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Svet_ta [14]
3 years ago
5

X-1/7>1/8 Solve using the addition principle.

Mathematics
1 answer:
amid [387]3 years ago
6 0
x- \frac{1}{7} \ \textgreater \ \frac{1}{8}  \\ x\ \textgreater \ \frac{1}{8}+ \frac{1}{7} \\ x\ \textgreater \ \frac{7+8}{56} \\ x\ \textgreater \ \frac{15}{56}
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The problem ask to calculate the corresponding distance on the new patio where as the the patio will have two long parallel side and an area of 360 square feet and the area of the similar patio is 250 square feet and its long parallel sides are 12.5 feet apart. To calculate this the ratio of the distance of the two polygon is P:Q so it means that the are is also P^2:Q^2 so the ratio of the area is  36:25 so the the distance of the new patio is 18 feet
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An arithmetic sequence is represented in the following table. Enter the missing term of the sequence.
bija089 [108]

a_1=-9\\\\a_2=a_1-8\to a_2=-9-8=-17\\\\a_3=a_2-8\to a_3=-17-8=-25\\\\a_4=a_3-8\to a_4=-25-8=-33\\\\\text{The general term of an arithmetic sequence:}\\\\a_n=a_1+(n-1)d\\\\a_1-first\ term\\d-common\ difference\\\\\text{We have}\ a_1=-9\ \text{and}\ d=-8.\ \text{Substitute:}\\\\a_n=-9+(n-1)(-8)=-9-8n+8=-8n-1\\\\\text{Put n = 43 and calculate}\ a_{43}\\\\a_{43}=-8(43)-1=-344-1=-345\\\\Answer:\ \boxed{?=-345}

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3 years ago
Find three consecutive integers such that that product of the first and third is one
Tema [17]

Answer:

n = the first

n + 1 = the 2nd

n + 2 = the 3rd

n(n + 2) - (n + 1) = 10(n + 2) + 1

n2 + 2n - n - 1 = 10n + 20 + 1

n2 - 9n - 22 = 0

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3 years ago
If a ball is thrown straight up into the air with an initial velocity of 70 ft/s, its height in feet after t seconds is given by
Vitek1552 [10]

Answer:

a) Average velocity at 0.1 s is 696 ft/s.

b) Average velocity at 0.01 s is 7536 ft/s.

c) Average velocity at 0.001 s is 75936 ft/s.

Step-by-step explanation:

Given : If a ball is thrown straight up into the air with an initial velocity of 70 ft/s, its height in feet after t seconds is given by y = 70t-16t^2.

To find : The average velocity for the time period beginning when t = 2 and lasting.  a. 0.1 s. , b. 0.01 s. , c. 0.001 s.

Solution :    

a) The average velocity for the time period beginning when t = 2 and lasting 0.1 s.

(\text{Average velocity})_{0.1\ s}=\frac{\text{Change in height}}{0.1}

(\text{Average velocity})_{0.1\ s}=\frac{h_{2.1}-h_2}{0.1}

(\text{Average velocity})_{0.1\ s}=\frac{(70(2.1)-16(2.1)^2)-(70(0.1)-16(0.1)^2)}{0.1}

(\text{Average velocity})_{0.1\ s}=\frac{69.6}{0.1}

(\text{Average velocity})_{0.1\ s}=696\ ft/s

b) The average velocity for the time period beginning when t = 2 and lasting 0.01 s.

(\text{Average velocity})_{0.01\ s}=\frac{\text{Change in height}}{0.01}

(\text{Average velocity})_{0.01\ s}=\frac{h_{2.01}-h_2}{0.01}

(\text{Average velocity})_{0.01\ s}=\frac{(70(2.01)-16(2.01)^2)-(70(0.01)-16(0.01)^2)}{0.01}

(\text{Average velocity})_{0.01\ s}=\frac{75.36}{0.01}

(\text{Average velocity})_{0.01\ s}=7536\ ft/s

c) The average velocity for the time period beginning when t = 2 and lasting 0.001 s.

(\text{Average velocity})_{0.001\ s}=\frac{\text{Change in height}}{0.001}  

(\text{Average velocity})_{0.001\ s}=\frac{h_{2.001}-h_2}{0.001}

(\text{Average velocity})_{0.001\ s}=\frac{(70(2.001)-16(2.001)^2)-(70(0.001)-16(0.001)^2)}{0.001}

(\text{Average velocity})_{0.001\ s}=\frac{75.936}{0.001}

(\text{Average velocity})_{0.001\ s}=75936\ ft/s

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How many 0.2 liter glasses of water are contained in a 3.6 liter pitcher
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3.6 / 0.2 = 18 liter glasses :))
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