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Delicious77 [7]

2 years ago

8

Find the distance between points P(1, 2) and Q(4, 8) to the nearest tenth. i neeeeeed help

1 answer:

Stella [2.4K]2 years ago

5 0

The answer is
the vector PQ has (4-1, 8-2) as components, so it is vecPQ= (3, 6)
using the definition
the distance PQ= sqrt ( 3²+6²)= sqrt(9+36)=sqrt(45)

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Enter your answer in the box.<br>The triangles are similar.<br>What is the value of x?

IrinaVladis [17]

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Step-by-step explanation:

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2 years ago

Find the area of a plane figure bounded by lines

natulia [17]

Answer: 4.5

<u>Step-by-step explanation:</u>

First, find the points of intersection by solving the system.

y = x² + 2x + 4

y = x + 6

Solve by substitution:

x² + 2x + 4 = x + 6 ⇒ x² + x - 2 = 0 ⇒ (x + 2)(x - 1) = 0 ⇒ x = -2, x = 1

Now, integrate from x = -2 to x = 1

$\int\limits^1_2 {(x+6)-(x^{2}+2x+4) } \,$ <em>the bottom of the integral is -2 </em>

= $\int\limits^1_2 {x+6-x^{2}-2x-4 } \,$

= $\int\limits^1_2 {-x^{2}-x+2 } \,$

= $\frac{-x^{3}}{3} - \frac{x^{2}}{2}+2x\int\limits^1_2 {} \,$

= $(\frac{-1^{3}}{3} - \frac{1^{2}}{2}+2(1)) - (\frac{-(-2)^{3}}{3} - \frac{(-2)^{2}}{2}+2(-2))$

= $(\frac{-1}{3} - \frac{1}{2} +2) - (\frac{8}{3} -\frac{4}{2} -4)$

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3 0

3 years ago

Given: m∠EYL=72°<br> Find: mEHL , mLVE

White raven [17]

Answer:

Part 1) The measure of arc EHL is $108\°$

Part 2) The measure of angle LVE is $54\°$

Step-by-step explanation:

step 1

Let

x-----> the measure of arc EHL

y----> the measure of arc EVL

we know that

The measurement of the outer angle is the semi-difference of the arcs it encompasses.

so

$m$

we have

$m$

substitute

$72\°=\frac{1}{2}(y-x)$

$144\°=(y-x)$

$y=144\°+x$ ------> equation A

Remember that

$x+y=360\°$ -----> equation B ( complete circle)

substitute equation A in equation B and solve for x

$x+(144\°+x)=360\°$

$2x=360\°-144\°$

$x=216\°/2=108\°$

Find the value of y

$y=144\°+x$

$y=144\°+108\°=252\°$

therefore

The measure of arc EHL is $108\°$

The measure of arc EVL is $252\°$

step 2

Find the measure of angle LVE

we know that

The inscribed angle measures half that of the arc comprising

Let

x-----> the measure of arc EHL

$m$

we have

$x=108\°$

substitute

$m$

3 0

2 years ago