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3 years ago

Simplify 6r · s · 4rt. 10rst

Mathematics

2 answers:

Andrew [12]3 years ago

5 0

Answer:

24r^2ts

Have a good day :)

Scrat [10]3 years ago

4 0

If the problem is 6r•s•4rt•10rst, then the answer is 240•r^3•s^2•t^2.

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HELPPP ASAP!!! GIVING BRAINLIEST!!!!

diamong [38]

Answer:

x = -4 and x = -2

Step-by-step explanation:

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2 years ago

Find the point(s) of intersection between the quadratic function y = x + 4x + 4 and the linear function y = 2x+4.

Andrews [41]

Answer:

(0, 4)

Step-by-step explanation:

To find the intersection of two lines, we want to find the value when they equal each other. To do this, we want to set the equations equal to each other.

First, let's simplify y = x + 4x + 4 by combining the x's.

y = 5x + 4

Now let's set the equations equal to each other. Since they both equal y, we can set the opposite sides equal to each other.

5x + 4 = 2x + 4

Now you want to combine the terms.

[subtract 4] 5x = 2x

[subtract 2x] 3x = 0

Now you want to isolate the x.

[divide by 3] x = 0

Now we want to find y by plugging x = 0 back into the equations.

y = 5(0) + 4

[multiply] y = 0 + 4

[add] y = 4

Check this with the other equation.

y = 2(0) + 4

[multiply] y = 0 + 4

[add] y = 4

Your answer is correct!

(0, 4)

4 0

2 years ago

Help on 3 I do not get it

Rama09 [41]

The answer is nine hope this helps :)

8 0

3 years ago

Tia, Veronica, Pam, and Lily are sisters. Tia is 8 years old and she is 2 years older than Pam. Pam is 5 years younger than Vero

JulijaS [17]

Answer:

Pam 6

Tia 8

Veronica 11

Lily 15

Step-by-step explanation:

Tia's age = 8

Pam's age = Tia's age - 2

8 - 2 = 6

Veronica = pam's age + 5

6 + 5 = 11

Lily's age = veronica's age + 4

11 + 4 = 15

3 0

3 years ago

Find dy/dx if y =x^3+5x+2/x²-1

stiks02 [169]

<u>Differentiate using the Quotient Rule</u> –

$\qquad$ $\pink{\twoheadrightarrow \sf \dfrac{d}{dx} \bigg[\dfrac{f(x)}{g(x)} \bigg]= \dfrac{ g(x)\:\dfrac{d}{dx}\bigg[f(x)\bigg] -f(x)\dfrac{d}{dx}\:\bigg[g(x)\bigg]}{g(x)^2}}\\$

According to the given question, we have –

f(x) = x^3+5x+2
g(x) = x^2-1

Let's solve it!

$\qquad$ $\green{\twoheadrightarrow \bf \dfrac{d}{dx}\bigg[ \dfrac{x^3+5x+2 }{x^2-1}\bigg]} \\$

$\qquad$ $\twoheadrightarrow \sf \dfrac{(x^2-1) \dfrac{d}{dx}(x^3+5x+2) - ( x^3+5x+2) \dfrac{d}{dx}(x^2-1)}{(x^2-1)^2 }\\$

$\qquad$ $\twoheadrightarrow \sf \dfrac{(x^2-1)(3x^2+5) - ( x^3+5x+2) 2x}{(x^2-1)^2 }\\$

$\qquad$ $\pink{\sf \because \dfrac{d}{dx} x^n = nx^{n-1} }\\$

$\qquad$ $\twoheadrightarrow \sf \dfrac{3x^4+5x^2-3x^2-5-(2x^4+10x^2+4x)}{(x^2-1)^2 }\\$

$\qquad$ $\twoheadrightarrow \sf \dfrac{3x^4+5x^2-3x^2-5-2x^4-10x^2-4x}{(x^2-1)^2 }\\$

$\qquad$ $\green{\twoheadrightarrow \bf \dfrac{x^4-8x^2-4x-5}{(x^2-1)^2 }}\\$

$\qquad$ $\pink{\therefore \bf{\green{\underline{\underline{\dfrac{d}{dx} \dfrac{x^3+5x+2 }{x^2-1}} = \dfrac{x^4-8x^2-4x-5}{(x^2-1)^2 }}}}}\\\\$

7 0

1 year ago