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2 years ago

What is the length, in units, of the line segment with endpoints at (1, 4) and (3, 7)?

Mathematics

2 answers:

uranmaximum [27]2 years ago

3 0

Answer:

√13

Step-by-step explanation:

GuDViN [60]2 years ago

3 0

Square root of 13 I hope this helps!

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Order the integers from least to greatest.<br><br><br><br> -7 2 6 -4 3

tangare [24]

-7, -4, 2, 3, 6.

Hoped I helped!

7 0

3 years ago

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Ann opened a new savings account with an initial deposit of $250. Which combination will result in a zero in balance in Ann's ac

Blababa [14]

C. (10x3)10 - (27.5x2)10 = 0

8 0

4 years ago

I really need some help .............

Effectus [21]

Answer:

Step-by-step explanation:

add the number of minutes for each day and then divide that number by the number of days

$35+60+65+60+85$ $=305$

$305 / 5 = 61$

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4 years ago

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Describe how to find all the points on a baseball field that are equidistant from second base and third base.

Zielflug [23.3K]

From the given diagram, the following statements are true:

AB = BC

3y = 5y - 22

3y - 5y = -22

-2y = -22

y = 11

Similarly, m<DBE = 50 degrees

m<BED = 90 - m<DBE

m<BED = 90 - 50

m<BED = 40 degrees

Since BE = BC = 3y

BE = BC = 3(11) = 33

Also, 7x - 2 = 40

7x = 40 + 2

7x = 42

x = 6

Get the measure of m<BEA

m<BEA = 7x - 2

m<BEA = 7(6) - 2

m<BEA = 40 degrees

Learn more on angles here: brainly.com/question/16281260

8 0

3 years ago

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