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3 years ago

What is the equation of the line with slope of ½ , and which passes through the point

Mathematics

1 answer:

Law Incorporation [45]3 years ago

6 0

Answer:

y = 1/2x - 7

Step-by-step explanation:

slope intercept form:

y = mx + b

y = -5

m(slope) = 1/2

x = 4

y = mx + b

substitute

-5 = 1/2*4 + b

-5 = 2 + b

-5 - 2 = b

-7 = b

y = 1/2x - 7

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Your gas tank has approximately 78.1% of fuel left. If the tank can old 30 gallons, how much is left?

andrey2020 [161]

Answer:

23.43 gallons.

Step-by-step explanation:

30*0.781 which is 23.43 gallons.

Hope this helps!

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4 0

3 years ago

Find the product of 2(5.9) mentally using the distributive property.

VARVARA [1.3K]

2(5.9) is 11.8

Hopefully i is right I did this in my head.

7 0

3 years ago

Write the word sentence as an inequality. Then solve the inequality.

LenaWriter [7]

Answer:

x/7<-3
x<-21

Step-by-step explanation:

x/7<-3
x<-3x7
x<-21

7 0

3 years ago

Hello people ~

Luden [163]

Cone details:

height: h cm

radius: r cm

Sphere details:

radius: 10 cm

================

From the endpoints (EO, UO) of the circle to the center of the circle (O), the radius is will be always the same.

Using Pythagoras Theorem

(a)

TO² + TU² = OU²

(h-10)² + r² = 10² [insert values]

r² = 10² - (h-10)² [change sides]

r² = 100 - (h² -20h + 100) [expand]

r² = 100 - h² + 20h -100 [simplify]

r² = 20h - h² [shown]

r = √20h - h² ["r" in terms of "h"]

(b)

volume of cone = 1/3 * π * r² * h

===========================

$\longrightarrow \sf V = \dfrac{1}{3} * \pi * (\sqrt{20h - h^2})^2 \ ( h)$

$\longrightarrow \sf V = \dfrac{1}{3} * \pi * (20h - h^2) (h)$

$\longrightarrow \sf V = \dfrac{1}{3} * \pi * (20 - h) (h) ( h)$

$\longrightarrow \sf V = \dfrac{1}{3} \pi h^2(20-h)$

To find maximum/minimum, we have to find first derivative.

(c)

First derivative

$\Longrightarrow \sf V' =\dfrac{d}{dx} ( \dfrac{1}{3} \pi h^2(20-h) )$

apply chain rule

$\sf \Longrightarrow V'=\dfrac{\pi \left(40h-3h^2\right)}{3}$

Equate the first derivative to zero, that is V'(x) = 0

$\Longrightarrow \sf \dfrac{\pi \left(40h-3h^2\right)}{3}=0$

$\Longrightarrow \sf 40h-3h^2=0$

$\Longrightarrow \sf h(40-3h)=0$

$\Longrightarrow \sf h=0, \ 40-3h=0$

$\Longrightarrow \sf h=0,\:h=\dfrac{40}{3}$

maximum volume: when h = 40/3

$\sf \Longrightarrow max= \dfrac{1}{3} \pi (\dfrac{40}{3} )^2(20-\dfrac{40}{3} )$

$\sf \Longrightarrow maximum= 1241.123 \ cm^3$

minimum volume: when h = 0

$\sf \Longrightarrow min= \dfrac{1}{3} \pi (0)^2(20-0)$

$\sf \Longrightarrow minimum=0 \ cm^3$

6 0

2 years ago

Read 2 more answers

Find -4(3)+(-7) What order should be preformed in this expression

Sergio [31]

Answer:

-19

Step-by-step explanation:

According to bedmas rules, we have to first solve the brackets.

-4(3)+(-7)

-12 + (-7)

Now, let's add

-12 + (-7)

-12 - 7

= -19

8 0

2 years ago