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3 years ago

Evaluate g(4): g(x)=8x^2+9x-7 A. 121 B. 135 C. 154 D. 157

1 answer:

7 0

The answer is: [D]: " 157" .
________________________________________________________
Explanation:
______________________________________________________

g(x) = 8x² + 9x − 7 ;

g(4) = 8(4²) + 9(4) − 7 ;

= 8(4*4) + 36 − 7 ;

= 8(16) + 36 − 7 ;

= 128 + 36 − 7 ;

= 164 − 7 ;

= 157 .
_______________________________________________________
The answer is: [D]: " 157" .
_______________________________________________________

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Alex sold 165 raffle tickets, which was about 70% of the number of tickets sold by his brother Zane. Approximately how many tick

Len [333]

Answer:

115.5 tickets sold.

Step-by-step explanation:

Finding the percent of something is the same as multiplying by that percent as a decimal. So, 70% = 0.70.

0.70 x 165 = 115.5 tickets sold.

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3 years ago

The plane x + y + z = 12 intersects paraboloid z = x^2 + y^2 in an ellipse.(a) Find the highest and the lowest points on the ell

emmasim [6.3K]

Answer:

Highest (-3,-3)

Lowest (2,2)

Farthest (-3,-3)

Closest (2,2)

Step-by-step explanation:

To solve this problem we will be using Lagrange multipliers.

Let us find out first the restriction, which is the projection of the intersection on the XY-plane.

From x+y+z=12 we get z=12-x-y and replace this in the equation of the paraboloid:

$\bf 12-x-y=x^2+y^2\Rightarrow x^2+y^2+x+y=12$

completing the squares:

$\bf x^2+y^2+x+y=12\Rightarrow (x+1/2)^2-1/4+(y+1/2)^2-1/4=12\Rightarrow\\\\\Rightarrow (x+1/2)^2+(y+1/2)^2=12+1/2\Rightarrow (x+1/2)^2+(y+1/2)^2=25/2$

and we want the maximum and minimum of the paraboloid when (x,y) varies on the circumference we just found. That is, we want the maximum and minimum of

$\bf f(x,y)=x^2+y^2$

subject to the constraint

$\bf g(x,y)=(x+1/2)^2+(y+1/2)^2-25/2=0$

Now we have

$\bf \nabla f=(\displaystyle\frac{\partial f}{\partial x},\displaystyle\frac{\partial f}{\partial y})=(2x,2y)\\\\\nabla g=(\displaystyle\frac{\partial g}{\partial x},\displaystyle\frac{\partial g}{\partial y})=(2x+1,2y+1)$

Let $\bf \lambda$ be the Lagrange multiplier.

The maximum and minimum must occur at points where

$\bf \nabla f=\lambda\nabla g$

that is,

$\bf (2x,2y)=\lambda(2x+1,2y+1)\Rightarrow 2x=\lambda (2x+1)\;,2y=\lambda (2y+1)$

we can assume (x,y)≠ (-1/2, -1/2) since that point is not in the restriction, so

$\bf \lambda=\displaystyle\frac{2x}{(2x+1)} \;,\lambda=\displaystyle\frac{2y}{(2y+1)}\Rightarrow \displaystyle\frac{2x}{(2x+1)}=\displaystyle\frac{2y}{(2y+1)}\Rightarrow\\\\\Rightarrow 2x(2y+1)=2y(2x+1)\Rightarrow 4xy+2x=4xy+2y\Rightarrow\\\\\Rightarrow x=y$

Replacing in the constraint

$\bf (x+1/2)^2+(x+1/2)^2-25/2=0\Rightarrow (x+1/2)^2=25/4\Rightarrow\\\\\Rightarrow |x+1/2|=5/2$

from this we get

x=-1/2 + 5/2 = 2 or x = -1/2 - 5/2 = -3

and the candidates for maximum and minimum are (2,2) and (-3,-3).

Replacing these values in f, we see that

f(-3,-3) = 9+9 = 18 is the maximum and

f(2,2) = 4+4 = 8 is the minimum

Since the square of the distance from any given point (x,y) on the paraboloid to (0,0) is f(x,y) itself, the maximum and minimum of the distance are reached at the points we just found.

We have then,

(-3,-3) is the farthest from the origin

(2,2) is the closest to the origin.

3 0

3 years ago

Express 150 as a product of PRIME FACTORS. Answer:

Flauer [41]

Answer:

Step-by-step explanation:

$150=2X3X5X5\\ \\ 150=2(3)5^2$

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2 years ago

Which correctly describes a cross section of the cube below? Check all that apply.

dmitriy555 [2]

Answer:

A cross section parallel to the base is a square measuring 4 cm by 4 cm.

A cross section perpendicular to the base through the midpoints of opposite sides is a square measuring 4 cm by 4 cm.

A cross section that passes through the entire bottom front edge and the entire top back edge is a rectangle measuring 4 cm by greater than 4 cm.

Step-by-step explanation:

Cross sections parallel and Perpendicular to the base are squares 4 × 4

The diagonal cross section will be a rectangle with base 4 and height

sqrt(4² + 4²) = 4sqrt(2) > 4

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3 years ago

Read 2 more answers

My teacher wouldn't help me.. and this is my last question. I need to know how to find the perimeter of the geometric figure (it

nika2105 [10]

This isn't really a geometry problem. It's just an addition of fractions.

You know that 'perimeter' means 'the distance all the way around'.
And you know the length of all the sides of the parallelogram.
All you need to do is add them up !

(13/12) + (3/13) + (13/12) + (3/13) = the perimeter

Notice that two of the sides are equal, and the other two sides are also equal.
So you can make the job a little easier if you add up the twelfths first

(13/12) + (13/12) = 26/12

and then add up the thirteenths ...

(3/13) + (3/13) = 6/13 .

Now, the perimeter looks a little bit less complicated.
It's just
(26/12) + (6/13) = the perimeter.

This is the tough part. Before you can add fractions, they need to have
a common denominator.

The smallest common denominator for 12ths and 13ths is 156 !
Change each fraction to (something over 156), and add um up.
I'll leave that part to you.

5 0

3 years ago