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k0ka [10]
3 years ago
11

1)

Mathematics
1 answer:
MrRissso [65]3 years ago
7 0

Answer:

Step-by-step explanation:

The diagram of the triangles are shown in the attached photo.

1) Looking at ∆AOL, to determine AL, we would apply the sine rule

a/SinA = b/SinB = c/SinC

21/Sin25 = AL/Sin 105

21Sin105 = ALSin25

21 × 0.9659 = 0.4226AL

AL = 20.2839/0.4226

AL = 50

Looking at ∆KAL,

AL/Sin55 = KL/Sin100

50/0.8192 = KL/0.9848

50 × 0.9848 = KL × 0.8192

KL = 49.24/0.8192

KL = 60

AK/Sin25 = AL/Sin 55

AKSin55 = ALSin25

AK × 0.8192 = 0.4226 × 50

AK = 21.13/0.8192

AK = 25.8

2) looking at ∆AOC,

Sin 18 = AD/AC = 18/AC

AC = 18/Sin18 = 18/0.3090

AC = 58.25

Sin 85 = AD/AB = 18/AB

AB = 18/Sin85 = 18/0.9962

AB = 18.1

To determine BC, we would apply Sine rule.

BC/Sin77 = 58.25/Sin85

BCSin85 = 58.25Sin77

BC = 58.25Sin77/Sin85

BC = 58.25 × 0.9744/0.9962

BC = 56.98

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Zina [86]

Answer:

y = \frac{1}{2}x + 89

Step-by-step explanation:

slope-intercept form: y  = mx + b

y = (x , y)

x = (x , y)

m = slope = 1/2

b = y-intercept = 89

Plug in the corresponding numbers & variables to the corresponding variables:

y = (1/2)x + 89

y = \frac{1}{2}x + 89  is your answer.

~

5 0
3 years ago
Read 2 more answers
f is an even function. a = A 2-column table with 4 rows. Column 1 is labeled x with entries negative 2, 0, 2, 3. Column 2 is lab
Lady_Fox [76]

The values of a and b are 4 and -3

<h3>How to solve for (a) and (b)?</h3>

To solve for a, we make use of the function f(x).

x    f(x)

2     4

0     5

2     a

3     7

Remove the x values 0 and 3

x    f(x)

2     4

2     a

The above table implies that:

f(2) = 4 and f(2) = a

Substitute 4 for f(2) in f(2) = a

4 = a

Rewrite as:

a = 4

To solve for b, we make use of the function g(x).

x    g(x)

2     b

0     0

2     -3

3     -4

Remove the x values 0 and 3

x    g(x)

2     b

2     -3

The above table implies that:

f(2) = b and f(2) = -3

Substitute -3 for f(2) in f(2) = b

-3 = b

Rewrite as:

b = -3

Hence, the values of a and b are 4 and -3

Read more about functions at:

brainly.com/question/13153717

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<u>Complete question</u>

Use the Symmetry of a Function to Find Coordinates

f is an even function             g is an odd function

x    f(x)                                        x             g(x)

2     4                                          2              b

0     5                                          0              0

2     a                                          2              -3

3     7                                          3              -4

Find a and b

4 0
2 years ago
Obtain the total salary?)
Nonamiya [84]

Answer:

F(d) = 30 + 0.50d

Step-by-step explanation:

Given

Charges = P8.00 ---- first 4 km

Additional = P0.50

Required

Write a function to address the scenario.

Represent the whole distance covered with d.

First,we need to determine the total charges for the first four hours.

Charges = 8.00 * 4

Charges = 32.00

Next, we determine the charges for additional distance.

Charges = 0.50 * (d - 4)

d - 4 is the remaining distance after the first 4.

Charges = 0.50d - 2

The function is then written as;

F(d) = 32 + 0.50d - 2

F(d) = 32 - 2 + 0.50d

F(d) = 30 + 0.50d

8 0
3 years ago
If m ACB = 48° and <br> m CAB = 47°, is line BC a tangent to circle A?
muminat

Answer: No, because ABC is not a right angle.


Step-by-step explanation: In triangle ABC, the three angles add up to 180

So angle B = 180 - 48 - 47 = 85.


7 0
3 years ago
D^2(y)/(dx^2)-16*k*y=9.6e^(4x) + 30e^x
MA_775_DIABLO [31]
The solution depends on the value of k. To make things simple, assume k>0. The homogeneous part of the equation is

\dfrac{\mathrm d^2y}{\mathrm dx^2}-16ky=0

and has characteristic equation

r^2-16k=0\implies r=\pm4\sqrt k

which admits the characteristic solution y_c=C_1e^{-4\sqrt kx}+C_2e^{4\sqrt kx}.

For the solution to the nonhomogeneous equation, a reasonable guess for the particular solution might be y_p=ae^{4x}+be^x. Then

\dfrac{\mathrm d^2y_p}{\mathrm dx^2}=16ae^{4x}+be^x

So you have

16ae^{4x}+be^x-16k(ae^{4x}+be^x)=9.6e^{4x}+30e^x
(16a-16ka)e^{4x}+(b-16kb)e^x=9.6e^{4x}+30e^x

This means

16a(1-k)=9.6\implies a=\dfrac3{5(1-k)}
b(1-16k)=30\implies b=\dfrac{30}{1-16k}

and so the general solution would be

y=C_1e^{-4\sqrt kx}+C_2e^{4\sqrt kx}+\dfrac3{5(1-k)}e^{4x}+\dfrac{30}{1-16k}e^x
8 0
3 years ago
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