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3 years ago

Emeline can type at a constant rate of 1/4 pages/minute.

Mathematics

1 answer:

Degger [83]3 years ago

4 0

Answer:

a. 1/4

b. No, she does not have enough time

c. 28 minutes

Step-by-step explanation:

Emeline can type at a constant rate of 1/4 pages/minute.

a.the ratio of the number of pages to the number of minutes = 1/4 that she types 1 page every 4 minutes.

b. For a five page article she will take 5×4 minutes = 20 minutes( as she took 4 minutes for 1 page ). She won,t be able to complete task before the dead line as she has got only 18 days.

c. Emeline has to type a 7-page article. so, she will take 7×4 minutes = 28 minutes to complete the task.

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A water balloon is tossed into the air. The function h(x)=-0.5(x-4)squared +9 gives the height, in the feet, of the balloon from

NNADVOKAT [17]

Answer:

No, the maximum height that the balloon can reach is 9ft.

Step-by-step explanation:

Hi, to answer this question we have to analyze the information given:

The function h(x) =-0.5(x-4)² +9, is a Quadratic Function in the Vertex form.

Vertex form: f (x) = a(x - h) 2 + k

Where:

(h, k) is the vertex of the parabola-

h is the horizontal shift (how far left, or right, the graph has shifted from x = 0).

k represents the vertical shift (how far up, or down, the graph has shifted from y = 0).

In this case a = -0.5, it means that the parabola opens downward and has a maximum point at the vertex.

So, the maximum height that the balloon can reach is k=9ft.

9 ∠ 12

The balloon will not hit the ceiling 12 ft above the pool.

Feel free to ask for more if needed or if you did not understand something.

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3 years ago

HELP SO I DON’T FAIL! WILL RECEIVE BRAINLIEST! Use a calculator to find the r-value of these data. Round the value to three deci

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The answers C. this is the true answer okkkkkkk

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3 years ago

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Help pl0x, Algebra 1

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A 4-ft vertical post casts a 20-in shadow at the same time a nearby cell phone tower casts a 115-ft shadow. How tall is the c

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Step-by-step explanation4363 :

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3 years ago

The article on the next page from the San Francisco Chronicle, January 27, 1996, describes very serious problems of subsidence,

Mekhanik [1.2K]

Answer:

246.3%

the complete question is found in the attached document

Step-by-step explanation:

1st step:

using 1936 data,

w1= 356% = 3.56(356/100) , H= 79 feet

specific gravity = 2.65

Sₓ= 100%= 1

initial void ratio(e₀)= (w1 x specific gravity)/Sₓ

=3.56 x 2.65/1 = 9.434

2nd step

using 1996 data

ΔH= 22ft

ΔH/H = Δe/(1 + e₀)

22/79 = Δe/(1+9.434)

0.278=Δe/10.434

Δe= 0.278 x 10.434

Δe= 2.905

Δe= e₀ - eₓ

eₓ= e₀-Δe

eₓ= 9.434 - 2.905

eₓ= 6.529

3rd step

calculating water content in 1996

eₓ =6.529, specific gravity= 2.65, Sₓ= 100%

W2 X 2.65 = 1 x 6.529

w2 = 6.529/2.65 = 2.463 = 246.3%

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4 years ago