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4 years ago

Need help ASAP please! thank you! 20 pts

Mathematics

1 answer:

densk [106]4 years ago

4 0

$y=-\dfrac{x}{4}+[?]\\\\\text{We know: a line intersects the point (-8, -1).}\\\text{Put the coordinates of the point into the equation of a line}\\(-8,\ -1)\to x=-8,\ y=-1\\\\-1=-\dfrac{-8}{4}+[?]\\\\-1=2+[?]\qquad|\text{subtract 2 from both sides}\\\\-3=[?]\\\\Answer:\ y=-\dfrac{x}{4}-3$

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otez555 [7]

4 + 2x - 7 = 12

2x - 7 + 4 = 12

2x -3 = 12

2x = 15

x = 7.5

2x - 7 = 7.5 - 7 = 0.5

PLS MARK BRAINLIEST

3 0

3 years ago

No links, provide explanation.

a_sh-v [17]

Answer:

C: 18

Step-by-step explanation:

6 0

3 years ago

What is the hcf of two coprime numbers

Karolina [17]

The HCF is the highest common factor of two numbers.

By definition, if two numbers are coprime, they have no common factors other than 1.

So, the highest (and only) common factor of two coprime numbers is 1.

8 0

4 years ago

The position equation is s = -16t^2 + vot + so where s is height of the object in feet, vo is the initial velocity of the object

Hunter-Best [27]

A. For this case, let us set the variables:

s = 0 (final destination on the ground)

t = unknown

vo = 0

so = 8000 ft

Using the equation, we calculate for t:

0 = -16 t^2 + 0t + 8000

t = 22.36 s

B. For this case:

s = unknown

t = 22.36 s

vo = 600 miles/hr = 880 ft/s

so = 0

Using the equation, we calculate for s:

s = -16*(22.36)^2 + 880*22.36 + 0

s = 11, 677.29 ft = 2.21 miles

6 0

4 years ago

A rectangular box with a volume of 684 ftcubed is to be constructed with a square base and top. The cost per square foot for the

SCORPION-xisa [38]

Answer:

The dimensions of the rectangular box is 36.23 ft×36.23 ft×4.345 ft.

Minimum cost=2046.16 cents

Step-by-step explanation:

Given that a rectangular box with a volume of 684 ft³.

The base and the top of the rectangular box is square in shape

Let the length and width of the rectangular box be x.

[since the base is square in shape, length=width]

and the height of the rectangular box be h.

The volume of rectangular box is = Length ×width × height

=(x²h) ft³

$x^2h=684\Rightarrow h=\frac{684}{x^2}$ (1)

The area of the base and top of rectangular box is = x² ft²

The surface area of the sides= 2(length+width) height

=2(x+x)h

=4xh ft²

The total cost to construct the rectangular box is

=[(x²×20)+(x²×15)+(4xh×1.5)] cents

=(20x²+15x²+6xh) cents

=(25x²+6xh) cents

Total cost= C(x).

C(x) is in cents.

∴C(x)=25x²+6xh

Putting $h=\frac{684}{x^2}$

$C(x)=25x^2+6x\times\frac{684}{x^2} \Rightarrow C(x)=25x^2+\frac{4104}{x}$

Differentiating with respect to x

$C'(x)=50x-\frac{4104}{x^2}$

To find minimum cost, we set C'(x)=0

$\therefore50x-\frac{4104}{x^2}=0\\\Rightarrow50x=\frac{4104}{x^2}\\\Rightarrow x^3=\frac{4104}{50}\Rightarrow x\approx 4.345$ ft.

Putting the value x in equation (1) we get

$h=\frac{684}{(4.345)^2}$

≈36.23 ft.

The dimensions of the rectangular box is 36.23 ft×36.23 ft×4.345 ft.

Minimum cost C(x)=[25(4.345)²+10(4.345)(36.23)] cents

=2046.16 cents

5 0

3 years ago