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2 years ago

A campfire burn 2 1/2 pounds of wood per hour. how many pounds of wood do you need to keep the camp firing burning for 6 hours?

Mathematics

1 answer:

Natalka [10]2 years ago

8 0

15 pounds of wood. Hope this helped :)

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Denzel earned money after school

Kryger [21]

Great for him. I'm so proud.

3 0

2 years ago

Write the polynomial equation for -1/5,-1/2,-4

ycow [4]

Answer:

(5x+1)(2x+1)(x+4)

Step-by-step explanation:

The factored from of a polynomial can be found from the zeros or x-intercepts of the graph.

The x-intercepts here are x= -1/5, x=-1/2 and x= -4. You can find the factors by finding the expression which solves for each value.

Since x = -1/5 then the factor is 5x+1.

Since x = -1/2 then the factor is 2x+1.

Since x = -4 then the factor is x+4.

So the factored form is (5x+1)(2x+1)(x+4).

5 0

2 years ago

Could someone help me? Which color is it.

wariber [46]

Answer:

(1,3)

Step-by-step explanation:

y = 2x+1

y = 4x-1

Since both equations are equal to y, we can set them equal to each other

2x+1 = 4x-1

Subtract 2x from each side

2x+1-2x = 4x-2x-1

1 = 2x-1

Add 1 to each side

1+1 = 2x-1+1

2 = 2x

Divide each side by 2

2/2 = 2x/2

1 =x

Now we need to find y

y = 2x+1

y = 2(1)+1

y =3

8 0

2 years ago

What is the volume of a sphere with a radius of 3 inches? 12π cubic inches 24π cubic inches 36π cubic inches 48π cubic inches

Anni [7]

Answer:

The answer is

<h2>36π cubic inches</h2>

Step-by-step explanation:

Volume of a sphere is given by

where r is the radius

From the question

r = 3 inches

Substitute the value into the above formula and solve for the volume

That's

<h3> $V = \frac{4}{3}( {3})^{3} \pi \\ = \frac{4}{3} \times 27\pi \\ = 4 \times 9\pi$ </h3>

We have the final answer as

<h3>36π cubic inches</h3>

Hope this helps you

4 0

2 years ago

Elizabeth attempts a field goal by kicking a football from the ground with an initial vertical

sammy [17]

$\bf ~~~~~~\textit{initial velocity} \\\\ \begin{array}{llll} ~~~~~~\textit{in feet} \\\\ h(t) = -16t^2+v_ot+h_o \end{array} \quad \begin{cases} v_o=\stackrel{64}{\textit{initial velocity of the object}}\\\\ h_o=\stackrel{0\qquad \textit{from the ground}}{\textit{initial height of the object}}\\\\ h=\stackrel{}{\textit{height of the object at "t" seconds}} \end{cases} \\\\[-0.35em] \rule{34em}{0.25pt}$

$\bf h(t)=-16t^2+64t+0\implies h(t)=-16t^2+64t\implies \stackrel{\textit{hits the ground}~\hfill }{0=-16t^2+64t} \\\\\\ 0=-16t(t-4)\implies t= \begin{cases} 0\\ \boxed{4} \end{cases}$

Check the picture below, it hits the ground at 0 feet, where it came from, the ground, and when it came back down.

5 0

2 years ago