All categories

2 years ago

The diagram shows a trapezium

Mathematics

1 answer:

wel2 years ago

4 0

Answer:

SEE THE ABOVE ATTACHMENT TO SOLVE THE VERY INTERESTING & INTELLIGENT QUESTION.

Step-by-step explanation:

BUT THERE IS A SLIGHT MISTAKE IN THE QUESTION ; THE 125° WILL BE 120°. AND NOW, SEE THE PROCESS CAREFULLY AND SOLVE AND GET YOUR ANSWER READY.

FOR SOLVING SUCH INTERESTING AND INTELLIGENT QUESTION L, PLEASE COMMENT ME, I'M THERE TO SOLVE AND HELP YOU.

----- WITH REGARDS AN INDIAN....!

You might be interested in

Math help please!!!! will be marking brainly

zhuklara [117]

The answer is C. 3/5

20 girls, 3 boys. 23/40 is about 60 percent, or 3/5.

I wrote the wrong answer on accident for whatever reason when I first sent this, I apologize.

7 0

3 years ago

Help please!!! Thank you

Anna11 [10]

Answer: ∠ABE=20°

Step-by-step explanation:

∵ AB=BC

∴ ΔABC is a isosceles triangle

∵ ∠A=30

∴ ∠C=30

∵ The interior angle sum of triangle is 180°

∵ ∠A+∠B+∠C=180

30+∠B+30=180

∠B+60=180

∠B=120 (∠ABC)

∵ ∠EBD+∠ABE+∠ABC=linear pair

∴∠EBD+∠ABE+∠ABC=180°

4x+2x+120=180

6x+120=180

6x=60

x=10

∠ABE=2x=2(10)=20°

Hope this helps!! :)

Please let me know if you have any question

6 0

3 years ago

Marine biologists have determined that when a shark detectsthe presence of blood in the water, it will swim in the directionin w

siniylev [52]

Solution :

a). The level curves of the function :

$C(x,y) = e^{-(x^2+2y^2)/10^4}$

are actually the curves

$e^{-(x^2+2y^2)/10^4}=k$

where k is a positive constant.

The equation is equivalent to

$x^2+2y^2=K$

$\Rightarrow \frac{x^2}{(\sqrt K)^2}+\frac{y^2}{(\sqrt {K/2})^2}=1, \text{ where}\ K = -10^4 \ln k$

which is a family of ellipses.

We sketch the level curves for K =1,2,3 and 4.

If the shark always swim in the direction of maximum increase of blood concentration, its direction at any point would coincide with the gradient vector.

Then we know the shark's path is perpendicular to the level curves it intersects.

b). We have :

$\triangledown C= \frac{\partial C}{\partial x}i+\frac{\partial C}{\partial y}j$

$\Rightarrow \triangledown C =-\frac{2}{10^4}e^{-(x^2+2y^2)/10^4}(xi+2yj),$ and

$\triangledown C$ points in the direction of most rapid increase in concentration, which means $\triangledown C$ is tangent to the most rapid increase curve.