  -2i-√5

5 0
3 years ago
Hiiii.. please help me with this limit question ​
Alenkasestr [34]

Answer:

π

Step-by-step explanation:

Solving without L'Hopital's rule:

lim(x→0) sin(π cos²x) / x²

Use Pythagorean identity:

lim(x→0) sin(π (1 − sin²x)) / x²

lim(x→0) sin(π − π sin²x) / x²

Use angle difference formula:

lim(x→0) [ sin(π) cos(-π sin²x) − cos(π) sin(-π sin²x) ] / x²

lim(x→0) -sin(-π sin²x) / x²

Use angle reflection formula:

lim(x→0) sin(π sin²x) / x²

Now we multiply by π sin²x / π sin²x.

lim(x→0) [ sin(π sin²x) / x² ] (π sin²x / π sin²x)

lim(x→0) [ sin(π sin²x) / π sin²x] (π sin²x / x²)

lim(x→0) [ sin(π sin²x) / π sin²x] lim(x→0) (π sin²x / x²)

π lim(x→0) [ sin(π sin²x) / π sin²x] [lim(x→0) (sin x / x)]²

Use identity lim(u→0) (sin u / u) = 1.

π (1) (1)²

π

Solving with L'Hopital's rule:

If we plug in x = 0, the limit evaluates to 0/0.  So using L'Hopital's rule:

lim(x→0) [ cos(π cos²x) (-2π cos x sin x) ] / 2x

lim(x→0) [ -π cos(π cos²x) sin(2x) ] / 2x

-π/2 lim(x→0) [ cos(π cos²x) sin(2x) ] / x

Again, the limit evaluates to 0/0.  So using L'Hopital's rule one more time:

-π/2 lim(x→0) [ cos(π cos²x) (2 cos(2x)) + (-sin(π cos²x) (-2π cos x sin x)) sin(2x) ] / 1

-π/2 lim(x→0) [ 2 cos(π cos²x) cos(2x) + π sin(π cos²x) sin²(2x) ]

-π/2 (-2)

π

8 0
3 years ago
Inequality’s of (0,3) (2,-3)
Inga [223]

Good question next quetionn

5 0
3 years ago
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