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2 years ago

13

Someone help and please make sure it’s right :)

1 answer:

pashok25 [27]2 years ago

4 0

Answer:

9

Step-by-step explanation:

$11x + 1 + 7x + 17 = 180 \: \: \: \: 18x + 18 = 180 \: \: \: \: \: \: \: 18x = 180 - 18 \: \: \: \: \: \: \: \:18x = 162 \: \: \: \: \: \: \: \: \: \: 162 \div 18 = 9$

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Can some one answer this?

timofeeve [1]

9514 1404 393

Answer:

(c) 2

Step-by-step explanation:

A graph shows the zeros are -5, -3, 0, 2. Of these, only 2 is on your list.

2 is a zero of the expression

4 0

2 years ago

18. What is the least common denominator of the following<br> fractions? 3/10 and 1/4

Len [333]

Answer:

20

The denominators of 3 /10 and 1 /4 are 10 and 4.

The factors of 10 are 2 × 5 and 2 × 2

Therfore the LCD should have two 2's and one 5.

This means the answer is:

2 × 2 × 5 = <u>20</u>

I hope this helps :)

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3 years ago

Solve the system of equations by using substitution.

Nat2105 [25]

Answer:

(x, y) = (-1, 1)

Step-by-step explanation:

$y = 2x +3 ----(1) \\\\y = x + 2 ----(2)\\\\substitute \ (2) \ in \ (1) : x + 2 = 2x + 3$

$x - 2x = 3 - 2\\\\-x = 1\\\\x = -1\\\\Substitute \ x = -1 \ in \ (2) : y = -1 + 2 => y = 1$

5 0

2 years ago

Is it a square, rhombus, or a rectangle? <br> Why?

Alex73 [517]

Answer:

It is a square.

Step-by-step explanation:

If you look at the space of the lines from point to point, it's a even amount of little squares, so therefore, it is a square.

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3 years ago

Read 2 more answers

Find the points on the curve where the tangent is horizontal or vertical. If you have a graphing device, graph the curve to chec

klasskru [66]

Answer:

There is a horizontal tangent at (0,-4)

The tangent is vertical at (-2,-3) and (2,-3).

Step-by-step explanation:

The given function is defined parametrically by the equations:

$x=t^3-3t$

and

$y=t^2-4$

The tangent function is given by:

$\frac{dy}{dx}=\frac{\frac{dy}{dt} }{\frac{dx}{dt} }$

$\implies \frac{dy}{dx}=\frac{2t}{3t^2-3}$

The tangent is vertical at when $\frac{dx}{dt}=0$

$\implies \frac{3t^2-3}{2t}=0$

$\implies 3t^2-3=0$

$\implies 3t^2=3$

$\implies t^2=1$

$\implies t=\pm1$

When t=1,

$x=1^3-3(1)=-2$ and $y=1^2-4=-3$

When t=-1,

$x=(-1)^3-3(-1)=2$ and $y=(-1)^2-4=-3$

The tangent is vertical at (-2,-3) and (2,-3).

The tangent is horizontal, when $\frac{dy}{dx}=0$ or $\frac{dy}{dt}=0$

$\implies 2t=0$

$\implies t=0$

When t=0,

$x=0^3-3(0)=0$ and $y=0^2-4=-4$

There is a horizontal tangent at (0,-4)

5 0

3 years ago