F(x)=x^2+3x+4 is this quadratic, linear or neither????

Find the derivative of y=sin^(2)(x)cos^(2)(x)

wariber [46]

$sin^{2}x \ cos^{2} x= \frac{1}{2} (1-cos2x) \times \frac{1}{2} (1+cos2x) \\ = \frac{1}{4} (1- cos^{2} 2x) \\ = \frac{1}{4} ( sin^{2} 2x +cos^{2} 2x- cos^{2} 2x) \\ = \frac{1}{4} ( sin^{2} 2x) \\ = \frac{1}{4} ( \frac{1}{2} (1-cos2(2x))) \\ = \frac{1}{8} (1-cos4x) \\ \frac{d}{dx} (sin^{2}x \ cos^{2} x)= \frac{1}{8} \frac{d}{dx} (1-cos4x) \\ = \frac{1}{8} [-4(-sin4x) \\ \frac{1}{2} sin4x$

5 0

3 years ago

What is the solution to this equation x + 14= -6

Inga [223]

Answer:

This is linear equation you're asked to find C I guess so collect like terms and find X

x=-6-14

x=-20

5 0

3 years ago

When a number cannot be divided evenly, the amount left over is called the

scoundrel [369]

Answer:

The remainder.

Step-by-step explanation:

I hope this helps.

8 0

3 years ago

5/6(2x+12)-8=4-(x+1)

alekssr [168]

For this case we have the following equation:

$\frac {5} {6} (2x + 12) -8 = 4- (x + 1)$

We apply distributive property on the left side of the equation:

$\frac {5 * 2} {6} x + \frac {5 * 12} {6} -8 = 4- (x + 1)\\\frac {10} {6} x + \frac {60} {6} -8 = 4- (x + 1)$

We simplify the left side of the equation:

$\frac {5} {3} x + 10-8 = 4- (x + 1)$

Different signs are subtracted and the major sign is placed.

$\frac {5} {3} x + 2 = 4- (x + 1)$

On the right side we must take into account that:

$- * + = -$

So:

$\frac {5} {3} x + 2 = 4-x-1\\\frac {5} {3} x + 2 = 3-x$

We add x to both sides of the equation:

$\frac {5} {3} x + x + 2 = 3\\\frac {5 * 1 + 3 * 1} {3} x + 2 = 3\\\frac {5 + 3} {3} x + 2 = 3\\\frac {8} {3} x + 2 = 3$

We subtract 2 from both sides of the equation:

$\frac {8} {3} x = 3-2\\\frac {8} {3} x = 1$

We multiply by 3 on both sides of the equation:

$8x = 3$

We divide by 8 on both sides of the equation:

$x = \frac {3} {8}$

Thus, the solution of the equation is:

$x = \frac {3} {8}$

Answer:

$x = \frac {3} {8}$

4 0

4 years ago

Could anyone help me with these?

lora16 [44]

Part A:

Derrick is incorrect.
A yard is equivalent to 3 feet.
Thus, 20 yards will be equivalent to 60 feet.

The width is 20 yard and 2 feet. It's the same as 60 feet and 2 feet, or 62 feet. Thus, Derrick is incorrect.

Part B:

Let's calculate the area of the parking lot.

We know the width is 62 feet from part A.
The length is 30 yards, or 90 feet.

Multiply the length and width to get the area in square feet.

$90 \times 62=5580$

The parking lot has an area of 5580 square foot.
It costs 2 dollars per square foot to repave the parking lot. Multiply the area by 2 dollars.

$5580 \times 2=11160$

It will cost $11,160 to repave the whole parking lot.

4 0

3 years ago

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