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2 years ago

Using the midpoint formula find BC if B(8,-7) and C(-4,-2)

Mathematics

1 answer:

arsen [322]2 years ago

6 0

Answer:

(2, -4.5)

Step-by-step explanation:

Used midpoint formula

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Which decimal is equivalent to 8/45?

Vilka [71]

Answer:

.1777... (7 is repeating)

6 0

3 years ago

I need help finding y

denpristay [2]

Answer:

y = 9.8995

Step-by-step explanation:

This question requires the use of basic trigonometry.

In this problem, we know one of the sides and two of the angles.

Since one of the angles is 45 degrees and the sum of all internal angles of a triangle is 180 degrees, the other angle can only be 45 degrees, making this an isosceles triangle - that is, the length of the other leg, x, is also 7.

You could use the soh cah toa mnemonic to figure it out, but here just using the a^2 + b^2 = c^2 should be sufficient since we know both legs of the triangle.

7^2 + 7^2 = 98, and the square root of 98 gives you about 9.8995, which sounds about right for this situation.

Hope this helped!

4 0

3 years ago

-4(j+k+g) plz help using box method

disa [49]

Answer:

-4j-4k-4g

Step-by-step explanation:

8 0

4 years ago

Please help with this question

ioda

Answer:

$25$

Step-by-step explanation:

$\frac{5^8 \times 5^{-2}}{5^4}$

$\frac{5^{8+-2}}{5^4}$

$\frac{5^{6}}{5^4}$

$5^{6-4}$

$5^2$

$=25$

3 0

3 years ago

At 8:00 am, here's what we know about two airplanes: Airplane #1 has an elevation of 80870 ft and is decreasing at the rate of 4

wel

Let's begin by listing out the information given to us:

8 am

airplane #1: x = 80870 ft, v = -450 ft/ min

airplane #2: x = 5000 ft, v = 900ft/min

We must note that the airplanes are moving at a constant speed. The equation for the airplanes is given by:

$\begin{gathered} E=x_1+vt----1 \\ E=x_2+vt----2 \\ where\colon E=elevation,ft;x=InitialElevation,ft; \\ v=velocity,ft\text{/}min;t=time,min \\ x_1=80,870ft,v=-450ft\text{/}min \\ E=80870-450t----1 \\ x_2=5,000ft,v=900ft\text{/}min \\ E=5000+900t----2 \end{gathered}$

We equate equations 1 & 2 to get the time both airlanes will be at the same elevation. We have:

$\begin{gathered} 5000+900t=80870-450t \\ \text{Add 450t to both sides, we have:} \\ 900t+450t+5000=80870-450t+450t \\ 1350t+5000=80870 \\ \text{Subtract 5000 from both sides, we have:} \\ 1350t+5000-5000=80870-5000 \\ 1350t=75870 \\ \text{Divide both sides by 1350, we have:} \\ \frac{1350t}{1350}=\frac{75870}{1350} \\ t=56.2min \\ \\ \text{After }56.2\text{ minutes, both airplanes will be at the same elevation} \end{gathered}$

The elevation at that time (when the elevations of the two airplanes are the same) is given by substituting the value of time into equations 1 & 2. We have:

$\begin{gathered} E_1=80870-450t \\ E_1=80870-450(56.2) \\ E_1=80870-25290 \\ E_1=55580ft \\ \\ E_2=5000+900t \\ E_2=5000+900(56.2) \\ E_2=5000+50580 \\ E_2=55580ft \\ \\ \therefore E_1\equiv E_2=55580ft \end{gathered}$

6 0

1 year ago