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3 years ago

Can I get help on 1, 2, and 3 ?? Please

Mathematics

1 answer:

Virty [35]3 years ago

8 0

Hi there! I’m happy to help you!

1. The length of one side of the park is 60 meters. 3,600 squares is 60.

2. The length of the edge of the container is 20. 8,000 cubed is 20.

3. The area of the base will definitely be greater than one square foot. 1,000 cubed is 10. 10 * 10 is 100. 100 square inches is equivalent to 8 1/3 square feet.

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what is the area of rectangle PQRS with vertices at P (0,2) Q (0,10) R (4,10) S (4,2)? A. 12 square units B. 24 C. 32 D. 40

KatRina [158]

D is the answer mate

8 0

3 years ago

Read 2 more answers

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

Help.... Other question got deleted.

Helen [10]

-2/25 is the slope
x1=4
hope this helps!

7 0

3 years ago

Value of x in log5x = 4logx5 is

olchik [2.2K]

Log5 x=4 logx 5
Ln x/ln5=4(ln5 / ln x) logx z=ln x / ln z
least common multiple=ln5 * ln x

ln²x=4ln²5
√(ln² x)=√(4 ln²5)
ln x=2ln 5
x=e^2ln5=25

Answer: x=25.

4 0

3 years ago

Find x?<br> In 3x - In(x - 4) = ln(2x - 1) +ln3

earnstyle [38]

Answer:

$x = \displaystyle \frac{5 + \sqrt{17}}{2}$ .

Step-by-step explanation:

Because $3\, x$ is found in the input to a logarithm function in the original equation, it must be true that $3\, x > 0$ . Therefore, $x > 0$ .

Similarly, because $(x - 4)$ and $(2\, x - 1)$ are two other inputs to the logarithm function in the original equation, they should also be positive. Therefore, $x > 4$ .

Let $a$ and $b$ represent two positive numbers (that is: $a > 0$ and $b > 0$ .) The following are two properties of logarithm:

$\displaystyle \ln (a) + \ln(b) = \ln\left(a \cdot b\right)$ .

$\displaystyle \ln (a) - \ln(b) = \ln\left(\frac{a}{b}\right)$ .

Apply these two properties to rewrite the original equation.

Left-hand side of this equation:

$\begin{aligned}&\ln(3\, x) - \ln(x - 4)= \ln\left(\frac{3\, x}{x -4}\right)\end{aligned}$

Right-hand side of this equation:

$\ln(2\, x- 1) + \ln(3) = \ln\left(3 \left(2\, x - 1\right)\right)$ .

Equate these two expressions:

$\begin{aligned}\ln\left(\frac{3\, x}{x -4}\right) = \ln(3(2\, x - 1))\end{aligned}$ .

The natural logarithm function $\ln$ is one-to-one for all positive inputs. Therefore, for the equality $\begin{aligned}\ln\left(\frac{3\, x}{x -4}\right) = \ln(3(2\, x - 1))\end{aligned}$ to hold, the two inputs to the logarithm function have to be equal and positive. That is:

$\displaystyle \frac{3\ x}{x - 4} = 3\, (2\, x - 1)$ .

Simplify and solve this equation for $x$ :

$x^2 - 5\, x + 2 = 0$ .

There are two real (but not rational) solutions to this quadratic equation: $\displaystyle \frac{5 + \sqrt{17}}{2}$ and $\displaystyle \frac{5 - \sqrt{17}}{2}$ .

However, the second solution, $\displaystyle \frac{5 - \sqrt{17}}{2}$ , is not suitable. The reason is that if $x = \displaystyle \frac{5 - \sqrt{17}}{2}$ , then $(x - 4)$ , one of the inputs to the logarithm function in the original equation, would be smaller than zero. That is not acceptable because the inputs to logarithm functions should be greater than zero.

The only solution that satisfies the requirements would be $\displaystyle \frac{5 + \sqrt{17}}{2}$ .

Therefore, $x = \displaystyle \frac{5 + \sqrt{17}}{2}$ .

7 0

3 years ago