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2 years ago

Robert is painting a landscape for art class. He starts by mixing green and

Mathematics

2 answers:

shutvik [7]2 years ago

6 0

Answer:

the tree because white paint is a strong mixing agent the more white you add to green will make it a lighter color

Kryger [21]2 years ago

6 0

Answer:

tree

Step-by-step explanation:

IXL trust me

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If (x - h) = 4 and (y - k) = 3, what is the radius of the circle above?

I am Lyosha [343]

(x-h)²+(y-k)²=r² is the equation of a circle with radius of r

so if x-h=4 and y-k=3 then

4²+3²=r²
16+9=r²
25=r²
sqrt both sides
5=r
radius is 5 units

5 0

3 years ago

If f(x) = 4+ 3x, what is f(-9)?

Vlad [161]

Answer:

-23

Step-by-step explanation:

8 0

3 years ago

Read 2 more answers

Evaluate the expression <br> 33(4a), if a = 2.

Alexandra [31]

Answer:

33(4a) =264

Step-by-step explanation:

Parentheses

Exponents

Multipication

Division

Addition

Subtraction

so, if we do parentheses first 4 times 2 will equal 8 then 33 x 8=264

you're welcome :D

6 0

3 years ago

For trapezoid JKLM, A and B are midpoints of the legs. Find AB.

romanna [79]

Answer:

Step-by-step explanation:

The length of the median of a trapezoid equals one-half the sum of the lengths of the bases.

$\begin{array}{rcl}AB & = & \frac{1}{2}(JK + LM)\\\\& = & \frac{1}{2}(19 + 39)\\\\& = & \frac{1}{2}\times 58\\\\& = & \mathbf{29}\\\end{array}$

8 0

4 years ago

Find the absolute maximum and minimum values of f(x,y)=xy−4x in the region bounded by the x-axis and the parabola y=16−x2.

skad [1K]

Answer:

The absolute maximum and minimum is $20\; \text{and} -20.$

Step-by-step explanation:

We first check the critical points on the interior of the domain using the

first derivative test.

$f_x=y-4=0$

$f_y=x=0$

The only solution to this system of equations is the point (0, 4), which lies in the domain.

$f_{xx}=0, \;f_{yy}=0\; \text{and}\; f_{xy}=-1$

$\Rightarrow f_{xx}f_{yy}-f_{xy}=o-1=-1$

$\therefore (0,4)$ is a saddle point.

Boundary points - $(4,0), (-4,0), (0,16)$

Along boundary $y=16-x^2$

$f=x(16-x^2)-4x$

$=16x-x^3-4x$

$\Rightarrow f^'=16-3x^2-4=0$

$\Rightarrow 3x^2=12$

$\Rightarrow x=\pm2,\;\;y=14$

Values of f(x) at these points.

$\begin{array}{}(4,0)=-16\\(-4,0)=16\\(0,16)=0\\(2,14)=20\\(-2,14)=-20\end{array}$

Therefore, the absolute maximum and minimum is $20\; \text{and} -20.$

6 0

4 years ago