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3 years ago

4x-2y=14

2 answers:

4 0

4x - 2y = 14
y = 0.5x -1

Get just one variable on one side for both equations.

4x - 2y = 14 ⇒ 4x = 2y + 14 ⇒ 4x - 14 = 2y
y = 0.5x - 1 ⇒ x - 2 = 2y

Use the transitive property and then solve for x.

x-2 = 4x-14
12 = 3x
x = 4

Edit: Apprently I didn't use substitution. Oops! Sorry...

mamaluj [8]3 years ago

3 0

4x-2(1/2x-1)=14
4x-x+2=14. 1=1/2(4)-1
3x+2=14. 1=2-1
3x=12. 1=1
X=4
Y=1/2(4)-1
Y=2-1 Y=1

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In the graph above, the coordinates of the vertices of QPR are Q(3, 0), P(5, 6), and R(7, 0). If AQPR is reflected across

ozzi

Answer:

D. (7, 0)

Step-by-step explanation:

The rule for a reflection over the y-axis is (x, y) → (x, -y)

This means that the x-values stay the same while the y-values change.

Q(x, y) → (x, -y)

Q(3, 0) → (3, 0)

Q'(3, 0)

P(x, y) → (x, -y)

P(5, 6) → (5, -6)

P'(5, -6)

R(x, y) → (x, -y)

R(7, 0) → (7, 0)

R'(7, 0)

Therefore, the correct answer is D.

Hope this helps!

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2 years ago

99 POINT QUESTION, PLUS BRAINLIEST!!!

sattari [20]

We know, that the area of the surface generated by revolving the curve y about the x-axis is given by:

$\boxed{A=2\pi\cdot\int\limits_a^by\sqrt{1+\left(y'\right)^2}\, dx}$

In this case a = 0, b = 15, $y=\dfrac{x^3}{15}$ and:

$y'=\left(\dfrac{x^3}{15}\right)'=\dfrac{3x^2}{15}=\boxed{\dfrac{x^2}{5}}$

So there will be:

$A=2\pi\cdot\int\limits_0^{15}\dfrac{x^3}{15}\cdot\sqrt{1+\left(\dfrac{x^2}{5}\right)^2}\, dx=\dfrac{2\pi}{15}\cdot\int\limits_0^{15}x^3\cdot\sqrt{1+\dfrac{x^4}{25}}\,\, dx=\left(\star\right)\\\\-------------------------------\\\\
\int x^3\cdot\sqrt{1+\dfrac{x^4}{25}}\,\,dx=\int\sqrt{1+\dfrac{x^4}{25}}\cdot x^3\,dx=\left|\begin{array}{c}t=1+\dfrac{x^4}{25}\\\\dt=\dfrac{4x^3}{25}\,dx\\\\\dfrac{25}{4}\,dt=x^3\,dx\end{array}\right|=\\\\\\$

$=\int\sqrt{t}\cdot\dfrac{25}{4}\,dt=\dfrac{25}{4}\int\sqrt{t}\,dt=\dfrac{25}{4}\int t^\frac{1}{2}\,dt=\dfrac{25}{4}\cdot\dfrac{t^{\frac{1}{2}+1}}{\frac{1}{2}+1}= \dfrac{25}{4}\cdot\dfrac{t^{\frac{3}{2}}}{\frac{3}{2}}=\\\\\\=\dfrac{25\cdot2}{4\cdot3}\,t^\frac{3}{2}=\boxed{\dfrac{25}{6}\,\left(1+\dfrac{x^4}{25}\right)^\frac{3}{2}}\\\\-------------------------------\\\\$

$\left(\star\right)=\dfrac{2\pi}{15}\cdot\int\limits_0^{15}x^3\cdot\sqrt{1+\dfrac{x^4}{25}}\,\, dx=\dfrac{2\pi}{15}\cdot\dfrac{25}{6}\cdot\left[\left(1+\dfrac{x^4}{25}\right)^\frac{3}{2}\right]_0^{15}=\\\\\\=
\dfrac{5\pi}{9}\left[\left(1+\dfrac{15^4}{25}\right)^\frac{3}{2}-\left(1+\dfrac{0^4}{25}\right)^\frac{3}{2}\right]=\dfrac{5\pi}{9}\left[2026^\frac{3}{2}-1^\frac{3}{2}\right]=\\\\\\=
\boxed{\dfrac{5\Big(2026^\frac{3}{2}-1\Big)}{9}\pi}$

Answer C.

3 0

3 years ago

Solve |x| < 13 A. {-13, 13} B. {x| -13< x < 13} C. {x|x < -13, x> 13}

Gemiola [76]

Answer:

the solution for x will be given by {x| -13 < x < 13}

Therefore the correct option is B.

Step-by-step explanation:

i) given that |x| < 13

ii) therefore x < 13 for positive values of x

iii) for negative values of x we have -x < 13

therefore x > -13

iv) from ii) and iii) above we can conclude that the solution for x will be given by {x| -13 < x < 13}

Therefore the correct option is B.

7 0

3 years ago

The skateboard ramp has a width of 20 and a height of 16. What is its slope ( not to scale) ?

slega [8]

Answer:

25.61 ?

Step-by-step explanation:

use Pythagoreans theorem

a^2+b^2=c^2

20^2+16^2=656

sqrt 656=25.61

4 0

3 years ago

2xy+3x+4x+3 if x =5 and y = -5

Stels [109]

Answer:

The answer is -12

Step-by-step explanation:

2xy + 3x + 4x + 3 if x = 5 and y = -5

Now,

2xy + 3x + 4x + 3

2(5)(-5) + 3(5) + 4(5) + 3

-50 + 15 + 20 + 3

38 - 50 = -12

Thus, The answer is -12

-TheUnknownScientist

6 0

3 years ago