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4 years ago

Evaluate the numerical expression 10 + (25-10) ÷5

Mathematics

2 answers:

LekaFEV [45]4 years ago

7 0

10+(25-10)÷5
10+15÷5
10+3
13

The answer is 13. Hope this helped :)

pashok25 [27]4 years ago

5 0

Your answer will be 13. :-)

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Consider the vector field f(x,y,z)=(2z+3y)i+(3z+3x)j+(3y+2x)kf(x,y,z)=(2z+3y)i+(3z+3x)j+(3y+2x)k.

natima [27]

$\dfrac{\partial f}{\partial x}=2z+3y\implies f(x,y,z)=2xz+3xy+g(y,z)$

$\dfrac{\partial f}{\partial y}=3x+\dfrac{\partial g}{\partial y}=3z+3x$
$\dfrac{\partial g}{\partial y}=3z\implies g(y,z)=3yz+h(z)$
$\implies f(x,y,z)=3xz+3xy+3yz+h(z)$

$\dfrac{\partial f}{\partial z}=3x+3y+\dfrac{\mathrm dh}{\mathrm dz}=3y+2x$
$\dfrac{\mathrm dh}{\mathrm dz}=-x$

But we assume $h$ is a function of $z$ alone, so there is no solution for $h$ and hence no scalar function $f$ such that $\nabla f=\mathbf f$ .

6 0

3 years ago

The shape is composed of squares and quarter circles. Select all the expressions that represent its perimeter. (Select all that

Phantasy [73]

Answer:

91+14π

Step-by-step explanation:

The question is incomplete. Find the complete question attached.

From the diagram shown;

It contains 4 quarter circles which is equal to a circle.

Perimeter of a circle = 2πr

r is the radius = length of one side of the square =7

Perimeter of the circle = 2πr

Perimeter of the circle = 2π(7)

Perimeter of the 4 quarter circles = 14π

Also in the diagram, there are 13 sides of the square present.

Perimeter of the squares = 13×length of one side of the square (Sum of all the sides)

Perimeter of the squares = 13×7 =91

Perimeter of the figure = Perimeter of the 4 quarter circles + Perimeter of the squares

Perimeter of the figure = 91+14π

6 0

3 years ago

Read 2 more answers

HELP PLEASE ASAP NO LINKS OR SCAMS

kozerog [31]

Answer:

a. 5

b. 32

c. 130

d. 26

a. 5.65234

b. 5652.34

c. 8.1253758

d. 8125375.8

Step-by-step explanation:

3 0

3 years ago

Identify the x-intercepts of the function below f(x)=x^2+12x+24

damaskus [11]

<u>ANSWER: </u>

x-intercepts of $\mathrm{x}^{2}+12 \mathrm{x}+24=0 \text { are }(-6+2 \sqrt{3}),(-6-2 \sqrt{3})$

<u>SOLUTION:</u>

Given, $f(x)=x^{2}+12 x+24$ -- eqn 1

x-intercepts of the function are the points where function touches the x-axis, which means they are zeroes of the function.

Now, let us find the zeroes using quadratic formula for f(x) = 0.

$X=\frac{-b \pm \sqrt{b^{2}-4 a c}}{2 a}$

Here, for (1) a = 1, b= 12 and c = 24

$X=\frac{-(12) \pm \sqrt{(12)^{2}-4 \times 1 \times 24}}{2 \times 1}$

$\begin{array}{l}{X=\frac{-12 \pm \sqrt{144-96}}{2}} \\\\ {X=\frac{-12 \pm \sqrt{48}}{2}} \\\\ {X=\frac{-12 \pm \sqrt{16 \times 3}}{2}} \\\\ {X=\frac{-12 \pm 4 \sqrt{3}}{2}} \\ {X=\frac{2(-6+2 \sqrt{3})}{2}, \frac{2(-6-2 \sqrt{3})}{2}} \\\\ {X=(-6+2 \sqrt{3}),(-6-2 \sqrt{3})}\end{array}$