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3 years ago

6

Please help me, im so confused it's "Anthony owns a $133,000 brick home in Area B. How much will it cost him to insure the home

each year with a homeowners insurance policy?" but area b says
brick | 0.42%
wood frame | 0.54%

1 answer:

ExtremeBDS [4]3 years ago

8 0

Hi The answer is wood frame

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D

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The area (A) of a kite is calculated as

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This can be represented as

Multiply the diagonals then divide by 2 → D

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A right triangle has an area of 98 square inches. If the triangle is not an Isosceles triangle, what are the all possible length

VladimirAG [237]

Answer:

The possible length of the triangle =

1) (1inches, 196inches)

2) (2incheq, 98inches)

3) (4inches , 49 inches)

4) (7 inches , 28 inches)

Step-by-step explanation:

We are told the above Triangle is not an Isosceles Triangle

Hence, we assume it is a right angle triangle

The area of a triangle is = 1/2 × Base × Height

= let us represent Base and Height = x

Hence:

1/2 × x × x = 98

x² /2 = 98

Cross Multiply

x² = 98 × 2

x² = 196

Step 2

We find the factors of 196

1× 196 = 196 (1, 196)

2 × 98 = 196 (2, 98)

4 ×49 = 196 (4, 49)

7 × 28 = 196 (7, 28)

Therefore, all the possible length of the triangle =

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2) (2incheq, 98inches)

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3 years ago

Find dy/dx by implicit differentiation for ysin(y) = xcos(x)

tatyana61 [14]

Answer:

$\frac{dy}{dx}=\frac{\cos(x)-x\sin(x)}{\sin(y)+y\cos(y)}$

Step-by-step explanation:

So we have:

$y\sin(y)=x\cos(x)$

And we want to find dy/dx.

So, let's take the derivative of both sides with respect to x:

$\frac{d}{dx}[y\sin(y)]=\frac{d}{dx}[x\cos(x)]$

Let's do each side individually.

Left Side:

We have:

$\frac{d}{dx}[y\sin(y)]$

We can use the product rule:

$(uv)'=u'v+uv'$

So, our derivative is:

$=\frac{d}{dx}[y]\sin(y)+y\frac{d}{dx}[\sin(y)]$

We must implicitly differentiate for y. This gives us:

$=\frac{dy}{dx}\sin(y)+y\frac{d}{dx}[\sin(y)]$

For the sin(y), we need to use the chain rule:

$u(v(x))'=u'(v(x))\cdot v'(x)$

Our u(x) is sin(x) and our v(x) is y. So, u'(x) is cos(x) and v'(x) is dy/dx.

So, our derivative is:

$=\frac{dy}{dx}\sin(y)+y(\cos(y)\cdot\frac{dy}{dx}})$

Simplify:

$=\frac{dy}{dx}\sin(y)+y\cos(y)\cdot\frac{dy}{dx}}$

And we are done for the right.

Right Side:

We have:

$\frac{d}{dx}[x\cos(x)]$

This will be significantly easier since it's just x like normal.

Again, let's use the product rule:

$=\frac{d}{dx}[x]\cos(x)+x\frac{d}{dx}[\cos(x)]$

Differentiate:

$=\cos(x)-x\sin(x)$

So, our entire equation is:

$=\frac{dy}{dx}\sin(y)+y\cos(y)\cdot\frac{dy}{dx}}=\cos(x)-x\sin(x)$

To find our derivative, we need to solve for dy/dx. So, let's factor out a dy/dx from the left. This yields:

$\frac{dy}{dx}(\sin(y)+y\cos(y))=\cos(x)-x\sin(x)$

Finally, divide everything by the expression inside the parentheses to obtain our derivative:

$\frac{dy}{dx}=\frac{\cos(x)-x\sin(x)}{\sin(y)+y\cos(y)}$

And we're done!

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3 years ago