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3 years ago

What is the value of x in this proportion?

Mathematics

1 answer:

MatroZZZ [7]3 years ago

3 0

X=1/9 I just took a test

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Pls help ! good bless you If f(x)=3x-2√x . find f'(1) , f'(4) and f'(a^2)

Alexxandr [17]

$DERIVATIVES \\ \\ \\ We're \: given \: the \: expression \: \: - \\ \\ f(x) \: = 3x - 2 \sqrt{x} \\ \\ f'(x)= 3 - 2( \frac{1}{2 \sqrt{x} } ) \\ \\ f'(x)=3 - \frac{1}{ \sqrt{x} } \\ \\ Now \: , \: We've \: to \: find \: the \: value \: \\ of \: f'(1) \: \: f'(4 ) \: \: and \: \: f'( {a}^{2} ) \\ \\ \\ f'(1) = 3 - \frac{1}{ \sqrt{1} } = 3 - 1 = 2 \\ \\ \\ f'(4) = 3 - \frac{1}{ \sqrt{4} } = 3 - \frac{1}{2} = \frac{5}{2} \\ \\ \\ f'( {a}^{2} ) = 3 - \frac{1}{ \sqrt{ {a}^{2} } } = 3 - \frac{1}{a} = \frac{3a - 1}{a}$

3 0

3 years ago

Using the following equation, find the center and radius: x2 −2x + y2 − 6y = 26 (5 points)

Lisa [10]

Answer:

Center: (1,3)

Radius: 6

Step-by-step explanation:

Hi there!

$x^2-2x + y^2 - 6y = 26$

Typically, the equation of a circle would be in the form $(x-h)^2+(y-k)^2=r^2$ where $(h,k)$ is the center and $r$ is the radius.

To get the given equation $x^2-2x + y^2 - 6y = 26$ into this form, we must complete the square for both x and y.

1) Complete the square for x

Let's take a look at this part of the equation:

$x^2-2x$

To complete the square, we must add to the expression the square of half of 2. That would be 1² = 1:

$x^2-2x+1$

Great! Now, let's add this to our original equation:

$x^2-2x+1+y^2-6y = 26$

We cannot randomly add a 1 to just one side, so we must do the same to the right side of the equation:

$x^2-2x+1+y^2-6y = 26+1\\x^2-2x+1+y^2-6y = 27$

Complete the square:

$(x-1)^2+y^2-6y = 27$

2) Complete the square for y

Let's take a look at this part of the equation $(x-1)^2+y^2-6y = 27$ :

$y^2-6y$

To complete the square, we must add to the expression the square of half of 6. That would be 3² = 9:

$y^2-6y+9$

Great! Now, back to our original equation:

$(x-1)^2+y^2-6y+9= 27$

Remember to add 9 on the other side as well:

$(x-1)^2+y^2-6y+9= 27+9\\(x-1)^2+y^2-6y+9= 36$

Complete the square:

$(x-1)^2+(y-3)^2= 36$

3) Determine the center and the radius

$(x-1)^2+(y-3)^2= 36$

$(x-h)^2+(y-k)^2=r^2$

Now, we can see that (1,3) is in the place of (h,k). 36 is also in the place of r², making 6 the radius.

I hope this helps!

3 0

3 years ago

Read 2 more answers

Find the area of the shaded region

goldfiish [28.3K]

Answer:

38x² - 17x - 3

Step-by-step explanation:

The area of the shaded region is calculated as

area of outer rectangle - area of white rectangle, that is

(8x + 1)(5x - 3) - 2x(x - 1) ← expand (8x - 1)(5x - 3) using FOIL

= 40x² - 24x + 5x - 3 - 2x² + 2x ← collect like terms

= 40x² - 17x - 3

5 0

4 years ago

A large fish tank at an aquarium needs to be emptied so that it can be cleaned. When its

VikaD [51]

Answer:

The draining time when only the big drain is opened is 2.303 hours.

The draining time when only the small drain is opened is 5.303 hours.

Step-by-step explanation:

From Physics, we know that volume flow rate ( $\dot V$ ), measured in liters per hour, is directly proportional to draining time ( $t$ ), measured in hours. That is:

$\dot V \propto \frac{1}{t}$

$\dot V = \frac{k}{t}$ (Eq. 1)

Where $k$ is the proportionality constant, measured in liters.

From statement, we have the following three expressions:

(i) Large and small drains are opened

$\dot V_{s}+\dot V_{l} = \frac{k}{2}$ (Eq. 2)

$\frac{\dot V_{s}+\dot V_{l}}{k} = \frac{1}{2}$

(ii) Only the small drain is opened

$\dot V_{s} = \frac{k}{t_{l}+3}$ (Eq. 3)

$\frac{\dot V_{s}}{k} = \frac{1}{t_{l}+3}$

(iii) Only the big drain is opened

$\dot V_{l} = \frac{k}{t_{l}}$ (Eq. 4)

$\frac{\dot V_{l}}{k} = \frac{1}{t_{l}}$

By applying (Eqs. 3, 4) in (Eq. 2) and making some algebraic handling, we find that:

$\frac{1}{t_{l}+3}+\frac{1}{t_{l}} = \frac{1}{2}$

$\frac{t_{l}+t_{l}+3}{t_{l}\cdot (t_{l}+3)} = \frac{1}{2}$

$2\cdot t_{l}+3 = t_{l}^{2}+3\cdot t_{l}$

$t_{l}^{2}-t_{l}-3 = 0$ (Eq. 5)

Whose roots are determined by the Quadratic Formula:

$t_{l,1}\approx 2.303\,h$ and $t_{l,2} \approx -1.302\,h$

Only the first roots offers a solution that is physically reasonable. Hence, the draining time when only the big drain is opened is 2.303 hours. And the time needed for the small drain is calculated by the following formula:

$t_{s} = 2.303\,h+3\,h$

$t_{s} = 5.303\,h$

The draining time when only the small drain is opened is 5.303 hours.

7 0

4 years ago

Help!! How do you calculate relative frequency?....A=Frequency times total

spayn [35]

Answer:

Step-by-step explanation:

5 0

3 years ago

Read 2 more answers