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maks197457 [2]
3 years ago
13

INTELLIGENT STUDENT IN TRIG?.Ill give brainlest​

Mathematics
1 answer:
faltersainse [42]3 years ago
8 0

Answers:

x = 6

y = 6*sqrt(2)

=======================

Explanation:

To find these values, you can use the general template of a 45-45-90 triangle or you can use trig ratios

tan(45) = 1

tan(angle) = opposite/adjacent

tan(45) = x/6

1 = x/6

6*1 = x

x = 6

-------

sin(45) = 1/sqrt(2)

sin(angle) = opposite/hypotenuse

sin(45) = 6/y

1/sqrt(2) = 6/y

1*y = sqrt(2)*6

y = 6*sqrt(2)

You might be interested in
Match each function with the corresponding function formula when h(x)=5-3x and g(x)=-3+5
Grace [21]

Answer:

k(x) = (3g + 5h)(x) ⇒ (1)

k(x) = (5h - 3g)(x) ⇒ (3)

k(x) = (h - g)(x) ⇒ (2)

k(x) = (g + h)(x) ⇒ (4)

k(x) = (5g + 3h)(x) ⇒ (5)

k(x) = (3h - 5g)(x) ⇒ (6)

Step-by-step explanation:

* To solve this problem we will substitute h(x) and g(x) in k(x) in the

  right column to find the corresponding function formula in the

  left column

∵ h(x) = 5 - 3x

∵ g(x) = -3^x + 5

- Lets start with the right column

# k(x) = (3g + 5h)(x)

∵ g(x) = -3^x + 5

∵ 3g(x) = 3[-3^x + 5] = [3 × -3^x + 3 × 5]

- Lets simplify 3 × -3^x

 take the negative out -(3 × 3^x), and use the rule a^n × a^m = a^(n+m)

∴ -3(3 × 3^x) = -(3^x+1)

∴ 3g(x) = -3^x+1 + 15

∵ h(x) = 5 - 3x

∵ 5h(x) = 5[5 - 3x] = [5 × 5 - 5 × 3x] = 25 - 15x

- Now substitute 3g(x) and 5h(x) in k(x)

∵ k(x) = (3g + 5h)(x)

∴ k(x) = -3^x+1 + 15 + 25 - 15x ⇒ simplify

∴ k(x) = 40 - 3^x+1 - 15x

∴ k(x) = 40 - 3^x+1 - 15x ⇒ k(x) = (3g + 5h)(x)

* k(x) = (3g + 5h)(x) ⇒ (1)

# k(x) = (5h - 3g)(x)

∵ 5h(x) = 25 - 15x

∵ 3g(x) = -3^x+1 + 15

∵ k(x) = (5h - 3g)(x)

∴ k(x) = 25 - 15x - (-3^x+1 + 15) = 25 -15x + 3^x+1 - 15 ⇒ simplify

∴ k(x) = 10 + 3^x+1 - 15x

∴ k(x) = 10 + 3^x+1 - 15x ⇒ k(x) = (5h - 3g)(x)

* k(x) = (5h - 3g)(x) ⇒ (3)

# k(x) = (h - g)(x)

∵ h(x) = 5 - 3x

∵ g(x) = -3^x + 5

∵ k(x) = (h - g)(x)

∴ k(x) = 5 - 3x - (-3^x + 5) = 5 - 3x + 3^x - 5 ⇒ simplify

∴ k(x) = 3^x - 3x

∴ k(x)= 3^x - 3x ⇒ k(x) = (h - g)(x)

* k(x) = (h - g)(x) ⇒ (2)

# k(x) = (g + h)(x)

∵ h(x) = 5 - 3x

∵ g(x) = -3^x + 5

∵ k(x) = (g + h)(x)

∴ k(x) = -3^x + 5 + 5 - 3x ⇒ simplify

∴ k(x) = 10 - 3^x - 3x

∴ k(x)= 10 - 3^x - 3x ⇒ k(x) = (g + h)(x)

* k(x) = (g + h)(x) ⇒ (4)

# k(x) = (5g + 3h)(x)

∵ g(x) = -3^x + 5

∵ 5g(x) = 5[-3^x + 5] = [5 × -3^x + 5 × 5] = 5(-3^x) + 25

∴ 5g(x) = -5(3^x) + 25

∵ h(x) = 5 - 3x

∵ 3h(x) = 3[5 - 3x] = [3 × 5 - 3 × 3x] = 15 - 9x

- Now substitute 5g(x) and 3h(x) in k(x)

∵ k(x) = (5g + 3h)(x)

∴ k(x) = -5(3^x) + 25 + 15 - 9x ⇒ simplify

∴ k(x) = 40 - 5(3^x) - 9x

∴ k(x) = 40 - 5(3^x) - 9x ⇒ k(x) = (5g + 3h)(x)

* k(x) = (5g + 3h)(x) ⇒ (5)

# k(x) = (3h - 5g)(x)

∵ 3h(x) = 15 - 9x

∵ 5g(x) = -5(3^x) + 25

∵ k(x) = (3h - 5g)(x)

∴ k(x) = 15 - 9x - [-5(3^x) + 25] = 15 - 9x + 5(3^x) - 25 ⇒ simplify

∴ k(x) = 5(3^x) - 9x - 10

∴ k(x) = 5(3^x) - 9x - 10 ⇒ k(x) = (3h - 5g)(x)

* k(x) = (3h - 5g)(x) ⇒ (6)

4 0
3 years ago
Determine whether each exponential equation is a decay or a growth.
kolezko [41]

Answer:

OHHH THE ANSWER IS 1 would be pie 7 and the last one is .9 ok thanks for 15 points

Step-by-step explanation:

STEP 1: GO GET A CALULATOR

STEP 2 GO F UR STEP SIS

5 0
3 years ago
Show your work<br> −17x = −204
pishuonlain [190]
You would divide -204 divide it by -17 and you would get 12
X=12
6 0
3 years ago
Determine a and k so the given points are on the graph of the function.
pshichka [43]

Step-by-step explanation:

Explanation:

Because

(2,6) is in the function, we can plug

6 as y

and

2 as x.

From that we get

6=a(2−1)2+k

6=a+k

Same for the next point:

12=a(3−1)2+k

12=22a+k

12=4a+k

Now we have this system of equations to solve:

a+k=6

4a+k=12

Solving it leads to the conclusion that

a=2,k=4 .

3 0
3 years ago
If p = 2 and q = 8, the value of p3 + V64 q ls____​
gavmur [86]

Answer:

518

Step-by-step explanation:

P = 2

Q = 8

p3 + 64q

Multiply "p" (which is 2) with 3

Multiply "q" (which is 8) with 8

Add the two numbers you get

And you have you answer

I hope this helps. :)

4 0
3 years ago
Read 2 more answers
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