Point e is the midpoint of ca and bd. which statements about the diagram are true?

Mathematics

1 answer:

defon2 years ago

8 0

Answer:

Step-by-step explanation: i don’t know to be honest

You might be interested in

Please help me on this question

Marrrta [24]

It is the right answer........

4 0

3 years ago

Are all squares simular true or false

enyata [817]

ANSWER:
true

STEP-BY-STEP EXPLANATION:

5 0

2 years ago

What is the result of isolating y2 in the equation below?

lyudmila [28]

STEP-BY-STEP SOLUTION:

( x + 4 )^2 + y^2 = 22

x^2 + 2 × x × 4 + 4^2 + y^2 = 22

x^2 + 8x + 16 + y^2 = 22

y^2 = 22 - x^2 - 8x - 16

y^2 = - x^2 - 8x + 6

FINAL ANSWER:

Therefore, the answer is:

D. y^2 = - x^2 - 8x + 6

Please mark as brainliest if you found this helpful! :)
Thank you <3

6 0

3 years ago

If f(x)=2−x12 and g(x)=x2−9, what is the domain of g(x)÷f(x)?

Keith_Richards [23]

$\bf \begin{cases}
f(x)=2-x^{12}\\
g(x)=x^2-9\\
g(x)\div f(x)=\frac{g(x)}{f(x)}
\end{cases}\implies \cfrac{x^2-9}{2-x^{12}}$

now, for a rational expression, the domain, or "values that x can safely take", applies to the denominator NOT becoming 0, because if the denominator is 0, then the rational turns to undefined.

now, what value of "x" makes this denominator turn to 0, let's check by setting it to 0 then.

$\bf 2-x^{12}=0\implies 2=x^{12}\implies \pm\sqrt[12]{2}=x\\\\
-------------------------------\\\\
\cfrac{x^2-9}{2-x^{12}}\qquad \boxed{x=\pm \sqrt[12]{2}}\qquad \cfrac{x^2-9}{2-(\pm\sqrt[12]{2})^{12}}\implies \cfrac{x^2-9}{2-\boxed{2}}\implies \stackrel{und efined}{\cfrac{x^2-9}{0}}$

so, the domain is all real numbers EXCEPT that one.

4 0

2 years ago

During a shark week

bearhunter [10]

Answer:

25 feet

Step-by-step explanation:

Multiply 2.5 and 10

8 0

2 years ago