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irina1246 [14]
3 years ago
15

Triangle PQR is a right triangle. If PQ = 8 , what is RQ?

Mathematics
2 answers:
slavikrds [6]3 years ago
8 0
The answer to this is B
seraphim [82]3 years ago
8 0

Answer:

The value of the RQ is 16 unit .

Option (B) is correct .

Step-by-step explanation:

Now by using the trignometric identity

cos \theta = \frac{Base}{Hypotenuse}

Base = RQ

Hypotenuse = PQ = 8 unit

\theta = 60^{\circ}

cos60^{\circ} = \frac{RQ}{PQ}

cos60^{\circ} = \frac{RQ}{8}

cos\ 60^{\circ} = \frac{1}{2}

\frac{8}{2}=RQ

RQ = 4

     = 4 unit

Therefore the value of the RQ is 4 unit .

Option (B) is correct .

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If you're good at logarithms please help me with question two d and h thanks
Alik [6]

Answer:

d. -0.301

h. 1.38

Step-by-step explanation:

Loga0.5 = loga2^-1

= -1(loga2)

       = -1 (0.301)

= -0.301

Loga24 = loga(2^3 x 3)

= loga2^3 + loga3

= 3 (loga2) + loga3

       = 3 (0.301) + (0.477)

       = 1.38

5 0
3 years ago
Find the value of kk for which the constant function x(t)=kx(t)=k is a solution of the differential equation 5t3dxdt+2x−2=05t3dx
uranmaximum [27]
Given the differential equation

5t^3 \frac{dx}{dt} +2x-2=0

The solution is as follows:

5t^3 \frac{dx}{dt} +2x-2=0 \\  \\ \Rightarrow5t^3 \frac{dx}{dt} =2-2x \\  \\ \Rightarrow \frac{5}{2-2x} dx= \frac{1}{t^3} dt \\  \\ \Rightarrow \int {\frac{5}{2-2x} } \, dx = \int {\frac{1}{t^3}} \, dt \\  \\ \Rightarrow- \frac{5}{2} \ln(2-2x)=- \frac{1}{2t^2} +A \\  \\ \Rightarrow\ln(2-2x)= \frac{1}{5t^2} +B\\  \\ \Rightarrow2-2x=Ce^{\frac{1}{5t^2}} \\  \\ \Rightarrow 2x=Ce^{\frac{1}{5t^2}}+2 \\  \\ \Rightarrow x=De^{\frac{1}{5t^2}}+1
3 0
3 years ago
Two lines that have the same y intercept and different slopes intersect at exactly one point.
jek_recluse [69]

Answer:

False. They will be parallel to each other

6 0
3 years ago
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vova2212 [387]
In order to see the formulas that you used in a spreadsheet, but doesn't want to see the results of the formulas, then just click on Show Formulas. That can be done by clicking the cell and click, show formula. Letter A.
5 0
3 years ago
Calculate limits x&gt;-infinity<br> -2x^5-3x+1
Lera25 [3.4K]

Given:

The limit problem is:

\lim_{x\to -\infty}(-2x^5-3x+1)

To find:

The value of the given limit problem.

Solution:

We have,

\lim_{x\to -\infty}(-2x^5-3x+1)

In the function -2x^5-3x+1, the degree of the polynomial is 5, which is an odd number and the leading coefficient is -2, which is a negative number.

So, the function approaches to positive infinity as x approaches to negative infinity.

\lim_{x\to -\infty}(-2x^5-3x+1)=\infty

Therefore, \lim_{x\to -\infty}(-2x^5-3x+1)=\infty.

3 0
2 years ago
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