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3 years ago

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, angle 2 =72°.

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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