All categories

3 years ago

PLS ANSWER RN What is the area of AJKL?

Mathematics

2 answers:

alex41 [277]3 years ago

7 0

Answer:

8.5

Step-by-step explanation:

slope of the line that contains KL

(y2 - y1)/(x2-x1)

(0-1)/(7-3)

m = -1/4

slope of the line that contains JK

(5-1)/(4-3)

m = 4

the linea are perpendicular so triangle is right

area = (leg1 x leg2)/2

KL = $\sqrt{(7-3)^2 + (0-1)^2} = \sqrt{16 + 1} = \sqrt{17}$

KJ = $\sqrt{(4-3)^2 + (5-1)^2}= \sqrt{1 + 16} = \sqrt{17}$

area = (√17)^2/2 = 17/2 = 8.5

swat323 years ago

6 0

8.5

….hope this helps!

You might be interested in

(y +12) (y - 18) (x) (z) x= y = Z =

Wittaler [7]

Answer:

YURRRRRRRRRRRRRRRRRRRRRRR

Step-by-step explanation:

3 0

3 years ago

The capacity of a cylindrical water tank is 539 litres. If its height is 1.4 meter then find the radius of the base.

natulia [17]

Answer:

350m

Step-by-step explanation:

capacity= volume/1000
but we know that volume of cylinder is πr²h so substitute it I'm the volume space
then substitute for π as 22/7, h as 1.4 , and also capacity as 539
539= ((22/7)×r²×1.4)/1000
when you simplify you get r as 350m

6 0

3 years ago

Given JKN = LMN and LKN and JMN. Prove: JKLM is a parallelogram.

inessss [21]

Answer:

JNM = KNL and JKL = LMJ

Step-by-step explanation:

6 0

3 years ago

Add the fractions and simplify the answer 1/2 +2 /3

IrinaVladis [17]

What we are going to do is add $\frac{1}{2}$ by $\frac{2}{3}$ . The first thing we do is make common denominators for $\frac{1}{2}$ and $\frac{2}{3}$ : $\frac{1*3}{2*3}= \frac{3}{6} \ and \ \frac{2*2}{3*2}= \frac{4}{6}$ . Now we add $\frac{3}{6}$ by $\frac{4}{6}$ : $\frac{3}{6}+ \frac{4}{6}= \frac{7}{6}$ . Now we make $\frac{7}{6}$ a mixed number: $\frac{7}{6}=1 \frac{1}{6}$ . Therefore, your answer is $\boxed{\boxed{1 \frac{1}{6}}}$ . Hoped I helped. :)