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antoniya [11.8K]
3 years ago
12

FIND THE MEASURE OF ANGLE 2 PLEASE HELP ASAP!

Mathematics
1 answer:
Bond [772]3 years ago
4 0

Answer:

We know that the figure has a symmetry

angle 1 = angle 2

angle 1 =180°-(18°+90°) = 180° - 108°=72°

Therefore, <em><u>angle 2 =72°</u></em><em><u>.</u></em>

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Need to know angle 1, angle 2, angle 3
wolverine [178]

Answer: Step-by-step explanation:

angle 1=41°

angle 2=angle 1=41°

angle 3=180-2×41=98

4 0
2 years ago
Another equation for 32x - 12.8
Law Incorporation [45]
You can use a calculator and put is where I’d can tell you the answer
8 0
3 years ago
HEEEEEEELLLLLLLPPPPPPPPPP PPPPPLLLLLLSSSSS
ladessa [460]

Answer:

d. -19, -3.7, -3/4, 6.05

Step-by-step explanation:

Hope this helps mate (´▽`ʃ♡ƪ)

7 0
1 year ago
Read 2 more answers
For A I somehow got 41.05 but its 2.95cm can someone show me the working and what I did wrong?
777dan777 [17]

Answer:

Step-by-step explanation:

a). tan(75°) = \frac{\text{Opposite side}}{\text{Adjacent side}}

                  = \frac{11}{k}

    k = \frac{11}{\text{tan}(75)}

    k = 2.947

    k = 2.95 cm

b). cos(52°) = \frac{16}{s}

    s = \frac{16}{\text{cos}(52)}

    s = 25.988

    s ≈ 25.99 cm

c). sin(5°) = \frac{\text{Opposite side}}{\text{Hypotenuse}}

               = \frac{16.1}{q}

    q = \frac{16.1}{\text{sin}(5)}

    q = 184.727

    q ≈ 184.73 cm

4 0
2 years ago
According to data from a medical​ association, the rate of change in the number of hospital outpatient​ visits, in​ millions, in
GaryK [48]

Answer:

a) f(t)=0.001155[\frac{2}{3}t(t-1980)^{3/2}-\frac{4}{15}(t-1980)^{5/2}]+264,034,000

b) f(t=2015) = 264,034,317.7

Step-by-step explanation:

The rate of change in the number of hospital outpatient​ visits, in​ millions, is given by:

f'(t)=0.001155t(t-1980)^{0.5}

a) To find the function f(t) you integrate f(t):

\int \frac{df(t)}{dt}dt=f(t)=\int [0.001155t(t-1980)^{0.5}]dt

To solve the integral you use:

\int udv=uv-\int vdu\\\\u=t\\\\du=dt\\\\dv=(t-1980)^{1/2}dt\\\\v=\frac{2}{3}(t-1980)^{3/2}

Next, you replace in the integral:

\int t(t-1980)^{1/2}=t(\frac{2}{3}(t-1980)^{3/2})- \frac{2}{3}\int(t-1980)^{3/2}dt\\\\= \frac{2}{3}t(t-1980)^{3/2}-\frac{4}{15}(t-1980)^{5/2}+C

Then, the function f(t) is:

f(t)=0.001155[\frac{2}{3}t(t-1980)^{3/2}-\frac{4}{15}(t-1980)^{5/2}]+C'

The value of C' is deduced by the information of the exercise. For t=0 there were 264,034,000 outpatient​ visits.

Hence C' = 264,034,000

The function is:

f(t)=0.001155[\frac{2}{3}t(t-1980)^{3/2}-\frac{4}{15}(t-1980)^{5/2}]+264,034,000

b) For t = 2015 you have:

f(t=2015)=0.001155[\frac{2}{3}(2015)(2015-1980)^{1/2}-\frac{4}{15}(2015-1980)^{5/2}]+264,034,000\\\\f(t=2015)=264,034,317.7

3 0
3 years ago
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