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3 years ago

10

Can solve real world problems involving embedded triangles:

1 answer:

Anuta_ua [19.1K]3 years ago

3 0

Answer:

It is 15 x=15

Step-by-step explanation:

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Pasta comes with tomato sauce and can be ordered with some, all, or none of these ingredients in the sauce: {onions, garlic, car

Radda [10]

Answer:

FORMULA USED-

NUMBER OF SUBSETS OF A SET WITH n ELEMENTS= 2^n

Here our set is { onions, garlic, carrots, brocoli, shrimp, mushrooms, zucchini, green pepper}

Different variations available for ordering pasta with tomato sauce= 28 = 256 ( As here n= 8)

8 0

3 years ago

Please please please help asap

blondinia [14]

Answer:

The answer is 22 cards are left in the stack.

Step-by-step explanation:

7 0

3 years ago

Read 2 more answers

<img src="https://tex.z-dn.net/?f=%20%28%7B%20%7Bx%7D%5E%7B2%7D%20%20-%204%7D%29%5E%7B5%7D%20%28%20%7B4x%20-%205%7D%29%5E%7B4%7D

Makovka662 [10]

Let $u=x^2-4$ and $v=4x-5$ . By the product rule,

$\dfrac{\mathrm d(u^5v^4)}{\mathrm dx}=\dfrac{\mathrm d(u^5)}{\mathrm dx}v^4+u^5\dfrac{\mathrm d(v^4)}{\mathrm dx}$

By the power rule, we have $(u^5)'=5u^4$ and $(v^4)'=4v^3$ , but $u,v$ are functions of $x$ , so we also need to apply the chain rule:

$\dfrac{\mathrm d(u^5)}{\mathrm dx}=5u^4\dfrac{\mathrm du}{\mathrm dx}$

$\dfrac{\mathrm d(v^4)}{\mathrm dx}=4v^3\dfrac{\mathrm dv}{\mathrm dx}$

and we have

$\dfrac{\mathrm du}{\mathrm dx}=2x$

$\dfrac{\mathrm dv}{\mathrm dx}=4$

So we end up with

$\dfrac{\mathrm d(u^5v^4)}{\mathrm dx}=10xu^4v^4+16u^5v^3$

Replace $u,v$ to get everything in terms of $x$ :

$\dfrac{\mathrm d((x^2-4)^5(4x-5)^4)}{\mathrm dx}=10x(x^2-4)^4(4x-5)^4+16(x^2-4)^5(4x-5)^3$

We can simplify this by factoring:

$10x(x^2-4)^4(4x-5)^4+16(x^2-4)^5(4x-5)^3=2(x^2-4)^4(4x-5)^3\bigg(5x(4x-5)+8(x^2-4)\bigg)$

$=2(x^2-4)^4(4x-5)^3(28x^2-57)$

7 0

3 years ago

20 POINTS I need your help

disa [49]

The answer to your question is 405

6 0

3 years ago

Read 2 more answers

Subtract the given function and indicate the domain of the difference

maxonik [38]

Answer:

The domain is $x\in (-\infty,\infty).$

Step-by-step explanation:

Given functions $f(x)=x^2+3x+1$ and $g(x)=2x^2-4x-1$

Subtract these two functions:

$f(x)-g(x)\\ \\=(x^2+3x+1)-(2x^2-4x-1)\\ \\=x^2+3x+1-2x^2+4x+1\\ \\=(x^2-2x^2)+(3x+4x)+(1+1)\\ \\=-x^2+7x+2$

Plot these difference on the coordinate plane (see attached diagram). This function is defined for all vlues of x, so the domain is $x\in (-\infty,\infty).$

6 0

4 years ago