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3 years ago

2.

Mathematics

1 answer:

zaharov [31]3 years ago

8 0

Answer:

(#1 starts at "has two pairs of parallel sides" and checkmark=c, while no checkmark=x)

Step-by-step explanation:

1. c

2. x

3. x

4. x

5. c

6. x

7. c

8. x

9. c

10. c

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Drag the tiles to the correct boxes to complete the pairs. Not all tiles will be used.

Katena32 [7]

Answer:

(a)

$5^{-3}=\frac{1}{125}$

(b)

$-5^{-3}=-\frac{1}{125}$

(c)

$(-5^{-3})^{-1}=-125$

(d)

$(-5^{-3})^{0}=1$

Step-by-step explanation:

(a)

$5^{-3}$

we can use property of exponent

$a^{-n}=\frac{1}{a^n}$

we get

$5^{-3}=\frac{1}{5^3}$

$5^{-3}=\frac{1}{5\times 5\times 5}$

$5^{-3}=\frac{1}{125}$ ........Answer

(b)

$-5^{-3}$

we can use property of exponent

$a^{-n}=\frac{1}{a^n}$

we get

$-5^{-3}=-\frac{1}{5^3}$

$-5^{-3}=-\frac{1}{5\times 5\times 5}$

$-5^{-3}=-\frac{1}{125}$ ........Answer

(c)

$(-5^{-3})^{-1}$

we can use property of exponent

$(a^{n})^m=a^{m\times n}$

we get

$(-5^{-3})^{-1}=(-5)^{-3\times -1}$

$(-5^{-3})^{-1}=(-5)^3$

$(-5^{-3})^{-1}=(-5)\times (-5)\times (-5)$

$(-5^{-3})^{-1}=-125$ ........Answer

(d)

$(-5^{-3})^{0}$

we can use property of exponent

$(a^{n})^m=a^{m\times n}$

we get

$(-5^{-3})^{0}=(-5)^{-3\times 0}$

$(-5^{-3})^{-1}=(-5)^0$

we can use property

$a^0=1$

$(-5^{-3})^{0}=1$ ........Answer

4 0

3 years ago

For the function y=-3+ cos (x + 4)], what is the minimum value?

Mrac [35]

Answer:

-4 is the mimumum of y=-3+cos(x+4)

Step-by-step explanation:

The minimum value of y=cos(x) is -1.

The minimum value of y=cos(x+4) is still -1; the +4 inside the cosine function only affected the horizontal shift.

The minimum value of y=-3+cos(x+4) is -3-1 which is -4. This brought the graph down 3 units so if the minimum was previously -1 and it got brought down 3 units then it's new minimum is -4.

5 0

3 years ago

Write the expression seven less than n

melamori03 [73]

Answer:

$n-7$

Step-by-step explanation:

If the expression is 'seven less than n', we are subtracting 'seven' from 'n'.

6 0

3 years ago

Can someone help me with this one please ? Thanks!

inysia [295]

I hope this helps you

4 0

3 years ago

3. At noon Lan is in his blue mini-van 300 km north of Makenna in her Bugatti. Lan heads south

LiRa [457]

Answer:

The rate at which the distance between them is changing at 2:00 p.m. is approximately 1.92 km/h

Step-by-step explanation:

At noon the location of Lan = 300 km north of Makenna

Lan's direction = South

Lan's speed = 60 km/h

Makenna's direction and speed = West at 75 km/h

The distance Lan has traveled at 2:00 PM = 2 h × 60 km/h = 120 km

The distance north between Lan and Makenna at 2:00 p.m = 300 km - 120 km = 180 km

The distance West Makenna has traveled at 2:00 p.m. = 2 h × 75 km/h = 150 km

Let 's' represent the distance between them, let 'y' represent the Lan's position north of Makenna at 2:00 p.m., and let 'x' represent Makenna's position west from Lan at 2:00 p.m.

By Pythagoras' theorem, we have;

s² = x² + y²

The distance between them at 2:00 p.m. s = √(180² + 150²) = 30·√61

ds²/dt = dx²/dt + dy²/dt

2·s·ds/dt = 2·x·dx/dt + 2·y·dy/dt

2×30·√61 × ds/dt = 2×150×75 + 2×180×(-60) = 900

ds/dt = 900/(2×30·√61) ≈ 1.92

The rate at which the distance between them is changing at 2:00 p.m. ds/dt ≈ 1.92 km/h

5 0

3 years ago