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gladu [14]
3 years ago
13

I need help on this on this onr too anyone mind helping?

Mathematics
1 answer:
V125BC [204]3 years ago
4 0
The correct answer should be 23 3/4 miles
I hope this helps
YOU'RE WELCOME :D
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The sum of 5 consecutive integers is 265. What is the fifth number in this sequence ?
Yanka [14]
Hello,
Let's assume n,n+1,n+2,n+3,n+4 the 5 numbers

n+(n+1)+...+(n+4)=5n+10=265
5n=265-10
5n=255
n=51
The 5th number is 51+4=55

6 0
3 years ago
The equation of a circle is (x + 4)2 + (y + 6)2 = 16. Determine the length of the radius.
Viefleur [7K]

Answer:

see explanation

Step-by-step explanation:

The equation of a circle in standard form is

(x - h)² + (y - k)² = r²

where (h, k) are the coordinates of the centre and r is the radius

(x + 4)² + (y + 6)² = 16 is in this form

with r² = 16 ⇒ r = \sqrt{16} = 4

---------------------------------------------------------------

given (h, k) = (5, 2) and r = 20, then

(x - 5)² + (y - 2)² = 400 ← represents the delivery area


5 0
3 years ago
17. If a + b = c, which of the following statements is true?
Scilla [17]

Answer:

c: c-a = b

.................

5 0
3 years ago
Read 2 more answers
Solve this problem using the substitution method 2y-1/2y=2
stealth61 [152]

Answer:

y=3/2and x =2

Step-by-step explanation:

6 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

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
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