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2 years ago

9

Find each unknown angle measurement.

1 answer:

Alex Ar [27]2 years ago

5 0

Answer:

angle(2) = 118 degree

angle(3) = 118 degree

angle(4) = 62 degree

Step-by-step explanation:

To find angle 2 :

$62 + angle{2} = 180 \\ angle{2} = 180 - 62 = 118 \: degree$

The rule (head to head):

angle(2) = angle(3) = 118 degree

And the last one is angle(4) which is the same thing:

angle(4) = angle(1) = 62 degree

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DENIUS [597]

Let BS be the event that the patient survives bypass surgery. Let H be the event that the heart damage will heal. Then P(BS) = 0.60, and also we have a conditional probability: GIVEN that the patient survives, the probability that the heart damage will heal is 0.5, that is P(H|BS) = 0.5 We want to know P(BS and H). Using the formula of the conditional probability: P(H and BS) = P(H|BS)·P(BS) = (0.6)(0.5) = 0.3

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3 years ago

What would x be equal to?

Vilka [71]

Answer:

x = $\frac{C-By}{A}$

Step-by-step explanation:

Given

Ax + By = C ( subtract By from both sides )

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3 years ago

PLZ help of u know the Answer for Sure.

anzhelika [568]

Answer:

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Step-by-step explanation:

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3 years ago

Will mark bianleast

Leni [432]

Answer:

(–5, 5)

Step-by-step explanation:

Start with the x-axis (left and right).

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Next is y-axis (up and down).

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3 years ago

Andre45 [30]

Answer:

$a_n=-3(3)^{n-1}$ ; {-3,-9, -27,- 81, -243, ...}

$a_n=-3(-3)^{n-1}$ ; {-3, 9,-27, 81, -243, ...}

$a_n=3(\frac{1}{2})^{n-1}$ ; {3, 1.5, 0.75, 0.375, 0.1875, ...}

$a_n=243(\frac{1}{3})^{n-1}$ ; {243, 81, 27, 9, 3, ...}

Step-by-step explanation:

The first explicit equation is

$a_n=-3(3)^{n-1}$

At n=1,

$a_1=-3(3)^{1-1}=-3$

At n=2,

$a_2=-3(3)^{2-1}=-9$

At n=3,

$a_3=-3(3)^{3-1}=-27$

Therefore, the geometric sequence is {-3,-9, -27,- 81, -243, ...}.

The second explicit equation is

$a_n=-3(-3)^{n-1}$

At n=1,

$a_1=-3(-3)^{1-1}=-3$

At n=2,

$a_2=-3(-3)^{2-1}=9$

At n=3,

$a_3=-3(-3)^{3-1}=-27$

Therefore, the geometric sequence is {-3, 9,-27, 81, -243, ...}.

The third explicit equation is

$a_n=3(\frac{1}{2})^{n-1}$

At n=1,

$a_1=3(\frac{1}{2})^{1-1}=3$

At n=2,

$a_2=3(\frac{1}{2})^{2-1}=1.5$

At n=3,

$a_3=3(\frac{1}{2})^{3-1}=0.75$

Therefore, the geometric sequence is {3, 1.5, 0.75, 0.375, 0.1875, ...}.

The fourth explicit equation is

$a_n=243(\frac{1}{3})^{n-1}$

At n=1,

$a_1=243(\frac{1}{3})^{1-1}=243$

At n=2,

$a_2=243(\frac{1}{3})^{2-1}=81$

At n=3,

$a_3=243(\frac{1}{3})^{3-1}=27$

Therefore, the geometric sequence is {243, 81, 27, 9, 3, ...}.

6 0

3 years ago