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3 years ago

A set of 12 windup toys costs $9. What is the cost per windup toy?

Mathematics

2 answers:

-BARSIC- [3]3 years ago

4 0

Answer:

0.75

Step-by-step explanation:

9 ÷ 12 = 0.75

hope this helps...

Ann [662]3 years ago

4 0

Answer:

9/12 = 3/4 = 0.75

Step-by-step explanation:

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T=2π√h+k /gh. Find k given that 2π2= 10

Arte-miy333 [17]

Answer:

Bne matematikçi değilim^o^

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2 years ago

Sean b= 5,79; c= 10,4,el angulo A= 54,46°, el angulo C mide ?

Rzqust [24]

Answer:

$C \approx 91.732^{\circ}$

Step-by-step explanation:

(This exercise is presented in Spanish and for that reason explanation will be held in such language)

El lado restante se determina por la Ley del Coseno:

$a = \sqrt{b^{2}+c^{2}-2\cdot b\cdot c \cdot \cos A}$

$a = \sqrt{5.79^{2}+10.4^{2}-2\cdot (5.79)\cdot (10.4)\cdot \cos 54.46^{\circ}}$

$a \approx 8.466$

Finalmente, el angulo C se halla por medio de la misma ley:

$\cos C = - \frac{c^{2}-a^{2}-b^{2}}{2\cdot a \cdot b}$

$\cos C = -\frac{10.4^{2}-8.466^{2}-5.79^{2}}{2\cdot (8.466)\cdot (5.79)}$

$\cos C = -0.030$

$C \approx 91.732^{\circ}$

3 0

3 years ago

Evaluate the integral by interpreting it in terms of areas Draw a picture of the region the integral

denis23 [38]

Answer:

Step-by-step explanation:

The picture is below of how to separate this into 2 different regions, which you have to because it's not continuous over the whole function. It "breaks" at x = 2. So the way to separate this is to take the integral from x = 0 to x = 2 and then add it to the integral for x = 2 to x = 3. In order to integrate each one of those "parts" of that absolute value function we have to determine the equation for each line that makes up that part.

For the integral from [0, 2], the equation of the line is -3x + 6;

For the integral from [2, 3], the equation of the line is 3x - 6.

We integrate then:

$\int\limits^2_0 {-3x+6} \, dx+\int\limits^3_2 {3x-6} \, dx$ and

$-\frac{3x^2}{2}+6x\left \} {{2} \atop {0}} \right. +\frac{3x^2}{2}-6x\left \} {{3} \atop {2}} \right.$ sorry for the odd representation; that's as good as it gets here!