Geometry Proof - Thanks for any help!

Which is the value of this expression when p=-2 and q=-1?

saul85 [17]

Answer:

D. 4

Step-by-step explanation:

$[(p^2) (q^{-3}) ]^{-2}.[(p)^{-3}(q)^5] ^{-2}\\\\=[(p^2) (q^{-3}) \times(p)^{-3}(q)^5 ]^{-2}\\\\=[(p^{2}) \times(p)^{-3} \times (q^{-3}) \times(q)^5 ]^{-2}\\\\=[(p^{2-3}) \times (q^{5-3}) ]^{-2}\\\\=[(p^{-1}) \times (q^{2}) ]^{-2}\\\\=(p^{-1\times (-2)}) \times (q^{2\times (-2) }) \\\\=p^{2}\times q^{-4} \\\\= \frac{p^2}{q^4}\\\\= \frac{(-2)^2}{(-1)^4}\\\\= \frac{4}{1}\\\\= 4$

8 0

3 years ago

Fing Using Addil

Alexxx [7]

Answer:

keiko- x

imelda- x+ 36

Sarah- x- 38

Step-by-step explanation:

keiko is x since we have not been given his number of boxes.
because imelda's number of boxes was 36 more than keiko's, you will add the x which is for keiko and 36 for imelda(x+36).
and sarah's number of boxes will be x-36 because we have been told her number of boxes is 38 fewer than  keiko's.

7 0

3 years ago

What is the zero in the quadratic function f(x)= 9x^2-54x-19?

Alex17521 [72]

Answer:

6.33... and 0.333...

Step-by-step explanation:

The quadratic formula is

$x=\frac{-b+\sqrt{b^2-4ac} }{2a}$ .

It is important because while some quadratics are factorable and can be solved not all are. The formula will solve all quadratic equations and can also give both real and imaginary solutions. Using the formula will require less work than finding the factors if factorable. We will substitute a=9, b=-54 and c=-19.

$x=\frac{-b+/-\sqrt{b^2-4ac} }{2a}\\x=\frac{-(-54)+/-\sqrt{(-54)^2-4(9)(-19)} }{2(9)}\\x=\frac{54+/-\sqrt{2916+684} }{18}$

We will now solve for the plus and the minus.

The plus,,,

$x=\frac{54+\sqrt{3600}}{18}\\x=\frac{54+60}{18}\\x=\frac{114}{18}\\x=6.3$

and the minus...

$x=\frac{54-\sqrt{3600}}{18}\\x=\frac{54-60}{18}\\x=\frac{-6}{18}\\x=-0.333...$

8 0

3 years ago

Here's another geometry thing I suck at...

solniwko [45]

A = 133; b = 31; c = 82; d = 64.
Opposite angles in an inscribed quadrilateral are supplementary; this means that d + 116 = 180. Subtracting 116 from both sides, we have d = 64.

By the same theorem, c + 98 = 180; subtracting 98 from both sides, we have c = 82.

Inscribed angles are equal to 1/2 the measure of the intercepted arc. Using this, we have

116 = 1/2(a+99)

Multiplying both sides by 2, we have
232 = a+99

Subtract 99 from both sides:
232 - 99 = a + 99 - 99
133 = a

We also have that
82 = 1/2(133+b)

Multiplying both sides by 2, we have:
164 = 133 + b

Subtract 133 from both sides:
164 - 133 = 133 + b - 133
31 = b

6 0

3 years ago

mr. graves gave her class 12 minutes to read. carrie read 5 1/2 pages in the that time. at what rate,in pages per hour, did carr

GarryVolchara [31]

27.5 pages per an hour

7 0

3 years ago

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