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2 years ago

HELP PLS! I CAN’T FAIL PRE-ALGEBRA!

Mathematics

1 answer:

Liono4ka [1.6K]2 years ago

8 0

Answer:

Q1. 2 - 5/ 8 - 0 = -3/8

Q2. 1- (-1)/6-2 = 2/4 simplified as 1/2

Q3. (-5)- (-2)/ (-1)- (-3) = -3/2

Use a ruler to help you graph a line whose slope is 1/3. Label this line “a”

*Answer- 1/3 is being shown for rise/run (Rise over Run, 1 being rise and 3 being run). this means that you go up one and right 3.

Step-by-step explanation:

y2-y1 *over* x2- x1. Also if you can simplify the end fraction do it but if not then leav it as it is. This is the equation that I used for the first 3 problems. Also I didnt quit understand the 1st word problem sorry! I hope this helps :v

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A trough has ends shaped like isosceles triangles, with width 5 m and height 7 m, and the trough is 12 m long. Water is being pu

Svet_ta [14]

Answer:

$\dfrac{dh}{dt}=21 \text{m/min}$

the rate of change of height when the water is 1 meter deep is 21 m/min

Step-by-step explanation:

First we need to find the volume of the trough given its dimensions and shape: (it has a prism shape so we can directly use that formula OR we can multiply the area of its triangular face with the length of the trough)

$V = \dfrac{1}{2}(bh)\times L$

here L is a constant since that won't change as the water is being filled in the trough, however 'b' and 'h' will be changing. The equation has two independent variables and we need to convert this equation so it is only dependent on 'h' (the height of the water).

As its an isosceles triangle we can find a relationship between b and h. the ratio between the b and h will be always be the same:

$\dfrac{b}{h} = \dfrac{5}{7}$

$b=\dfrac{5}{7}h$ this can be substituted back in the volume equation

$V = \dfrac{5}{14}h^2L$

the rate of the water flowing in is:

$\dfrac{dV}{dt} = 6$

The question is asking for the rate of change of height (m/min) hence that can be denoted as: $\frac{dh}{dt}$

Using the chainrule:

$\dfrac{dh}{dt}=\dfrac{dh}{dV}\times \dfrac{dV}{dt}$

the only thing missing in this equation is dh/dV which can be easily obtained by differentiating the volume equation with respect to h

$V = \dfrac{5}{14}h^2L$

$\dfrac{dV}{dh} = \dfrac{5}{7}hL$

reciprocating

$\dfrac{dh}{dV} = \dfrac{7}{5hL}$

plugging everything in the chain rule equation:

$\dfrac{dh}{dt}=\dfrac{dh}{dV}\times \dfrac{dV}{dt}$

$\dfrac{dh}{dt}=\dfrac{7}{5hL}\times 6$

$\dfrac{dh}{dt}=\dfrac{42}{5hL}$

L = 12, and h = 1 (when the water is 1m deep)

$\dfrac{dh}{dt}=\dfrac{42}{5(1)(12)}$

$\dfrac{dh}{dt}=21 \text{m/min}$

the rate of change of height when the water is 1 meter deep is 21 m/min

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3 years ago

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Komok [63]

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3 years ago

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Answer:

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Step-by-step explanation:

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JulsSmile [24]

Answer:

the correct choice is marked

Step-by-step explanation:

Let x represent the smaller number. Then the larger number is 8x, and the difference is ...

8x -x = 280

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<em>Additional comment</em>

Effectively, we have solved for the multiplier (40) that gives the ratio with 320 on top and a difference between top and bottom of 280:

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