I’m better with square angles :((

Mathematics

2 answers:

Ratling [72]3 years ago

8 0

Another easy way to do this is: break it up into two sections. Break it into a triangle and a rectangle.

The triangle would have the dimensions: 20, 6, and 25.

To find the area take A x B x 1/2 = Area 20 x 6 = 120 1/2 x 120 = 60

Then, the dimensions of the rectangle would be: 30 x 20.

30 x 20 = 600 Then, 600 + 60 = 660 There's your answer!

AfilCa [17]3 years ago

6 0

1/2 (H) (B1+B2)

1/2 (20) (36+30)

10 (96)

660

This is what I got and my work^

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A loaf of bread costs $0.90. Each loaf has 15 slices. How much does each slice of bread cost?

scoray [572]

Answer:

Each slice of bread costs $0.06.

Step-by-step explanation:

To find how much each slice of bread costs, you need to divide the cost (0.90) by how many slices there are (15). When you divide the two, you get 0.06. To check your answer you can do 0.06x15 and that will equal $0.90.

8 0

2 years ago

Read 2 more answers

How to differentiate ?

Bas_tet [7]

Use the power, product, and chain rules:

$y = x^2 (3x-1)^3$

• product rule

$\dfrac{\mathrm dy}{\mathrm dx} = \dfrac{\mathrm d(x^2)}{\mathrm dx}\times(3x-1)^3 + x^2\times\dfrac{\mathrm d(3x-1)^3}{\mathrm dx}$

• power rule for the first term, and power/chain rules for the second term:

$\dfrac{\mathrm dy}{\mathrm dx} = 2x\times(3x-1)^3 + x^2\times3(x-1)^2\times\dfrac{\mathrm d(3x-1)}{\mathrm dx}$

• power rule

$\dfrac{\mathrm dy}{\mathrm dx} = 2x\times(3x-1)^3 + x^2\times3(3x-1)^2\times3$

Now simplify.

$\dfrac{\mathrm dy}{\mathrm dx} = 2x(3x-1)^3 + 9x^2(3x-1)^2 \\\\ \dfrac{\mathrm dy}{\mathrm dx} = x(3x-1)^2 \times (2(3x-1) + 9x) \\\\ \boxed{\dfrac{\mathrm dy}{\mathrm dx} = x(3x-1)^2(15x-2)}$

You could also use logarithmic differentiation, which involves taking logarithms of both sides and differentiating with the chain rule.

On the right side, the logarithm of a product can be expanded as a sum of logarithms. Then use other properties of logarithms to simplify

$\ln(y) = \ln\left(x^2(3x-1)^3\right) \\\\ \ln(y) = \ln\left(x^2\right) + \ln\left((3x-1)^3\right) \\\\ \ln(y) = 2\ln(x) + 3\ln(3x-1)$

Differentiate both sides and you end up with the same derivative:

$\dfrac1y\dfrac{\mathrm dy}{\mathrm dx} = \dfrac2x + \dfrac9{3x-1} \\\\ \dfrac1y\dfrac{\mathrm dy}{\mathrm dx} = \dfrac{15x-2}{x(3x-1)} \\\\ \dfrac{\mathrm dy}{\mathrm dx} = \dfrac{15x-2}{x(3x-1)} \times x^2(3x-1)^3 \\\\ \dfrac{\mathrm dy}{\mathrm dx} = x(15x-2)(3x-1)^2$