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The mass of an ant is 1/10 of the load, which it can carry in one try. What is the mass of the ant, if in one time it can carry

Viefleur [7K]

If you call $m$ the mass of the ant and $l$ the load, we have the equation

$m = \cfrac{l}{10}$

In fact, the mass of the ant is one tenth of the load, which is exactly what this equation states.

Since we are given the load, we simply need to plug its value in the equation to deduce the mass of the ant:

$m = \cfrac{\frac{7}{250}}{10} = \cfrac{7}{250}\cdot\cfrac{1}{10} = \cfrac{7}{2500}$

8 0

3 years ago

Line A passes through the points (-1, 4) and (4, 9). Line B passes through the points (4, -4) and (-3, 24). Find the point where

exis [7]

A calculator can help

7 0

2 years ago

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Based on the table, which best predicts the end behavior of the graph of f(x)?

Yuki888 [10]

Answer: A. As x → ∞, f(x) → ∞, and as x → –∞, f(x) → ∞.

6 0

3 years ago

Read 2 more answers

Solve the following equation by completing the square. 3x^2-3x-5=13

mr Goodwill [35]

we'll start off by grouping some

$\bf 3x^2-3x-5=13\implies (3x^2-3x)-5=13\implies 3(x^2-x)-5=13 \\\\\\ 3(x^2-x)=18\implies (x^2-x)=\cfrac{18}{3}\implies (x^2-x)=6\implies (x^2-x+~?^2)=6$

so we have a missing guy at the end in order to get the a perfect square trinomial from that group, hmmm, what is it anyway?

well, let's recall that a perfect square trinomial is

$\bf \qquad \textit{perfect square trinomial} \\\\ (a\pm b)^2\implies a^2\pm \stackrel{\stackrel{\text{\small 2}\cdot \sqrt{\textit{\small a}^2}\cdot \sqrt{\textit{\small b}^2}}{\downarrow }}{2ab} + b^2$

so we know that the middle term in the trinomial, is really 2 times the other two without the exponent, well, in our case, the middle term is just "x", well is really -x, but we'll add the minus later, we only use the positive coefficient and variable, so we'll use "x" to find the last term.

$\bf \stackrel{\textit{middle term}}{2(x)(?)}=\stackrel{\textit{middle term}}{x}\implies ?=\cfrac{x}{2x}\implies ?=\cfrac{1}{2}$

so, there's our fellow, however, let's recall that all we're doing is borrowing from our very good friend Mr Zero, 0, so if we add (1/2)², we also have to subtract (1/2)²

$\bf \left( x^2 -x +\left[ \cfrac{1}{2} \right]^2-\left[ \cfrac{1}{2} \right]^2 \right)=6\implies \left( x^2 -x +\left[ \cfrac{1}{2} \right]^2 \right)-\left[ \cfrac{1}{2} \right]^2=6 \\\\\\ \left(x-\cfrac{1}{2} \right)^2=6+\cfrac{1}{4}\implies \left(x-\cfrac{1}{2} \right)^2=\cfrac{25}{4}\implies x-\cfrac{1}{2}=\sqrt{\cfrac{25}{4}} \\\\\\ x-\cfrac{1}{2}=\cfrac{\sqrt{25}}{\sqrt{4}}\implies x-\cfrac{1}{2}=\cfrac{5}{2}\implies x=\cfrac{5}{2}+\cfrac{1}{2}\implies x=\cfrac{6}{2}\implies \boxed{x=3}$

6 0

3 years ago

The perimeter of the base of a regular quadrilateral prism is 60 cm, the area of a lateral face is 105 cm2. Find: the volume of

Harman [31]

Since the base is a regular quadrilateral, each of its 4 sides must have length
s = P/4
s = (60 cm)/4 = 15 cm

The area of one lateral face is the product of side length and height.
A = s×h
105 cm² = (15 cm)×h
Then the height of the prism is
h = (105 cm²)/(15 cm) = 7 cm

The area of the base is then
B = s²
B = (15 cm)² = 225 cm²

The volume of the prism is the product of its base area and height.
V = Bh
V = (225 cm²)×(7 cm) = 1575 cm³

The volume is 1575 cm³.

5 0

2 years ago