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Sedaia [141]
2 years ago
10

Layana’s house is located at (2 and two-thirds, 7 and one-third) on a map. The store where she works is located at (–1 and one-t

hird, 7 and one-third). What is the distance from Layana’s home to the store?
4 units

8 and two-thirds units

10 units

14 and two-thirds units
Mathematics
2 answers:
NeTakaya2 years ago
3 0

We have been given that Layana’s house is located at (2\frac{2}{3}, 7\frac{1}{3}) on a map. The store where she works is located at (-1\frac{1}{3}, 7\frac{1}{3}).

We are asked to find the distance from Layana’s home to the store

We will use distance formula to solve our given problem.

Let us convert our given coordinates in improper fractions.

2\frac{2}{3}\Rightarrow \frac{8}{3}

7\frac{1}{3}\Rightarrow \frac{22}{3}

-1\frac{1}{3}\Rightarrow -\frac{4}{3}

Now we will use distance formula to solve our given problem.

D=\sqrt{(x_2-x_1)^2+(y_2-y_1)^2}

Upon substituting coordinates of our given point in above formula, we will get:

D=\sqrt{(\frac{22}{3}-\frac{22}{3})^2+(\frac{8}{3}-(-\frac{4}{3}))^2}

D=\sqrt{(0)^2+(\frac{8}{3}+\frac{4}{3})^2}

D=\sqrt{0+(\frac{8+4}{3})^2}

D=\sqrt{(\frac{12}{3})^2}

D=\sqrt{(4)^2}

D=4

Therefore, the distance from Layana's home to the store is 4 units and option A is the correct choice.

Liula [17]2 years ago
3 0

Answer:

A

Step-by-step explanation:

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Ben and Landon took turns driving on their recent 820 mile road trip. Ben averaged 60 miles per hour while landon averaged 56 mi
Furkat [3]

Landon drove for 5 hours.

<u>Step-by-step explanation</u>:

Given that,

Total distance of the road trip = 820 miles

Speed of Ben = 60 miles per hour

Speed of Landon = 56 miles per hour

<u><em>Step 1 </em></u><em>:</em>

Let 'x' be the number of miles Ben drove

Let 'y' be the number of miles Landon drove

<u><em>Step 2</em></u><em> :</em>

The trip took them 14 hours to complete 820 miles together.

x + y = 14

60x + 56y = 820

<u><em>Step 3 </em></u><em>:</em>

Solve the linear system x+ y= 14 and 60x+ 56y= 820

x + y = 14

x = 14 - y

<u><em>Step 4 </em></u><em>:</em>

Substitute x= 14-y in the second equation,

60(14-y) + 56y = 820

840 - 60y + 56y = 820

<u><em>Step 5</em></u><em> :</em>

Keeping y terms on one side and constants on other side,

60y - 56y = 840 - 820

4y = 20

y = 20/4

y = 5

Since 'y' is the number of miles Landon drove, the time Landon drove is 5 hours.

3 0
2 years ago
Consider a regenerative vapor power cycle with two feedwater heaters, a closed one and an open one. Steam enters the first turbi
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Answer:

Consider a regenerative vapor power cycle with two feedwater heaters, a closed one and an open one, and reheat. Steam enters the first turbine stage at 12 MPa, 480∘C, and expands to 2 MPa. Some steam is extracted at 2 MPa and fed to the closed feedwater heater. The remainder is reheated at 2 MPa to 440∘C and then expands through the second-stage turbine to 0.3 MPa, where an additional amount is extracted and fed into the open feedwater heater operating at 0.3 MPa. The steam expanding through the third-stage turbine exits at the condenser pressure of 6 kPa. Feedwater leaves the closed heater at 210∘C, 12 MPa, and condensate exiting as saturated liquid at 2 MPa is trapped into the open feedwater heater. Saturated liquid at 0.3 MPa leaves the open feedwater heater. Assume all pumps and turbine stages operate isentropically. Determine for the cycle

a. Draw the cycle on a T-S diagram using the same numbering in the schematic

b. Determine the thermal efficiency of the cycle.

c. Determine the mass flow rate of steam entering the first turbine of the cycle.

(i) Thermal efficiency of the cycle = 43.185 %

(ii) The mass flow rate of steam =93.66 kg/h

Step-by-step explanation:

So we have at

For Point 1 on the T-S diagram we have

p₁ = 80 bar,  

t₁ = 480 °C,

From the super-heated steam tables we have

h₁ = 3349.6 kJ/kg, s₁ = 6.6613 kJ/kg·K

Point 2

p₂ = 20 bar

s₁ = s₂  =with x₂ = (6.6613 -6.6409)/(6.6849-6.6409) = 0.464

therefore h₂ =2953.1 + 0.464×(2977.1 - 2953.1) = 2964.22 kJ/kg

Point 3 on the T-S diagram we have

p₃ = 3 bar again s₁ = s₃  so we go to 3 bar on the steam tables and look up s = 6.6613 kJ/kg·K which is on the saturated steam tables

and x₃ is given as (6.6613 -1.6717)/(6.9916-1.6717) = 0.9379 and

h₃ = 561.43 + x₃×2163.5 = 2590.6 kJ/kg

Point 4

p₄ = 0.08 bar, s₁ = s₄, x₄ = 0.7949 and h₄ = 2083.45 kJ/kg

Point 5  

p₅ = 0.08 bar, h_{f5}= 173.84 kJ/kg

Point 6

Here h₆ is given by  h_{f5} plus the work done to move the water to the open heater therefore h₆ =

= 173.84 kJ/kg + 0.00100848×(3 - 0.08) × 100

= 173.84 kJ/kg + 0.29447616 kJ/kg = 174.13 kJ/kg

Point 7

p₇ = 3 bar, and h_{f7} = 561.43 kJ/kg

Point 8

Here again work is done to convey the fluid t constant pressure thus

h₈ = h_{f7} + v_{f7}× (p₈ - p₇)

561.43 kJ/kg + 0.00107317×(80 - 3)×100 = 569.69 kJ/kg

Point 9

p₉ = 80 bar  and T₉  = 205°C

By interpolating the values on the subcooled teperature tables we get

x₉ = 0.5 and h₉ =  854.94 + 0.5× (899.79 - 854.94) = 877.365 kJ/kg

Point 10

p₁₀ =  20 bar, h₁₀   = h_{f10} = 908.50 kJ/kg

point 11

Here h₁₁ = h₁₀ = 908.50 KJ/kg

For the closed feed water heater, energy and mass flow rate balance gives

m₁ × (h₂ - h₁₀) + (h₈ - h₉) = 0

Therefore m₁ = \frac{ (h_{9}  - h_8)}{(h_{2} - h_{10})}  = 0.14967

while the open water heater we get

m₂×h₃+(1-m₁-m₂)×h₆+m₁×h₁₁ - h₇ = 0

from where m₂ = 0.11479

W_{T} = (h₁-h₂) + (1 - m₁)(h₂ - h₃) +(1 - m₁ - m₂)(h₃ - h₄)

= 1076.11 kJ/kg

W_{p} = (h₈ - h₇) + (1 - m₁ - m₂)×(h₆ - h₅)

= 8.4733 kJ/kg

Q = h₁ -h₉ = 2472.235 kJ/kg

Efficiency = η = \frac{W_{T} - W_{P} }{Q} = 43.185 %

(ii)W_{cycle} = m_1*(W_T -W_P)

m'₁ = 100×10³/1066.63 = 93.66 kg/h

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Shelley sells 5 bone-shaped treats for $3.50. How much should she charge for a package of 12 treats?
kondaur [170]
She will be charged $42
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A psychologist is collecting data on the time it takes to learn a certain task. For 50 randomly selected adult subjects, the sam
tatuchka [14]

Answer: (15.263,\ 17.537)

Step-by-step explanation:

According to the given information, we have

Sample size : n= 50

\overline{x}=16.40

s=4.00

Since population standard deviation is unknown, so we use t-test.

Critical value for  95 percent confidence interval  :

t_{n-1,\alpha/2}=t_{49, 0.025}= 2.009575\approx2.010

Confidence interval : \overline{x}\pm t_{n-1, \alpha/2}\dfrac{s}{\sqrt{n}}

16.40\pm (2.010)\dfrac{4}{\sqrt{50}}\\\\=16.40\pm1.13702770415\\\\=16.40\pm1.1370\\\\=(16.40-1.1370,\ 16.40+1.1370)\\\\=(15.263,\ 17.537)

Required 95% confidence interval :  (15.263,\ 17.537)

8 0
2 years ago
Three percent of Jennie’s skin cells were burned when she escaped from a fire. If 3.7×10’(of her skin cells were burned then, ho
lora16 [44]

Answer:

The quantity of Jennie's skin cell were not burned is 119.601 × 10^{10}  .

Step-by-step explanation:

Given as :

The percentage of Jennie's skin cell were burned = 3 %

The quantity of Jennie's skin cell were burned = 3.7 × 10^{10}

Let The quantity of Jennie's skin cell were not burned = x

Let Total quantity of whole skin cell = n

<u>Now, According to question</u>

The quantity of Jennie's skin cell were burned = Total quantity of whole skin cell × percentage of Jennie's skin cell were burned

i.e 3.7 × 10^{10} = n × 3 %

Or, 3.7 × 10^{10} = n × \dfrac{3}{100}

Or, n =  3.7 × 10^{10} × \dfrac{100}{3}

Or, n = \frac{3.7\times 10^{12}}{3}

i.e n = 1.233 × 10^{12}

Or, Total quantity of whole skin cell = n = 1.233 × 10^{12}

<u>Now, Again</u>

∵ percentage of Jennie's skin cell were burned = 3 %

So, percentage of Jennie's skin cell were not burned = 100 % - 3 % = 97 %

so , The quantity of Jennie's skin cell were not burned = 97 % of Total quantity of whole skin cell

Or, x = \frac{97}{100} × n

Or, x =  \frac{97}{100} × 1.233 × 10^{12}

∴ x = 119.601 × 10^{10}

So, The quantity of Jennie's skin cell were not burned = x = 119.601 × 10^{10}

Hence,  The quantity of Jennie's skin cell were not burned is 119.601 × 10^{10}  . Answer

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