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3 years ago

6

a national park charges $26 per adult and $16 per child for rafting down one of their two rivers. Write an algebraic expression

that can be used to represent the total cost for a adults and c children to raft down the wild river

1 answer:

blondinia [14]3 years ago

5 0

26a+16c would be the answer

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Which is a true statement concerning outliers for the data set summarized by the box-and-whisker plot shown?

Ray Of Light [21]

B :^) hope this helps!

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3 years ago

Solve for x.<br> - 30 – 5x - 1) + 7 = - 110

ololo11 [35]

Answer:

$x = \frac{86}{5} = 17 \frac{1}{5} = 17.2$

Step-by-step explanation:

$( - 30 - 5x - 1) + 7 = - 110$

$( - 31 - 5x) + 7 = - 110$

$- 31 - 5x + 7 = - 110$

$- 24 - 5x = - 110$

$- 5x = - 110 + 24$

$- 5x = - 86$

$\boxed{\green{x = \frac{86}{5} = 17 \frac{1}{5} = 17.2}}$

5 0

3 years ago

Please help me. (20 points)

alexandr402 [8]

Hello from MrBillDoesMath!

Answer:

75

Discussion:

A plane triangle has 180 degrees. Using this in the diagram gives

x + 81 + angle BAC = 180 (***)

(The diagram is a bit fuzzy but I think the above numbers are accurate)

The diagram also shows that (156 + angle BAC )= 180 as a straight line has 180 degrees. So

angle BAC = 180 - 156 = 24

Substituting this in (***) above gives

x + 81 + 24 = 180 =>

x + 105 = 180 => (as 81 + 24 = 105)

x = 180 107 = 75

Check: Does

x + 81 + angle BAC = 180 ?

75 + 81 + 24 = 180. Yes

Thank you,

MrB

7 0

3 years ago

If elephant grass grows 2 1/5 inches a day, how many inches will it grow in 5 days?

Sonja [21]

I hope this helps you

5 0

3 years ago

Read 2 more answers

Find an equation for the plane that is tangent to the surface z equals ln (x plus y )at the point Upper P (1 comma 0 comma 0 ).

alexira [117]

Let $f(x,y,z)=z-\ln(x+y)$ . The gradient of $f$ at the point (1, 0, 0) is the normal vector to the surface, which is also orthogonal to the tangent plane at this point.

So the tangent plane has equation

$\nabla f(1,0,0)\cdot(x-1,y,z)=0$

Compute the gradient:

$\nabla f(x,y,z)=\left(\dfrac{\partial f}{\partial x},\dfrac{\partial f}{\partial y},\dfrac{\partial f}{\partial z}\right)=\left(-\dfrac1{x+y},-\dfrac1{x+y},1\right)$

Evaluate the gradient at the given point:

$\nabla f(1,0,0)=(-1,-1,1)$

Then the equation of the tangent plane is

$(-1,-1,1)\cdot(x-1,y,z)=0\implies-(x-1)-y+z=0\implies\boxed{z=x+y-1}$

7 0

4 years ago