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3 years ago

What is the number property of 17+0

Mathematics

1 answer:

vladimir1956 [14]3 years ago

8 0

Answer:

The identity property

Step-by-step explanation:

anything added to 0 is itself.

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3. 24 ten thousands =<br>THOUSANDS<br>ONES<br>Hundreds Tens Ones Hundreds Tens Ones

Liula [17]

huh ??????????? what u trying to say ???

5 0

3 years ago

Is (8,-16) a function

koban [17]

Answer:

No, 8-16 is not a function

8-15a is a function

7 0

3 years ago

A circular swimming pool has a diameter of 40 meters. The depth of the pool is constant along west-east lines and increases line

Nataliya [291]

Answer:

Volume is $2000\pi\ m^{3}$

Solution:

As per the question:

Diameter, d = 40 m

Radius, r = 20 m

Now,

From north to south, we consider this vertical distance as 'y' and height, h varies linearly as a function of y:

iff

h(y) = cy + d

Then

when y = 1 m

h(- 20) = 1 m

1 = c.(- 20) + d = - 20c + d (1)

when y = 9 m

h(20) = 9 m

9 = c.20 + d = 20c + d (2)

Adding eqn (1) and (2)

d = 5 m

Using d = 5 in eqn (2), we get:

$c = \frac{1}{5}$

Therefore,

$h(y) = \frac{1}{5}y + 5$

Now, the Volume of the pool is given by:

$V = \int h(y)dA$

where

A = $r\theta$

$A = rdr\ d\theta$

Thus

$V = \int (\frac{1}{5}y + 5)dA$

$V = \int_{0}^{2\pi}\int_{0}^{20} (\frac{1}{5}rsin\theta + 5) rdr\ d\theta$

$V = \int_{0}^{2\pi}\int_{0}^{20} (\frac{1}{5}r^{2}sin\theta + 5r}) dr\ d\theta$

$V = \int_{0}^{2\pi} (\frac{1}{15}20^{3}sin\theta + 1000) d\theta$

$V = [- 533.33cos\theta + 1000\theta]_{0}^{2\pi}$

$V = 0 + 2\pi \times 1000 = 2000\pi\ m^{3}$

7 0

3 years ago

How many minutes equal 4.8 days? Enter your answer in the box.

vitfil [10]

1hour=60min
24hour=?
24hour=1440 min

1d=1440 min
4.8d=?
4.8 days =6956 hope you like my explanation

6 0

3 years ago

The graph of F(x) can be stretched vertically and flipped over the x axis to produce the graph of G(x) if F(x)=x^2 which of the

ladessa [460]

Answer:

g(x) = -5x²

(option B)

Step-by-step explanation:

we know that our original graph, f(x) = x² is a parabola.

So, we can consider what happens when we adjust the function/equation of a parabola.

when we "vertically stretch" a parabola, we are increasing the value of x.

think of it this way: the steepness of a slope is rise over run. If we rise ten, and run one, that's going be a lot more steep than if we rise 1, run 1.

Let's say our x = 5

if f(x)=x²

f(5) = 25

> y value / steepness is 25

f(x) = 3x²

f(5) = 75

> y value / steepness is 75

So, we are looking for an equation with an increase in x present.

When a parabola has been flipped over the x-axis, we know that the original equation now includes a negative

suppose that x = 1

if y = x² ; y = 1² = 1

if y = -(x²) ; y = -(1²) ; y = -1

So, when we set x to be negative, we make our y-values end up as negative also (which makes the graph look as if it has been flipped upside-down)

This means that we are looking for a function with a negative x value.

So, we are looking for a negative x-value that is multiplied by a number >1

The graph that fits our requirements is g(x) = -5x²

hope this helps!!

6 0

2 years ago