Slope = (y2-y1)/(x2-x1)
The points: (2,0) and (0,3)
(3-0)/(0-2) = 3/-2 = -3/2
The solution is -3/2
<h3>Answer:</h3>
a) l = 5 + 2w
b) P = 2(l +w) = 2(5 +2w +w)
c) w = 8 cm, l = 21 cm
<h3>Explanation:</h3>
a) Using w for width, twice the width is 2w, and 5 more than that is 5+2w. This is your expression for l:
... l = 5 + 2w
b) The perimeter is the sum of the lengths of all sides, so is the sum of twice the length and twice the width.
... P = 2(l +w) = 2(5 +2w + w) . . . . substituting for l using the expression of (a)
c) Substituting the given perimeter, we have ...
... 58 = 2(5 +3w) = 10 + 6w . . . . . collect terms, eliminate parentheses
... 48 = 6w . . . . . . . . . . . . . . . . . . subtract 10
... 8 = w . . . . . . . . divide by 6
... l = 5 + 2·8 = 21 . . . . . find length using the formula for it
Dimensions are in centimeters, so we can write the solution as ...
- width = 8 cm
- length = 21 cm
Answer:
3971.570
Step-by-step explanation:
Answer:
The answer is ΔJMK ≈ ΔMLK ≈ ΔJLM ⇒ answer (A)
Step-by-step explanation:
* Lets start with the equal angles i the three triangles
- In ΔJMK
∵ m∠JKM = 90°
∴ m∠KJM + m∠KMJ = 90 ⇒ (1)
- IN ΔMLK
∵ m∠MKL = 90°
∴ m∠KML + m∠KLM = 90° ⇒ (2)
∵ m∠KMJ + m∠KML = 90° ⇒ (3)
- From (1) , (2) , (3)
∴ m∠KJM = m∠KML
∴ m∠KMJ = m∠KLM
* Now lets check the condition of similarity in the 3 triangles
- At first ΔJMK and ΔMLK
- In triangles JMK , MLK
∵ m∠KJM = m∠KML
∵ m∠KMJ = m∠KLM
∵ m∠JKM = m∠MKL
∴ ΔJMK ≈ ΔMLK ⇒ (4)
- At second ΔJMK and ΔJLM
∵ m∠KJM = m∠MJL
∵ m∠KMJ = m∠MLJ
∵ m∠JKM = m∠JML
∴ ΔJMK ≈ ΔJLM ⇒ (5)
* If two triangles are similar to one triangle, then they are
similar to each other
- From (4) and(5)
∴ ΔJMK ≈ ΔMLK ≈ ΔJLM
270 + 63 = 333
If you did it in your head, it would look like this:
9(30) + 9(7)
270 + 63 = 333
