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enyata [817]
3 years ago
8

Which ordered pair is in the solution set of 3x − 2y ≥ -1?​​​

Mathematics
1 answer:
andrew-mc [135]3 years ago
4 0

Answer:

G

Step-by-step explanation:

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The family ate 1/4 of banana bread and 3/4 of blueberry bread. What fraction of that bread was left over
Sidana [21]
1/4 +3/4=4/4 bread was left over.
neither 1/4-3/4 which is =-1/2
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Which of these points lies on a circle centered at A(3, 3) and passing through B(6, 5)? A. C(1, 6) B. D(6, 0) C. E(0, 3) D. F(3,
sleet_krkn [62]
 we are given with a circle with the center at (3,3) and one that passes through (6,5). The equation of a circle is (x-h)^2 + (y-k)^2 = r^2. In this case h = 3 and k = 3. Substituting, h and k and x = 6 and y = 5, r^2 is obtained to be 13. Among these points, A. point (1,6) fits to the equation. 
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Darlene earns $12.10/h. She works for 33 h one week. How much did she earn?
marusya05 [52]

Answer:

399.30

Step-by-step explanation:

12.10 times 33 hours equals 399.30

6 0
3 years ago
Read 2 more answers
Iliana performed an experiment by spinning a spinner a set number of times and noting the color on which the spinner landed.
tatiyna
An experimental probability is given by the number of observations divided by the number of trials.

From the given table it can be seen that the color with the highest frequency is red with 9 observations while that number of trials is given by the sum of the individual observations, i.e 7 + 9 + 6 + 8 = 30

Therefore, the experimental probability of the highest frequeny is given by 9 / 30 = 0.3
7 0
4 years ago
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(1 point) Find the length traced out along the parametric curve x=cos(cos(4t))x=cos⁡(cos⁡(4t)), y=sin(cos(4t))y=sin⁡(cos⁡(4t)) a
Mazyrski [523]

The length of a curve C given parametrically by (x(t),y(t)) over some domain t\in[a,b] is

\displaystyle\int_C\mathrm ds=\int_a^b\sqrt{\left(\frac{\mathrm dx}{\mathrm dt}\right)^2+\left(\frac{\mathrm dy}{\mathrm dt}\right)^2}\,\mathrm dt

In this case,

x(t)=\cos(\cos4t)\implies\dfrac{\mathrm dx}{\mathrm dt}=-\sin(\cos4t)(-\sin4t)(4)=4\sin4t\sin(\cos4t)

y(t)=\sin(\cos4t)\implies\dfrac{\mathrm dy}{\mathrm dt}=\cos(\cos4t)(-\sin4t)(4)=-4\sin4t\cos(\cos4t)

So we have

\displaystyle\left(\frac{\mathrm dx}{\mathrm dt}\right)^2+\left(\frac{\mathrm dy}{\mathrm dt}\right)^2=16\sin^24t\sin^2(\cos4t)+16\sin^24t\cos^2(\cos4t)=16\sin^24t

and the arc length is

\displaystyle\int_0^1\sqrt{16\sin^24t}\,\mathrm dt=4\int_0^1|\sin4t|\,\mathrm dt

We have

\sin(4t)=0\implies4t=n\pi\implies t=\dfrac{n\pi}4

where n is any integer; this tells us \sin(4t)\ge0 on the interval \left[0,\frac\pi4\right] and \sin(4t) on \left[\frac\pi4,1\right]. So the arc length is

=\displaystyle4\left(\int_0^{\pi/4}\sin4t\,\mathrm dt-\int_{\pi/4}^1\sin4t\,\mathrm dt\right)

=-\cos(4t)\bigg_0^{\pi/4}-\left(-\cos(4t)\bigg_{\pi/4}^1\right)

=(\cos0-\cos\pi)+(\cos4-\cos\pi)=\boxed{3+\cos4}

7 0
3 years ago
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