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3 years ago

14

Mason claims that y = 3x2- 6 is a function, but not a linear one. Select the statement that supports Mason’s claim.

1 answer:

Sladkaya [172]3 years ago

5 0

Answer:

B

Step-by-step explanation:

hope this helps!

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Simplify. (3x^2 + 3) - (6x + 4) + (3x^2 - 5x)

r-ruslan [8.4K]

Answer:

6x^2 -11x -1

Step-by-step explanation:

(3x^2 + 3) - (6x + 4) + (3x^2 - 5x)

Distribute the minus sign

(3x^2 + 3) - 6x - 4) + (3x^2 - 5x)

Combine like terms

3x^2 + 3x^2 - 6x - 5x -4+3

6x^2 -11x -1

3 0

2 years ago

Which graph is correct

ELEN [110]

C is the best answer

6 0

2 years ago

Find the area of a circle with a Radius = 4 ft. Use 3.14 for and round to 2 decimal places.

bagirrra123 [75]

Answer:

50.24 ft^2

Step-by-step explanation:

Well formula for area of a circle is pi(radius)^2

so the radius is 4 so you square that it’ll be 16

then do 16(3.14)

you’ll get 50.24

6 0

2 years ago

Read 2 more answers

Can someone please help me with these 7 questions please?

yarga [219]

$(1)\ (-xy)^3(xz)$

Expand

$(-xy)^3(xz) = (-x)^3* y^3*(xz)$

$(-xy)^3(xz) = -x^3* y^3*xz$

Rewrite as:

$(-xy)^3(xz) = -x^3*x* y^3*z$

Apply law of indices

$(-xy)^3(xz) = -x^4y^3z$

$(2)\ (\frac{1}{3}mn^{-4})^2$

Expand

$(\frac{1}{3}mn^{-4})^2 =(\frac{1}{3})^2m^2n^{-4*2}$

$(\frac{1}{3}mn^{-4})^2 =\frac{1}{9}m^2n^{-8$

$(3)\ (\frac{1}{5x^4})^{-2}$

Apply negative power rule of indices

$(\frac{1}{5x^4})^{-2}= (5x^4)^2$

Expand

$(\frac{1}{5x^4})^{-2}= 5^2x^{4*2}$

$(\frac{1}{5x^4})^{-2}= 25x^{8$

$(4)\ -x(2x^2 - 4x) - 6x^2$

Expand

$-x(2x^2 - 4x) - 6x^2 = -2x^3 + 4x^2 - 6x^2$

Evaluate like terms

$-x(2x^2 - 4x) - 6x^2 = -2x^3 -2x^2$

Factor out x^2

$-x(2x^2 - 4x) - 6x^2 = (-2x-2)x^2$

Factor out -2

$-x(2x^2 - 4x) - 6x^2 = -2(x+1)x^2$

$(5)\ \sqrt{\frac{4y}{3y^2}}$

Divide by y

$\sqrt{\frac{4y}{3y^2}} = \sqrt{\frac{4}{3y}}$

Split

$\sqrt{\frac{4y}{3y^2}} = \frac{\sqrt{4}}{\sqrt{3y}}$

$\sqrt{\frac{4y}{3y^2}} = \frac{2}{\sqrt{3y}}$

Rationalize

$\sqrt{\frac{4y}{3y^2}} = \frac{2}{\sqrt{3y}} * \frac{\sqrt{3y}}{\sqrt{3y}}$

$\sqrt{\frac{4y}{3y^2}} = \frac{2\sqrt{3y}}{3y}$

$(6)\ \frac{8}{3 + \sqrt 3}$

Rationalize

$\frac{8}{3 + \sqrt 3} = \frac{3 - \sqrt 3}{3 - \sqrt 3}$

$\frac{8}{3 + \sqrt 3} = \frac{8(3 - \sqrt 3)}{(3 + \sqrt 3)(3 - \sqrt 3)}$

Apply different of two squares to the denominator

$\frac{8}{3 + \sqrt 3} = \frac{8(3 - \sqrt 3)}{3^2 - (\sqrt 3)^2}$

$\frac{8}{3 + \sqrt 3} = \frac{8(3 - \sqrt 3)}{9 - 3}$

$\frac{8}{3 + \sqrt 3} = \frac{8(3 - \sqrt 3)}{6}$

Simplify

$\frac{8}{3 + \sqrt 3} = \frac{4(3 - \sqrt 3)}{3}$

$(7)\ \sqrt{40} - \sqrt{10} + \sqrt{90}$

Expand

$\sqrt{40} - \sqrt{10} + \sqrt{90} =\sqrt{4*10} - \sqrt{10} + \sqrt{9*10}$

Split

$\sqrt{40} - \sqrt{10} + \sqrt{90} =\sqrt{4}*\sqrt{10} - \sqrt{10} + \sqrt{9}*\sqrt{10}$

Evaluate all roots

$\sqrt{40} - \sqrt{10} + \sqrt{90} =2*\sqrt{10} - \sqrt{10} + 3*\sqrt{10}$

$\sqrt{40} - \sqrt{10} + \sqrt{90} =2\sqrt{10} - \sqrt{10} + 3\sqrt{10}$

$\sqrt{40} - \sqrt{10} + \sqrt{90} =4\sqrt{10}$

$(8)\ \frac{r^2 + r - 6}{r^2 + 4r -12}$

Expand

$\frac{r^2 + r - 6}{r^2 + 4r -12}=\frac{r^2 + 3r-2r - 6}{r^2 + 6r-2r -12}$

Factorize each

$\frac{r^2 + r - 6}{r^2 + 4r -12}=\frac{r(r + 3)-2(r + 3)}{r(r + 6)-2(r +6)}$

Factor out (r+3) in the numerator and (r + 6) in the denominator

$\frac{r^2 + r - 6}{r^2 + 4r -12}=\frac{(r -2)(r + 3)}{(r - 2)(r +6)}$

Cancel out r - 2

$\frac{r^2 + r - 6}{r^2 + 4r -12}=\frac{r + 3}{r +6}$

$(9)\ \frac{4x + 8}{x^2} \cdot \frac{x}{x^2 - 5x - 14}$

Cancel out x

$\frac{4x + 8}{x^2} \cdot \frac{x}{x^2 - 5x - 14} = \frac{4x + 8}{x} \cdot \frac{1}{x^2 - 5x - 14}$

Expand the numerator of the 2nd fraction

$\frac{4x + 8}{x^2} \cdot \frac{x}{x^2 - 5x - 14} = \frac{4x + 8}{x} \cdot \frac{1}{x^2 - 7x+2x - 14}$

Factorize

$\frac{4x + 8}{x^2} \cdot \frac{x}{x^2 - 5x - 14} = \frac{4x + 8}{x} \cdot \frac{1}{x(x - 7)+2(x - 7)}$

Factor out x - 7

$\frac{4x + 8}{x^2} \cdot \frac{x}{x^2 - 5x - 14} = \frac{4x + 8}{x} \cdot \frac{1}{(x + 2)(x - 7)}$

Factor out 4 from 4x + 8

$\frac{4x + 8}{x^2} \cdot \frac{x}{x^2 - 5x - 14} = \frac{4(x + 2)}{x} \cdot \frac{1}{(x + 2)(x - 7)}$

Cancel out x + 2

$\frac{4x + 8}{x^2} \cdot \frac{x}{x^2 - 5x - 14} = \frac{4}{x} \cdot \frac{1}{(x - 7)}$

$\frac{4x + 8}{x^2} \cdot \frac{x}{x^2 - 5x - 14} = \frac{4}{x(x - 7)}$

$(10)\ (3x^3 + 15x^2 -21x) \div 3x$

Factorize

$(3x^3 + 15x^2 -21x) \div 3x = 3x(x^2 + 5x -7) \div 3x$

Cancel out 3x

$(3x^3 + 15x^2 -21x) \div 3x = x^2 + 5x -7$

$(11)\ \frac{m}{6m + 6} - \frac{1}{m+1}$

Take LCM

$\frac{m}{6m + 6} - \frac{1}{m+1} = \frac{m(m + 1) - 1(6m + 6)}{(6m + 6)(m + 1)}$

Expand

$\frac{m}{6m + 6} - \frac{1}{m+1} = \frac{m^2 + m- 6m - 6}{(6m + 6)(m + 1)}$

$\frac{m}{6m + 6} - \frac{1}{m+1} = \frac{m^2 - 5m - 6}{(6m + 6)(m + 1)}$

$(12)\ \frac{\frac{1}{y - 3}}{\frac{2}{y^2 - 9}}$

Rewrite as:

$\frac{\frac{1}{y - 3}}{\frac{2}{y^2 - 9}} = \frac{1}{y - 3} \div \frac{2}{y^2 - 9}$

Express as multiplication

$\frac{\frac{1}{y - 3}}{\frac{2}{y^2 - 9}} = \frac{1}{y - 3} * \frac{y^2 - 9}{2}$

Express y^2 - 9 as y^2 - 3^2

$\frac{\frac{1}{y - 3}}{\frac{2}{y^2 - 9}} = \frac{1}{y - 3} * \frac{y^2 - 3^2}{2}$

Express as difference of two squares

$\frac{\frac{1}{y - 3}}{\frac{2}{y^2 - 9}} = \frac{1}{y - 3} * \frac{(y - 3)(y+3)}{2}$

$\frac{\frac{1}{y - 3}}{\frac{2}{y^2 - 9}} = \frac{1}{1} * \frac{(y+3)}{2}$

$\frac{\frac{1}{y - 3}}{\frac{2}{y^2 - 9}} = \frac{y+3}{2}$

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brainly.com/question/4372544

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2 years ago

The doctor order 100mg of Rocephin. Available is 1g/1mL. How many mL is to be administered?

Leokris [45]

Answer:

0.1 ML

Step-by-step explanation:

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3 years ago