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Nina [5.8K]
3 years ago
15

Which property of equality would be used to solve 3x=81

Mathematics
1 answer:
Makovka662 [10]3 years ago
5 0

Answer:

Division

Step-by-step explanation:

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In a 30-60-90 triangle, what is the length of the hypotenuse when the shorter leg is 5 cm?
zheka24 [161]

Shorter Leg=x

Hypotenuse=2x

Shorter Leg=5cm

Hypotenuse= 10cm

3 0
3 years ago
The radius of a circle is 1 foot. What is the length of a 45 degree arc
Elodia [21]
Circumference = 2 x PI x r
circumference = 3.14 x 2 x 12 inches = 75.36 inches

45/360 = 0.125

75.36 * 0.125 = 9.42 inches
7 0
3 years ago
How will the area of the circle change if it is dilated by a scale factor of 1/2? The area will be 2 times greater than the orig
coldgirl [10]
We know that

[scale factor]=1/2
[dilated area of circle]=[original area of circle]*[scale factor]²
so
[dilated area of circle]=[original area of circle]*[1/2]²
[dilated area of circle]=[original area of circle]*[1/4]

therefore

the answer is
<span>The area will be 1/4 the original</span>
7 0
3 years ago
Read 2 more answers
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

1.

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}

2.

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}

3.

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
Can you help me with these 3?
miskamm [114]

Step-by-step explanation:

1. 5,11,17,23,29,35

Common difference is 6

3. 8,21,34, 47,60

Common difference is 13

5. 5,10,15,20,25

Common difference is 5

3 0
3 years ago
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