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qwelly [4]
3 years ago
7

What is the area, in square centimeters, of the shaded part of the rectangle

Mathematics
1 answer:
mojhsa [17]3 years ago
8 0

Answer:

I honestly dont know I just need the points

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What is the the total cost of a book that is $45.60 with a sales tax of 7.5%
sineoko [7]
3.42 multiply 45.60 by 0.075


6 0
3 years ago
If y varies directly as x and y = 8 when x = 4, what is the value of y when x = 16? a. 32 b. 16 c. 64 d. 28
ddd [48]

By finding the proportional relation between x and y, we will see that when x = 16, we have y = 32.

<h3>How to get the value of y when x = 16?</h3>

If y varies directly with x, then we can write the relation as:

y = k*x

Where k is the constant of proportionality.

First, we know that when x = 4, we have y = 8, replacing that we get:

8 = k*4

Solving for k, we get:

k = 8/4 = 2.

Then the relation between x and y is:

y = 2*x

When x = 16, we have:

y = 2*16 = 32

Then the correct option is a.

If you want to learn more about proportional relations:

brainly.com/question/12242745

#SPJ1

5 0
2 years ago
Factor: 6x^2 + 11x + 3<br><br> Write each factor in standard form.
yKpoI14uk [10]
The answer is: (2x+3) (3x+1)
8 0
3 years ago
Find the local maximum and minimum values and saddle point(s) of the function. If you have three-dimensional graphing software,
vichka [17]

Answer:

A(-1,0) is a local maximum point.

B(-1,0)  is a saddle point

C(3,0)  is a saddle point

D(3,2) is a local minimum point.

Step-by-step explanation:

The given function is  

f(x,y)=x^3+y^3-3x^2-3y^2-9x

The first partial derivative with respect to x is  

f_x=3x^2-6x-9

The first partial derivative with respect to y is  

f_y=3y^2-6y

We now set each equation to zero to obtain the system of equations;

3x^2-6x-9=0

3y^2-6y=0

Solving the two equations simultaneously, gives;

x=-1,x=3  and y=0,y=2

The critical points are

A(-1,0), B(-1,2),C(3,0),and D(3,2).

Now, we need to calculate the discriminant,

D=f_{xx}(x,y)f_{yy}(x,y)-(f_{xy}(x,y))^2

But, we would have to calculate the second partial derivatives first.

f_{xx}=6x-6

f_{yy}=6y-6

f_{xy}=0

\Rightarrow D=(6x-6)(6y-6)-0^2

\Rightarrow D=(6x-6)(6y-6)

At A(-1,0),

D=(6(-1)-6)(6(0)-6)=72\:>\:0 and f_{xx}=6(-1)-6=-18\:

Hence A(-1,0) is a local maximum point.

See graph

At B(-1,2);

D=(6(-1)-6)(6(2)-6)=-72\:

Hence, B(-1,0) is neither a local maximum or a local minimum point.

This is a saddle point.

At C(3,0)

D=(6(3)-6)(6(0)-6)=-72\:

Hence, C(3,0) is neither a local minimum or maximum point. It is a saddle point.

At D(3,2),

D=(6(3)-6)(6(2)-6)=72\:>\:0 and f_{xx}=6(3)-6=12\:>\:0

Hence D(3,2) is a local minimum point.

See graph in attachment.

3 0
3 years ago
How do you solve this one 10x^2-25=x^2
solong [7]
10x^2 - 25 = x^2
Take the square root of each side.

\sqrt{x^2} =  \sqrt{10x^2 - 25}
x = 10x - 5i




4 0
3 years ago
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