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3 years ago

How to find an area of this figure?: thanks!

Mathematics

2 answers:

mixas84 [53]3 years ago

7 0

First find the area of the triangle, then find the area of the rectangle. Finally add them together and you will get your answer.

bearhunter [10]3 years ago

3 0

The answer to that shape is 216

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The least common multiple of two numbers is 3780, and the greatest common divisor is 18. given that one of the numbers is 180, w

garri49 [273]

To solve these GCF and LCM problems, factor the numbers you're working with into primes:
3780 = 2*2*3*3*3*5*7
180 = 2*2*3*3*5
We know that the LCM of the two numbers, call them A and B, = 3780 and that A = 180. The greatest common factor of 180 and B = 18 so B has factors 2*3*3 in common with 180 but no other factors in common with 180. So, B has no more 2's and no 5's
Now, LCM(180,B) = 3780. So, A or B must have each of the factors of 3780. B needs another factor of 3 and a factor of 7 since LCM(A,B) needs for either A or B to have a factor of 2*2, which A has, and a factor of 3*3*3, which B will have with another factor of 3, and a factor of 7, which B will have.
So, B = 2*3*3*3*7 = 378.

8 0

2 years ago

Is this true or false need answer asap

VLD [36.1K]

Answer: True

Step-by-Step Explanation:

=> 2x + 3y = -7 (Eq. 1)
=> -x = 2y (Eq. 2)
=> x = 1, y = -3

Substitute values of ‘x’ and ‘y’ in Eq. 1 :-

=> 2x + 3y = -7
= 2(1) + 3(-3) = -7
= 2 + -9 = -7
=> -7 = -7
=> LHS = RHS

Therefore, it is a Solution.

3 0

2 years ago

May used 2/7 of a ribbon to make bows for her cousins. Now, she has 10/21 of a meter of ribbon left.

leva [86]

Answer:

Step-by-step explanation:

1+1=21

4 0

2 years ago

Read 2 more answers

Simplest form of 39/27

Deffense [45]

Both the top and bottom are divisible by three, and 13/9ths cannot be simplified anymore

8 0

3 years ago

Use Lagrange multipliers to find the maximum and minimum values of (i) f(x,y)-81x^2+y^2 subject to the constraint 4x^2+y^2=9. (i

sp2606 [1]

i. The Lagrangian is

$L(x,y,\lambda)=81x^2+y^2+\lambda(4x^2+y^2-9)$

with critical points whenever

$L_x=162x+8\lambda x=0\implies2x(81+4\lambda)=0\implies x=0\text{ or }\lambda=-\dfrac{81}4$

$L_y=2y+2\lambda y=0\implies2y(1+\lambda)=0\implies y=0\text{ or }\lambda=-1$

$L_\lambda=4x^2+y^2-9=0$

If $x=0$ , then $L_\lambda=0\implies y=\pm3$ .
If $y=0$ , then $L_\lambda=0\implies x=\pm\dfrac32$ .
Either value of $\lambda$ found above requires that either $x=0$ or $y=0$ , so we get the same critical points as in the previous two cases.

We have $f(0,-3)=9$ , $f(0,3)=9$ , $f\left(-\dfrac32,0\right)=\dfrac{729}4=182.25$ , and $f\left(\dfrac32,0\right)=\dfrac{729}4$ , so $f$ has a minimum value of 9 and a maximum value of 182.25.

ii. The Lagrangian is

$L(x,y,z,\lambda)=y^2-10z+\lambda(x^2+y^2+z^2-36)$

with critical points whenever

$L_x=2\lambda x=0\implies x=0$ (because we assume $\lambda\neq0$ )

$L_y=2y+2\lambda y=0\implies 2y(1+\lambda)=0\implies y=0\text{ or }\lambda=-1$

$L_z=-10+2\lambda z=0\implies z=\dfrac5\lambda$

$L_\lambda=x^2+y^2+z^2-36=0$

If $x=y=0$ , then $L_\lambda=0\implies z=\pm6$ .
If $\lambda=-1$ , then $z=-5$ , and with $x=0$ we have $L_\lambda=0\implies y=\pm\sqrt{11}$ .

We have $f(0,0,-6)=60$ , $f(0,0,6)=-60$ , $f(0,-\sqrt{11},-5)=61$ , and $f(0,\sqrt{11},-5)=61$ . So $f$ has a maximum value of 61 and a minimum value of -60.

5 0

3 years ago