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jasenka [17]
3 years ago
14

PLEASEEE I DONT KNOW HOW TO DO ITTTT. THE 35.8 IS RIGHT BUT I CANT GET MEASURE OF ANGLE C AND A I GOT 46.2 for A AND 70.8 for C

and it’s WRONG. PLEASE HELP!

Mathematics
1 answer:
Rashid [163]3 years ago
3 0

Answer:

Hope it helps....!!!!!

Step-by-step explanation:

AB = c = 38

BC = a = 29

AC = b

Angle ABC = 63 degrees

Solving for AC   "b":

Cosine rule: c^2 = a^2 * b^2 -2ab * cos C

38^2 = 29^2 * b^2 - (2* 29) * b * (cos 38)

1444 = 841 * b^2 - 58 * b * 0.955

(1444 + 58)/0.955 = b^2 * b

1572.77486911 = b^3

11.62935 = b

11.63 = b (rounded to two decimal places)

Now solving for angle A:

Sine rule: a/sinA = b/sinB

29/sinA = 11.63/sin(63)

sinA/29 = sin(63)/11.63

sin A = (sin(63)/11.63) * 29

sin A = 0.41731

A = sin^-1 (0.41731)

A = 24 degrees 39 minutes 53 seconds

Now solving for angle C:

Sine rule: c/sinC = b/sinB

38/sinC = 11.63/sin(63)

sinC/38 = sin(63)/11.63

sin C = (sin(63)/11.63) * 38

sin C = 0.54682

C = sin^-1 (0.54682)

C = 33 degrees 8 minutes 56 seconds

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A shipping container contains seven complex electronic systems. Unknown to the purchaser, two are defective. Two of the seven ar
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     \large\boxed{\large\boxed{10/21}}

Explanation:

There are only two mutually exclusive possibilities for each complex electronic system: being <em>defective</em> or being <em>not defective</em>:

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  • Nmber of defective systems: 2
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Please help!
crimeas [40]

Answer:

The zeros are:

x =4, x=2, x = 5

  • The function has three distinct real zeros.

Hence, option (B) is true.

Step-by-step explanation:

Given the expression

h\left(x\right)=\left(x-4\right)^2\left(x^2-7x+\:10\right)

Let us determine the zeros of the function by putting h(x) = 0 and solving the expression

0=\left(x-4\right)^2\left(x^2-7x+10\right)

switch sides

\left(x-4\right)^2\left(x^2-7x+10\right)=0

as

x^2-7x+\:10=\left(x-2\right)\left(x-5\right)

so

\left(x-4\right)^2\left(x-2\right)\left(x-5\right)=0

Using the zero factor principle

  • \mathrm{If}\:ab=0\:\mathrm{then}\:a=0\:\mathrm{or}\:b=0\:\left(\mathrm{or\:both}\:a=0\:\mathrm{and}\:b=0\right)

so

x-4=0\quad \mathrm{or}\quad \:x-2=0\quad \mathrm{or}\quad \:x-5=0

x =4, x=2, x = 5

Thus, the zeros are:

x =4, x=2, x = 5

It is clear that there are three zeros and all the zeros are distinct real numbers.

Therefore,

  • The function has three distinct real zeros.

Hence, option (B) is true.

4 0
3 years ago
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