A triangle weighs 3 grams, and a circle weighs 6 grams. Use this applet to help find the weight of a square.

Mathematics

1 answer:

lesya [120]3 years ago

7 0

Answer: 3.75 g

Step-by-step explanation:

Assume the weight of the square is x.

The hanger is in balance so the left side is equal to the right side.

Equation therefore is:

Triangle + Square + 3 * circle = 5* square + circle

3 + x + (3 * 6) = 5x + 6

3 + x + 18 = 5x + 6

5x - x = 3 + 18 - 6

4x = 15

x = 15/4

x = 3.75 g

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Snowcat [4.5K]

Answer:

v= -2

Step-by-step explanation:

2v + 7 = 3

(-7 both sides)

2v = -4

(divide bu 2 on both sides)

v = -2

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3 0

3 years ago

Leno4ka [110]

Answer:

$\frac{s^2-25}{(s^2+25)^2}$

Step-by-step explanation:

Let's use the definition of the Laplace transform and the identity given: $\mathcal{L}[t \cos 5t]=(-1)F'(s)$ with $F(s)=\mathcal{L}[\cos 5t]$ .

Now, $F(s)=\int_0 ^{+ \infty}e^{-st}\cos(5t) dt$ . Using integration by parts with u=e^(-st) and dv=cos(5t), we obtain that $F(s)=\frac{1}{5}\sin(5t)e^{-st} |_{0}^{+\infty}+\frac{s}{5}\int_0 ^{+ \infty}e^{-st}\sin(5t) dt=\int_0 ^{+ \infty}e^{-st}\sin(5t) dt$ .

Using integration by parts again with u=e^(-st) and dv=sin(5t), we obtain that

$F(s)=\frac{s}{5}(\frac{-1}{5}\cos(5t)e^{-st} |_{0}^{+\infty}-\frac{s}{5}\int_0 ^{+ \infty}e^{-st}\sin(5t) dt)=\frac{s}{5}(\frac{1}{5}-\frac{s}{5}\int_0^{+ \infty}e^{-st}\sin(5t) dt)=\frac{s}{5}-\frac{s^2}{25}F(s)$ .