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3 years ago

what is as big as an elephant but does not weigh as much as an elephant ill give a brainiest to the first correct answer

Mathematics

2 answers:

Inga [223]3 years ago

6 0

An elephants shadow

SOVA2 [1]3 years ago

4 0

an elephant's shadow??

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PLEASE HELP I WILL MARK BRANLIEST

Finger [1]

Answer:

SAS

Step-by-step explanation:

We must prove that triangles ABC and EDC are congruent.

Since BD bisects AE, then AC is congruent with CE.

Since AE bisects BD, then BC is congruent with CD

Angle C is 90° in the triangle EDC and is also 90° in triangle BCA because they are vertical angles.

Being two sides and the included angle congruent, then both triangles are similar by the SAS theorem.

Answer: SAS

4 0

2 years ago

Which of the following is true?

gtnhenbr [62]

Answer:

C..... I think it is a answer

6 0

2 years ago

Is 2/3 equal to 2/4 ? if not is it greater

Dimas [21]

Answer:

Equivalent fraction of a given fraction is got by multiplying or dividing its numerator and denominator by the same whole number. For example, if we multiply the numerator and denominator of 2/3 by 4 we get. 2/3 = 2×4 / 3×4 = 8/12 which is an equivalent fraction of 2/3.

Step-by-step explanation:

5 0

2 years ago

Troy’s canoe is 18 feet long. Ariel’s canoe is the length of Troy’s canoe.

tankabanditka [31]

Well the answer i think is going to be 18

3 0

3 years ago

Read 2 more answers

This is a question on my partial fractions homework, but no matter what I try I can't figure it out..

Ierofanga [76]

$\dfrac{x^2+x+1}{(x+1)^2(x+2)}=\dfrac{a_1x+a_0}{(x+1)^2}+\dfrac b{x+2}$
$\implies\dfrac{x^2+x+1}{(x+1)^2(x+2)}=\dfrac{(a_1x+a_0)(x+2)+b(x+1)^2}{(x+1)^2(x+2)}$
$\implies x^2+x+1=(a_1+b)x^2+(2a_1+a_0+2b)x+(2a_0+b)$
$\implies\begin{cases}a_1+b=1\\2a_1+a_0+2b=1\\2a_0+b=1\end{cases}\implies a_1=-2,a_0=-1,b=3$

So you have

$\displaystyle\int_0^2\frac{x^2+x+1}{(x+1)^2(x+2)}\,\mathrm dx=-2\int_0^2\frac x{(x+1)^2}\,\mathrm dx-\int_0^2\frac{\mathrm dx}{(x+1)^2}+3\int_0^2\frac{\mathrm dx}{x+2}$
$=\displaystyle-2\int_1^3\dfrac{x-1}{x^2}\,\mathrm dx-\int_0^2\frac{\mathrm dx}{(x+1)^2}+3\int_0^2\frac{\mathrm dx}{x+2}$

where in the first integral we substitute $x\mapsto x+1$ .

$=\displaystyle-2\int_1^3\left(\frac1x-\frac1{x^2}\right)\,\mathrm dx-\frac1{1+x}\bigg|_{x=0}^{x=2}+3\ln|x+2|\bigg|_{x=0}^{x=2}$
$=-2\left(\ln|x|+\dfrac1x\right)\bigg|_{x=1}^{x=3}-\dfrac23+3(\ln4-\ln2)$
$=-2\left(\ln3+\dfrac13-1\right)-\dfrac23+3\ln2$
$=\dfrac23+\ln\dfrac89$

4 0

3 years ago