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3 years ago

If A and B are nonzero integers with the same signs is the product negative or postitive

1 answer:

3 0

Remember.

positive number x positive number = positive number
positive number x negative number = negative number
negative number x positive number = negative number
negative number x negative number = positive number

So if both numbers have the same sign, either both positive or both negative, then the product will be positive.

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Solve the equation 2/5x - 1 = 9. Complete the solution and check your solution.

sukhopar [10]

2/5x - 1 = 9
+1 +1
10
—— ——
2/5 2/5

x = 50

6 0

3 years ago

Write and simplify an expression to represent the perimeter of the triangle shown. Then find the value of x if the perimeter is

Pavlova-9 [17]

Answer:

Step-by-step explanation:

x +4+2x+1+3x-2 = perimeter of triangle.

(6x +3)ft = perimeter

given perimeter = 45feet

hence,

6x + 3 = 45

6x = 42

x = 7

Hope it helped!! plz mark brainliest!! have a good day !! :)

7 0

3 years ago

A company made $4 million in the second quarter. This is 1/3 more than it made in the first quarter

Paraphin [41]

It made 3 million in the first quarter

8 0

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!

5 0

3 years ago

PLSSSS!!!!! HELP!!!! ME!!!!! I !!!!! WILL!!!! MARK U!!!!!!!!!!!!

lidiya [134]

Answer:

i want to see david not mark

Step-by-step explanation:

duck what is plant? Been wonder tht=at fir long tim. Seen mary sheelly at the frank book!! You seen? HMMMM???!?! Im tired of this instagram ad go away! Its find out!!! Me too!!!

7 0

3 years ago

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