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3 years ago

15

If

x-formula"> and $y < 0$ ,where is the point (x,y) located?
a) quadrant I
b) quadrant II
c) quadrant III
d) quadrant IV

2 answers:

Dmitry_Shevchenko [17]3 years ago

7 0

Answer:

Quadrant III

Step-by-step explanation:

x < 0 is the area to the left of the y-axis, and y < 0 is the area beneath the x-axis. This point is therefore located in Quadrant III.

miss Akunina [59]3 years ago

5 0

Answer:

c) quadrant III

Step-by-step explanation:

I will be quick in this question:

$\text {For }y$

$\text {For }x$

What both have in common?

So, the answer is <u>Quadrant III</u><u>.</u>

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Which is the value of the expression (StartFraction (10 Superscript 4 Baseline) (5 squared) Over (10 cubed) (5 cubed)) cubed?

Flura [38]

Answer:

The value to the given expression is 8

Therefore $\left[\frac{(10^4)(5^2)}{(10^3)(5^3)}\right]^3=8$

Step-by-step explanation:

Given expression is (StartFraction (10 Superscript 4 Baseline) (5 squared) Over (10 cubed) (5 cubed)) cubed

Given expression can be written as below

$\left[\frac{(10^4)(5^2)}{(10^3)(5^3)}\right]^3$

To find the value of the given expression:

$\left[\frac{(10^4)(5^2)}{(10^3)(5^3)}\right]^3=\frac{((10^4)(5^2))^3}{((10^3)(5^3))^3}$

( By using the property ( $(\frac{a}{b})^m=\frac{a^m}{b^m}$ )

$=\frac{(10^4)^3(5^2)^3}{(10^3)^3(5^3)^3}$

( By using the property $(ab)^m=a^mb^m$ )

$=\frac{(10^{12})(5^6)}{(10^9)(5^9)}$

( By using the property $(a^m)^n=a^{mn}$ )

$=(10^{12})(5^6)(10^{-9})(5^{-9})$

( By using the property $\frac{1}{a^m}=a^{-m}$ )

$=(10^{12-9})(5^{6-9})$ (By using the property $a^m.b^n=a^{m+n}$ )

$=(10^3)(5^{-3})$

$=\frac{10^3}{5^3}$ ( By using the property $a^{-m}=\frac{1}{a^m}$ )

$=\frac{1000}{125}$

$=8$

Therefore $\left[\frac{(10^4)(5^2)}{(10^3)(5^3)}\right]^3=8$

Therefore the value to the given expression is 8

3 0

3 years ago

Read 2 more answers

I had a summer job cooking two items at a bbq place. I sell burgers for $9.00 and sell chicken sandwiches for $11.25. Because I

pishuonlain [190]

Answer:

He sold 180 chicken sandwiches and 150 burguers.

Step-by-step explanation:

Since 330 items were sold, then the amount of burguers and chicken sandwiches sold must be equal to this value. From this information we have:

burguers + sandwiches = 330

We know that each burguer cost $9 and each chicken sandwich cost $11.25 and that we made a total of $3,375. So we have:

9*burguers + 11.25*sandwiches = 3,375

From the first equation we have:

burguers = 330 - sandwiches

Applying it on the second equation:

9*(330 - sandwiches) + 11.25*sandwiches = 3,375

2,970 - 9*sandwiches + 11.25*sandwiches = 3,375

2.25*sandwiches = 3,375 - 2,970

2.25*sandwiches = 405

sandwiches = 405/2.25 = 180

So,

burguers = 330 - 180 = 150

He sold 180 chicken sandwiches and 150 burguers.

8 0

3 years ago

Hey:) can you please help me posted picture of question

Kryger [21]

To expand the given expression we proceed as follows:
(6x²-2x-6)(8x²+7x+8)
=6x²(8x²+7x+8)-2x(8x²+7x+8)-6(8x²+7x+8)
=48x⁴+42x³+48x²-16x³-14x²-16x-48x²-42x-48
putting like terms together:
48x⁴+(42x³-16x³)+(48x²-48x²)+(-16x-42x)-48
=48x⁴+26x³+0x²-58x-48
hence the answer is:
48x⁴+26x³-58x-48

6 0

3 years ago

Sean and Greg share a 24-ounce bucket of clay. By the end of the week, Sean has used 3 8 of the bucket, and Greg has used 1 4 of

kozerog [31]

Answer:

9

Step-by-step explanation:

Sean and Greg share a 24-ounce bucket of clay

Sean uses 3/8 of the bucket

= 3/8 ×24

= 9 ounce

Greg uses 1/4 if the bucket

= 1/4 ×24

= 6 ounce

= 9+6

= 15 ounce

Therefore the quantity left in the bucket can be calculated as follows

= 24-15

= 9

Hence 9 ounces are left in the bucket

4 0

3 years ago

The amount A of the radioactive element radium in a sample decays at a rate proportional to the amount of radium present. Given

slavikrds [6]

Answer:

a) $\frac{dm}{dt} = -k\cdot m$ , b) $m(t) = m_{o}\cdot e^{-\frac{t}{\tau} }$ , c) $m(t) = 10\cdot e^{-\frac{t}{2438.155} }$ , d) $m(300) \approx 8.842\,g$

Step-by-step explanation:

a) Let assume an initial mass m decaying at a constant rate k throughout time, the differential equation is:

$\frac{dm}{dt} = -k\cdot m$

b) The general solution is found after separating variables and integrating each sides:

$m(t) = m_{o}\cdot e^{-\frac{t}{\tau} }$

Where $\tau$ is the time constant and $k = \frac{1}{\tau}$

c) The time constant is:

$\tau = \frac{1690\,yr}{\ln 2}$

$\tau = 2438.155\,yr$

The particular solution of the differential equation is:

$m(t) = 10\cdot e^{-\frac{t}{2438.155} }$

d) The amount of radium after 300 years is:

$m(300) \approx 8.842\,g$

4 0

3 years ago

Read 2 more answers