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2 years ago

The diameter of a circle is 14cm. Which TWO statements are correct? A

Mathematics

1 answer:

Reika [66]2 years ago

3 0

Answer:

(a) and (e)

Step-by-step explanation:

Given

$d = 14cm$

Required

Select the true statements

First, is a statement about the area.

The area is calculated as:

$Area= \pi*\frac{d^2}{4}$

So, we have:

$Area= 3.14*\frac{14^2}{4}$

$Area= 3.14*49$

(a) is true

Next, is a statement about the circumference.

This is calculated as:

$C = \pi d$

So, we have:

$C = 3.14 * 14$

(e) is true

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Which inequality is true? A.195 > 14.1 B.205 < 14.1 C.208 14.1 D.218 - 14.1

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Answer:

A. 195 > 14.1

Step-by-step explanation:

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3 years ago

A special filter can process 34 gallon of liquid in 25 minutes. If the filter processes a liquid at a constant rate, what is the

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Answer:

1.36

Step-by-step explanation:

34/25=1.36 wait nvm I messed up

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3 years ago

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Help pls! Which rational function has zeros at x = -1 and x = 6 ?

bagirrra123 [75]

Answer:

$f(x) = \frac{x^{2} -5x-6}{x^{2} -4}$

Step-by-step explanation:

Let's factorize the 1st option :

⇒ $f(x) = \frac{x^{2} -5x-6}{x^{2} -4}$

⇒ $f(x) = \frac{x^{2}+x-6x-6 }{(x+2)(x-2)}$

⇒ $f(x) = \frac{(x+1)(x-6)}{(x+2)(x-2)}$

===============================================================

Finding the zeros :

⇒ Factors of numerator should be equated to 0

x + 1 = 0 ⇒ x = -1
x - 6 = 0 ⇒ x = 6

=============================================================

Solution :

$f(x) = \frac{x^{2} -5x-6}{x^{2} -4}$

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2 years ago

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Solve using elimination x+y-2z=8 5x-3y+z=-6 -2x-y+4z=-13

Free_Kalibri [48]

So here is your answer with LaTeX issued format interpretation. Full process elucidated briefly, below:

$\begin{alignedat}{3}x + y - 2z = 8 \\ 5x - 3y + 2 = - 6 \\ - 2x - y + 4z = - 13 \end{alignedat}$

For this equation to get obtained under the impression of those variables we have to eliminate them individually for moving further and simplifying the linear equation with three variables along the axis.

Multiply the equation of x + y - 2z = 8 by a number with a value of 5; Here this becomes; 5x + 5y - 10z = 40; So:

$\begin{alignedat}{3}5x + 5y - 10z = 40 \\ 5x - 3y + z = - 6 \\ - 2x - y + 4z = - 13 \end{alignedat}$

Pair up the equations in a way to eliminate the provided variable on our side, that is; "x":

5x - 3y + z = - 6

-

5x + 5y - 10z = 40
______________

- 8y + 11z = - 46

Therefore, we are getting.

$\begin{alignedat}{3}5x + 5y - 10z = 40 \\ - 8y + 11z = - 46 \\ - 2x - y + 4z = - 13 \end{alignedat}$

Multiply the equation of 5x + 5y - 10z = - 40 by a number with a value of 2; Here this becomes; 10x + 10y - 20z = 80.

Multiply the equation of - 2x - y + 4z = - 13 by a number with a value of 5; Here this becomes; - 10x - 5y + 20z = - 65; So:

$\begin{alignedat}{3}10x + 10y - 20z = 80 \\ - 8y + 11z = - 46 \\ - 10x - 5y + 20z = - 65 \end{alignedat}$

Pair up the equations in a way to eliminate the provided variables on our side, that is; "x" and "z":

- 10x - 5y + 20z = - 65

+
10x + 10y - 20z = 80
__________________

5y = 15

$\begin{alignedat}{3}10x + 10y - 20z = 80 \\ - 8y + 11z = - 46 \\ 5y = 15 \end{alignedat}$

Multiply the equation of - 8y + 11z = - 46 by a number with a value of 5; Here this becomes; - 40y + 55z = - 230.

Multiply the equation of 5y = 15 by a number with a value of 8; Here this becomes; 40y = 120; So:

$\begin{alignedat}{3}10x + 10y - 20z = 80 \\ - 40y + 55z = - 690 \\ 40y = 120 \end{alignedat}$

Pair up the equations in a way to eliminate the provided variables on our side, that is; "y":

40y = 120

+

- 40y + 55z = - 230
_________________

55z = - 110

$\begin{alignedat}{3}10x + 10y - 20z = 80 \\ - 40y + 55z = - 230 \\ 55z = - 110 \end{alignedat}$

Solving for the variable of 'z':

$\mathsf{55z = - 110}$

$\bf{\dfrac{55z}{55} = \dfrac{-110}{55}}$

Cancel out the common factor acquired on the numerator and denominator, that is, "55":

$z = - \dfrac{\overbrace{\sout{110}}^{2}}{\underbrace{\sout{55}}_{1}}$

$\boxed{\mathbf{z = - 2}}$

Solving for variable "y":

$\mathbf{\therefore \quad - 40y - 55 \big(- 2 \big) = - 230}$

$\mathbf{- 40y - 55 \times 2 = - 230}$

$\mathbf{- 40y - 110 = - 230}$

$\mathbf{- 40y - 110 + 110 = - 230 + 110}$

Adding the numbered value as 110 into this equation (in previous step).

$\mathbf{- 40y = - 120}$

Divide by - 40.

$\mathbf{\dfrac{- 40y}{- 40} = \dfrac{- 120}{- 40}}$

$\mathbf{y = \dfrac{- 120}{- 40}}$

$\boxed{\mathbf{y = 3}}$

Solve for variable "x":

$\mathbf{10x + 10y - 20z = 80}$

$\mathbf{Since, \: z = - 2; \quad y = 3}$

$\mathbf{10x + 10 \times 3 - 20 \times (- 2) = 80}$

$\mathbf{10x + 10 \times 3 + 20 \times 2 = 80}$

$\mathbf{10x + 30 + 20 \times 2 = 80}$

$\mathbf{10x + 30 + 40 = 80}$

$\mathbf{10x + 70 = 80}$

$\mathbf{10x + 70 - 70 = 80 - 70}$

$\mathbf{10x = 10}$

Divide by this numbered value $\mathbf{10}$ to get the final value for the variable "x".

$\mathbf{\dfrac{10x}{10} = \dfrac{10}{10}}$

The numbered values in the numerator and the denominator are the same, on both the sides. This will mean the "x" variable will be left on the left hand side and numbered values "10" will give a product of "1" after the division is done. On the right hand side the numbered values get divided to obtain the final solution for final system of equation for variable "x" as "1".

$\boxed{\mathbf{x = 1}}$

Final solutions for the respective variables in the form of " (x, y, z) " is:

$\boxed{\mathbf{\underline{\Bigg(1, \: \: 3, \: \: - 2 \Bigg)}}}$

Hope it helps.

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3 years ago

Read 2 more answers

Which of the following statements about the polynomial function F(x)=-x^3-6x^2+1

kifflom [539]

Answer:

Step-by-step explanation:

The graph begins up and ends down. This means as x heads into negative infinity the graph continues up, up and up towards infinity. While x heads to positive infinity on the right side of the graph, the graph heads down, down, and down towards negative infinity. See graph below.

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3 years ago