I need some help with this problem;-;

Mathematics

1 answer:

Fittoniya [83]3 years ago

8 0

Ok, firstly you need to plug in any number you can star by 1, 0 whatever, then add 2 numbers more, only to do accurate the graph and finally add the given number (3) then simplify:

y= 2(3)+4 = 10

B) $10 dollars after 3 days of late.

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a cone with a radius 3 units is shown below. it’s volume is 57 cubic units. find the height of the cone

ipn [44]

Given:

Given that the radius of the cone is 3 units.

The volume of the cone is 57 cubic units.

We need to determine the height of the cone.

Height of the cone:

The height of the cone can be determined using the formula,

$V=\frac{1}{3} \pi r^2 h$

Substituting the values, r = 3 and V = 57, we get;

$57=\frac{1}{3} (3.14) (3)^2 h$

Simplifying the terms, we get;

$57=\frac{1}{3} (3.14) (9) h$

Multiplying both sides of the equation by 3, we get;

$171= (3.14) (9) h$

$171=28.26 h$

Dividing both sides of the equation by 28.26, we get;

$6.05=h$

Thus, the height of the cone is 6.05 units.

8 0

4 years ago

ASAP WILL MARK BRAINLYEST answer these two please!

Trava [24]

$19.\\y=3x^2\\a=3,\ b=0,\ c=0\\\\\dfrac{-b}{2a}=\dfrac{-0}{2(3)=0\to x=0\\\\3x^2=0\ \ \ \ |:3\\\\x^2=0\to x=0$

$20.\\y=x^2-x-2\\a=1,\ b=-1,\ c=-2\\\\\dfrac{-b}{2a}=\dfrac{-(-1)}{2(1)}=\dfrac{1}{2}\to x=\dfrac{1}{2}\\\\x^2-x-2=0\\\\x^2-2x+x-2=0\\\\x(x-2)+1(x-2)=0\\\\(x-2)(x+1)=0\iff x-2=0\ \vee\ x+1=0\\\\x=2\ \vee\ x=-1$