Ask question

Ask question

All categories

3 years ago

12

Use the area to find the length of jk

1 answer:

enyata [817]3 years ago

6 0

The length of Jk= 2.5 m^2

You might be interested in

A bakery made 9 cakes using 3 bags of flour

bearhunter [10]

Answer:

3

Step-by-step explanation:

because you would divide 9 by 3 because 9 cakes and then 3 bags of flower 1 bag of flower to 3 cakes

5 0

3 years ago

If (2i/2+i)-(3i/3+i)=a+bi, then a=<br>A. 1/10<br>B. -10<br>C. 1/50<br>D. -1/10

Verizon [17]

Answer:

Option A is correct.

Step-by-step explanation:

We are given:

$\frac{2i}{2+i}-\frac{3i}{3+i} = a+bi$

We need to find the value of a.

The LCM of (2+i) and (3+i) is (2+i)(3+i)

$=\frac{2i(3+i)}{(2+i)(3+i)}-\frac{3i(2+i)}{(2+i)(3+i)}\\=\frac{6i+2i^2}{(2+i)(3+i)}-\frac{6i+3i^2}{(2+i)(3+i)}\\=\frac{6i+2i^2-(6i+3i^2)}{(2+i)(3+i)}\\=\frac{6i+2i^2-6i-3i^2)}{5+5i}\\=\frac{-i^2}{5+5i}\\i^2=-1\\=\frac{-(-1)}{5+5i}\\=\frac{1}{5+5i}$

Now rationalize the denominator by multiplying by 5-5i/5-5i

$=\frac{1}{5+5i}*\frac{5-5i}{5-5i} \\=\frac{5-5i}{(5+5i)(5-5i)}\\=\frac{5-5i}{(5+5i)(5-5i)}\\(a+b)(a-b)= a^2-b^2\\=\frac{5(1-i)}{(5)^2-(5i)^2}\\=\frac{5(1-i)}{25+25}\\=\frac{5(1-i)}{50}\\=\frac{1-i}{10}\\=\frac{1}{10}-\frac{i}{10}$

We are given

$\frac{2i}{2+i}-\frac{3i}{3+i} = a+bi$

Now after solving we have:

$\frac{1}{10}-\frac{i}{10}=a+bi$

So value of a = 1/10 and value of b = -1/10

So, Option A is correct.

8 0

3 years ago

Geometry i need help with this question pls help

Alexus [3.1K]

Answer:

X its X

Step-by-step explanation:

8 0

2 years ago

Triangle PQR is rotated 180° clockwise about point M. The image of point P is P' and the image of point R is R'. What are the co

Mama L [17]

Answer: C= (1,2)

Step-by-step explanation:

7 0

2 years ago

Consider a Triangle ABC like the one below. Suppose that C = 98, A = 74, and b = 11 (figure is not drawn to scale.) solve the tr

Degger [83]

Answer:

$A=73.8\°$

$B=8.2\°$

$c=76.3\ units$

Step-by-step explanation:

step 1

Find the measure of side c

Applying the law of cosines

$c^{2}= a^{2}+b^{2}-2(a)(b)cos(C)$

substitute the given values

$c^{2}= 74^{2}+11^{2}-2(74)(11)cos(98\°)$

$c^{2}=5,823.5738$

$c=76.3\ units$

step 2

Find the measure of angle A

Applying the law of sine

$\frac{a}{sin(A)}=\frac{c}{sin(C)}$

substitute the given values

$\frac{74}{sin(A)}=\frac{76.3}{sin(98\°)}$

$sin(A)=(74)sin(98\°)/76.3$

$A=arcsin((74)sin(98\°)/76.3)$

$A=73.8\°$

step 3

Find the measure of angle B

we know that

The sum of the internal angles of a triangle must be equal to 180 degrees

so

$A+B+C=180\°$

substitute the given values

$73.8\°+B+98\°=180\°$

$171.8\°+B=180\°$

$B=180\°-171.8\°=8.2\°$

5 0

2 years ago