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3 years ago

Y= -1/8x + 1/4 Find Slope- X-intercept- Y-intercept-

Mathematics

1 answer:

zavuch27 [327]3 years ago

8 0

slope = -1/8

y intercept = 1/4

x intercept set y=0 and solve for x

0 = 1/8 x + 1/4

subtract 1/4 from each side

-1/4=1/8x

multiply both sides by 8

-2 = x

x intercept = 2

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Serggg [28]

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3 years ago

Question 1 Part A One of Maria and Josie’s early tasks is printing and laminating signs to hang at local businesses around town.

maks197457 [2]

Answer:

Using the table

Maria does 1/4 of the job in 1 hour since it takes her 4 hours to complete the job , she works x hours and completes x/4 of the task

Josie does 1/2 of the job in 1 hour since it takes her 2 hours to complete the job , she works x hours and completes x/2

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3 years ago

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Using trig how do you answer this

Aneli [31]

Answer:

62.73° or 1.09 radians

Step-by-step explanation:

SOH CAH TOA

sin = $\frac{opposite}{hypotenuse}$ , so sin (?) = $\frac{8}{9}$

this means that ? = $sin^{-1}(\frac{8}{9})$ or ? = arcsin( $\frac{8}{9}$ ), depending on what calculator you have

plugging in $sin^{-1}(\frac{8}{9})$ or arcsin( $\frac{8}{9}$ ) into calculator will get you

62.73° or 1.09 radians

use the corresponding answer depending on if the question asks for degrees or radians

3 0

3 years ago

Read 2 more answers

Gravel is being dumped from a conveyor belt at a rate of 20 ft3 /min and its coarseness is such that it forms a pile in the shap

pantera1 [17]

Answer:

The height of the pile is increasing at the rate of $\mathbf{ \dfrac{20}{56.25 \pi} \ \ \ \ \ ft/min}$

Step-by-step explanation:

Given that :

Gravel is being dumped from a conveyor belt at a rate of 20 ft³ /min

i.e $\dfrac{dV}{dt}= 20 \ ft^3/min$

we know that radius r is always twice the diameter d

i.e d = 2r

Given that :

the shape of a cone whose base diameter and height are always equal.

then d = h = 2r

h = 2r

r = h/2

The volume of a cone can be given by the formula:

$V = \dfrac{\pi r^2 h}{3}$

$V = \dfrac{\pi (h/2)^2 h}{3}$

$V = \dfrac{1}{12} \pi h^3$

$V = \dfrac{ \pi h^3}{12}$

Taking the differentiation of volume V with respect to time t; we have:

$\dfrac{dV}{dt }= (\dfrac{d}{dh}(\dfrac{\pi h^3}{12})) \times \dfrac{dh}{dt}$

$\dfrac{dV}{dt }= (\dfrac{\pi h^2}{4} ) \times \dfrac{dh}{dt}$

we know that:

$\dfrac{dV}{dt}= 20 \ ft^3/min$

So;we have:

$20= (\dfrac{\pi (15)^2}{4} ) \times \dfrac{dh}{dt}$

$20= 56.25 \pi \times \dfrac{dh}{dt}$

$\mathbf{\dfrac{dh}{dt}= \dfrac{20}{56.25 \pi} \ \ \ \ \ ft/min}$

The height of the pile is increasing at the rate of $\mathbf{ \dfrac{20}{56.25 \pi} \ \ \ \ \ ft/min}$

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4 years ago

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Novay_Z [31]

Answer:

22.5 repeated

Step-by-step explanation:

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3 years ago

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