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3 years ago

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Please helpp!!!! I need it right now!!!!!!

1 answer:

marta [7]3 years ago

3 0

Because the image is two reflected triangles, the sides will be the same meaning ∆ AMC ≈ ∆ BMC must be true. M is also the same for both, due to it being the reflected side.

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At a robotics competition, robots pick up and then throw a ball. The distances, in meters, that eight robots can throw a ball ar

jenyasd209 [6]

The outlier in the given data set and the reason we know it is an outlier are;

<u><em>The outlier is 9.8</em></u>

<em><u>It is an outlier because it is the distance that is less than 10.</u></em>

Outliers in mathematics are simply values that lie outside of the other values in a set of data. This means they are far larger than the given set of values or far smaller than the given set of values.

Now, in this question, we see that the distances given in meters are;

10.1 m, 10.1 m, 10.2 m, 10.2 m, 10.3 m, 10.1 m, 10.2 m, 9.8 m

From these set of data values, we can see that all but 1 of the values are more than 10. This number that is less than 10 is the odd one out of these set of data values and therefore it is the outlier.

Read more at; brainly.com/question/3631910

7 0

1 year ago

A rectangular field has an area of 162 m2.

Fed [463]

Answer:Length: 27m

Step-by-step explanation:

the formula of area of rectangle is : width*length
so we just replace the values in the formula
162=6*Length
Length=162/6
Length=27m

3 0

2 years ago

Read 2 more answers

Given that 110ₓ = 40₅ , find the value of x.

adell [148]

Answer:

is therr more to that question

Step-by-step explanation:

4 0

3 years ago

Read 2 more answers

Denver, Engle and Fido are all dogs who eat differing amounts of dog food.

alekssr [168]

Answer:

25%

Step-by-step explanation:

If Denver gets 8/19 of the food, 11/19 of the food is left between Engle and Fido.

If the food is shared in a ratio of 6 : 2 between the pair, Engle gets 6/8 and Fido gets 2/8.

Thus, 2/8 = 25%.

3 0

3 years ago

Read 2 more answers

A tank contains 1600 L of pure water. Solution that contains 0.04 kg of sugar per liter enters the tank at the rate 2 L/min, and

goldfiish [28.3K]

Let S(t) denote the amount of sugar in the tank at time t. Sugar flows in at a rate of

(0.04 kg/L) * (2 L/min) = 0.08 kg/min = 8/100 kg/min

and flows out at a rate of

(S(t)/1600 kg/L) * (2 L/min) = S(t)/800 kg/min

Then the net flow rate is governed by the differential equation

$\dfrac{\mathrm dS(t)}{\mathrm dt}=\dfrac8{100}-\dfrac{S(t)}{800}$

Solve for S(t):

$\dfrac{\mathrm dS(t)}{\mathrm dt}+\dfrac{S(t)}{800}=\dfrac8{100}$

$e^{t/800}\dfrac{\mathrm dS(t)}{\mathrm dt}+\dfrac{e^{t/800}}{800}S(t)=\dfrac8{100}e^{t/800}$

The left side is the derivative of a product:

$\dfrac{\mathrm d}{\mathrm dt}\left[e^{t/800}S(t)\right]=\dfrac8{100}e^{t/800}$

Integrate both sides:

$e^{t/800}S(t)=\displaystyle\frac8{100}\int e^{t/800}\,\mathrm dt$

$e^{t/800}S(t)=64e^{t/800}+C$

$S(t)=64+Ce^{-t/800}$

There's no sugar in the water at the start, so (a) S(0) = 0, which gives

$0=64+C\impleis C=-64$

and so (b) the amount of sugar in the tank at time t is

$S(t)=64\left(1-e^{-t/800}\right)$

As $t\to\infty$ , the exponential term vanishes and (c) the tank will eventually contain 64 kg of sugar.

7 0

3 years ago