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Liula [17]
3 years ago
11

What is the value of X A 15 B 21 C 26 D 105

Mathematics
2 answers:
kaheart [24]3 years ago
7 0

Answer:

B)21

Step-by-step explanation:

∠KN=∠ML

∠KN=90°+15°

=105°

∠ML=105°

5x°=105°

x°=105÷5

x°=21°

Anni [7]3 years ago
6 0

Answer:

5x=90+15

Add, 90 + 15 = 105

5x=105

Divide both sides by 5

\frac{5x}{5}=\frac{105}{5}

x=21°

\textbf{OAmalOHopeO}

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An auto dealership sells minivans and sedans. In January, they sold 10 minivans and 20 sedans. In February, they sold 7 minivans
zaharov [31]

Answer:

january = 10/20 = 1/2

februry = 7/14 = 1/2

so, we conclude that the auto dealership didn't have a lower ratio, since the ratio is equal in both months.

<em>hope it helps :)</em>

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8 0
3 years ago
The time required to finish a test in normally distributed with a mean of 70 minutes and a standard deviation of 10 minutes. Wha
zalisa [80]
Note that 60 minutes is 1 standard deviation away from the mean and from recalling the 68-95-99.7 rule, the area that will remain is (100 - 68) = 32%. However, we only want the leftmost portion of this area, so the answer is 32%/2 = 16%. 

<span>Choose D.</span>
3 0
3 years ago
A given field mouse population satisfies the differential equation dp dt = 0.5p − 410 where p is the number of mice and t is the
ohaa [14]

Answer:

a) t = 2 *ln(\frac{82}{5}) =5.595

b) t = 2 *ln(-\frac{820}{p_0 -820})

c) p_0 = 820-\frac{820}{e^6}

Step-by-step explanation:

For this case we have the following differential equation:

\frac{dp}{dt}=\frac{1}{2} (p-820)

And if we rewrite the expression we got:

\frac{dp}{p-820}= \frac{1}{2} dt

If we integrate both sides we have:

ln|P-820|= \frac{1}{2}t +c

Using exponential on both sides we got:

P= 820 + P_o e^{1/2t}

Part a

For this case we know that p(0) = 770 so we have this:

770 = 820 + P_o e^0

P_o = -50

So then our model would be given by:

P(t) = -50e^{1/2t} +820

And if we want to find at which time the population would be extinct we have:

0=-50 e^{1/2 t} +820

\frac{820}{50} = e^{1/2 t}

Using natural log on both sides we got:

ln(\frac{82}{5}) = \frac{1}{2}t

And solving for t we got:

t = 2 *ln(\frac{82}{5}) =5.595

Part b

For this case we know that p(0) = p0 so we have this:

p_0 = 820 + P_o e^0

P_o = p_0 -820

So then our model would be given by:

P(t) = (p_o -820)e^{1/2t} +820

And if we want to find at which time the population would be extinct we have:

0=(p_o -820)e^{1/2 t} +820

-\frac{820}{p_0 -820} = e^{1/2 t}

Using natural log on both sides we got:

ln(-\frac{820}{p_0 -820}) = \frac{1}{2}t

And solving for t we got:

t = 2 *ln(-\frac{820}{p_0 -820})

Part c

For this case we want to find the initial population if we know that the population become extinct in 1 year = 12 months. Using the equation founded on part b we got:

12 = 2 *ln(\frac{820}{820-p_0})

6 = ln (\frac{820}{820-p_0})

Using exponentials we got:

e^6 = \frac{820}{820-p_0}

(820-p_0) e^6 = 820

820-p_0 = \frac{820}{e^6}

p_0 = 820-\frac{820}{e^6}

8 0
3 years ago
1. In order to join a dancing club, there is a $30 startup fee and a $4 monthly fee. Write an equation in slope-intercept form t
weqwewe [10]

Answer:

y =mx+b

y= 4x+30

Step-by-step explanation:

4 0
3 years ago
Circle O has tangent AB and secant AC as shown. arc BC measures 4x+2-. arc BE measures 2x-10. What is the measure of arc CB
dangina [55]

Answer:

b. 160°

Step-by-step explanation:

angle A = (far arc - near arc)/2

far arc is  arc BC and near arc is arc BE. Replacing into the equation:

∠A = (arc BC - arc BE)/2

50 = [4x + 20 - (2x - 10)]/2

50*2 = 2x + 30

x = (100 - 30)/2

x = 35°

Replacing this information into arc BC equation:

arc BC = 4(35) + 20 = 160°

7 0
3 years ago
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