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3 years ago

HELP QUICK PLEASE 15 points!!!

Mathematics

1 answer:

Reil [10]3 years ago

3 0

Answer:

I think the ans is option 'a'

Final Answer:

Corresponding Angles Theorum; ∠AGF and ∠EHD are congruent

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Combine like terms: 3p2q2-3p2q3+4p2q3-3p2q2+pq PLEASE HELP!!! ASAP!!!

ch4aika [34]

Answer:

p²q³ + pq and pq(pq² + 1)

Step-by-step explanation:

Given

3p²q² - 3p²q³ +4p²q³ -3p²q² + pq

Required

Collect like terms

We start by rewriting the expression

3p²q² - 3p²q³ +4p²q³ -3p²q² + pq

Collect like terms

3p²q² -3p²q² - 3p²q³ +4p²q³ + pq

Group like terms

(3p²q² -3p²q²) - (3p²q³ - 4p²q³ ) + pq

Perform arithmetic operations on like terms

(0) - (-p²q³) + pq

- (-p²q³) + pq

Open bracket

p²q³ + pq

The answer can be further simplified

Factorize p²q³ + pq

pq(pq² + 1)

Hence, 3p²q² - 3p²q³ +4p²q³ -3p²q² + pq is equivalent to p²q³ + pq and pq(pq² + 1)

7 0

3 years ago

32 is 90% of what number?

spayn [35]

Answer:

$\large\boxed{32\ is\ 90\%\ of\ 35\dfrac{5}{9}}$

Step-by-step explanation:

<h3>METHOD 1:</h3>

$\begin{array}{ccc}32&-&90\%\\\\x&-&100\%\end{array}\qquad\text{cross multiply}\\\\\\90x=3200\qquad\text{divide both sides by 90}\\\\x=\dfrac{3200}{90}\\\\x=\dfrac{320}{9}\\\\x=35\dfrac{5}{9}$

<h3>METHOD 2:</h3>

$\begin{array}{cccc}32&-&90\%&\text{divide both sides by 9}\\\\\dfrac{32}{9}&-&10\%\end{array}\\\\\\90\%+10\%=100\%\to32+\dfrac{32}{9}=32+3\dfrac{5}{9}=35\dfrac{5}{9}$

3 0

3 years ago

Find f(x) and g(x) so the function can be expressed as y = f(g(x)). y= <img src="https://tex.z-dn.net/?f=%5Cfrac%7B2%7D%7Bx%5E2%

lisabon 2012 [21]

Answer:

One possible answer is:

f(x) = (2/x) + 3 and g(x) = x².

Step-by-step explanation:

Explanation:

We are to write this equation as y = f(g(x)). This means we want it to be a composite of functions; in f(x), we take the value of g(x) and use in place of x.

If we let g(x) = x², this means everywhere we see an x in f(x), we will replace it with x². To make our equation y = 2/x² + 3, working backward we would substitute x for x²; this would give us f(x) = 2/x + 3.

6 0

3 years ago

Evaluating integral by net area

N76 [4]

The graph of $f(x)=2$ is a horizontal line. So the area under $f(x)$ on [0, 8] is equal to the area of a rectangle with length 8 and height 2, or 16.

The graph of $g(x)=\sqrt{64-x^2}$ is the upper half of a circle with radius $\sqrt{64}=8$ . It's symmetric about $x=0$ , so on the interval [0, 8], we're considering a quarter of the circle. The area of such a sector is $\frac{\pi 8^2}4=16\pi$ .