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Jlenok [28]
3 years ago
8

Can someone help me on this with the steps? thanks!

Mathematics
2 answers:
soldi70 [24.7K]3 years ago
8 0

Answer:

8h + 15 = 39

Step-by-step explanation:

The base fee is 15 so you shouldn't multiply the hours by it.

The question states 8 per hour which means that you should multiply the 8 by hours and add on the base fee which is 15.

Brilliant_brown [7]3 years ago
4 0

Answer:

8h+15=39

im pretty sure this is your answer. let me know if it was or not

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What is square root of 250 form of 6
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30√10


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Y = 5 - 2x<br> inputs are : -1, 1, 3, and 5.<br> I need the outputs :)
IRINA_888 [86]
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8 0
3 years ago
Can u help me
fiasKO [112]

The diagram shows that FG = IJ are congruent segments due to the double tickmarks. These are the hypotenuses of the right triangles.

So that takes care of the "H" in "HL".

The L stands for "leg", so we need to have a congruent pair of corresponding legs.

Due to the orientation of the triangles, it's not clear how the corresponding legs match up, but if any of the following is true

  • EF = IK
  • EF = JK
  • EG = IK
  • EG = JK

then we'd have enough to use HL.

4 0
3 years ago
Find the local maximum and minimum values and saddle point(s) of the function.
iogann1982 [59]
f(x,y)=9e^y(y^2-x^2)
\nabla f=\left\langle-18xe^y,9ye^y(2+y)\right\rangle

Critical points occur where the gradient is zero. This is guaranteed whenever x=0 and either y=0 or y=-2.

The Hessian matrix for this function looks like

H(x,y)=\begin{bmatrix}f_{xx}&f_{xy}\\f_{yx}&f_{yy}\end{bmatrix}=\begin{bmatrix}-18e^y&-18xe^y\\-18xe^y&9e^y(2-x^2+4y+y^2)\end{bmatrix}

and has determinant

|H(x,y)|=-162e^{2y}(2+x^2+4y+y^2)

Maxima occur whenever the determinant is positive and f_{xx}. Minima occur whenever both the determinant and f_{xx} are positive. Saddle points occur whenever the determinant is negative.

At (0,0), you have a saddle point since the determinant reduces to -324, so (0,0,0) is the saddle point.

At (0,-2), the determinant is \dfrac{324}{e^4}>0 and f_{xx}(0,-2)=-\dfrac{18}{e^2}, so \left(0,-2,\dfrac{36}{e^2}\right) is a local maximum.

No other critical points remain, so you're done.
4 0
2 years ago
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