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Zigmanuir [339]
3 years ago
9

Help I’m timed I’ll mark u brainliesttt

Mathematics
1 answer:
Zigmanuir [339]3 years ago
4 0
It would be 5sqrt2

a^2 + b^2 = c^2
5^2 + 5^2 = c^2
25 + 25 = c^2
50 = c^2
c = sqrt 50
c = 5sqrt2
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Morgan has a loyalty card good for a discount at her local grocery store. The item she wants to buy is priced at $43, before dis
rewona [7]

Answer:

20%

43*0.8=34.4

Hope This Helps!!!

6 0
3 years ago
Calculate the area of triangle ABC with altitude CD, given A (6, -2), B (1, 3), C (5, 5), and D (2, 2).
Flura [38]

Answer:

15 units

Step-by-step explanation:

I just took this geometry test with the same question. Its 15

3 0
3 years ago
Read 2 more answers
6. The value of (tan 1° tan2º tan 3º ... tan 89°) is​
Thepotemich [5.8K]

Answer:

1

Step-by-step explanation:

Using the trigonometric identities

tan(90 - x) = cotx , cotx = \frac{1}{tanx}

Given

tan1tan2tan3....................... tan87tan88tan89

= tan1tan2tan3............... tan(90-3)tan(90-2)(tan90 - 1)

= tan1tan2tan3.............. cot3cot2cot1

= tan1cot1tan2cot2tan3cot3 ........................

= 1 × 1 × 1 ×....................... × 1

= 1

4 0
3 years ago
Use lagrange multipliers to find the shortest distance, d, from the point (4, 0, −5 to the plane x y z = 1
Varvara68 [4.7K]
I assume there are some plus signs that aren't rendering for some reason, so that the plane should be x+y+z=1.

You're minimizing d(x,y,z)=\sqrt{(x-4)^2+y^2+(z+5)^2} subject to the constraint f(x,y,z)=x+y+z=1. Note that d(x,y,z) and d(x,y,z)^2 attain their extrema at the same values of x,y,z, so we'll be working with the squared distance to avoid working out some slightly more complicated partial derivatives later.

The Lagrangian is

L(x,y,z,\lambda)=(x-4)^2+y^2+(z+5)^2+\lambda(x+y+z-1)

Take your partial derivatives and set them equal to 0:

\begin{cases}\dfrac{\partial L}{\partial x}=2(x-4)+\lambda=0\\\\\dfrac{\partial L}{\partial y}=2y+\lambda=0\\\\\dfrac{\partial L}{\partial z}=2(z+5)+\lambda=0\\\\\dfrac{\partial L}{\partial\lambda}=x+y+z-1=0\end{cases}\implies\begin{cases}2x+\lambda=8\\2y+\lambda=0\\2z+\lambda=-10\\x+y+z=1\end{cases}

Adding the first three equations together yields

2x+2y+2z+3\lambda=2(x+y+z)+3\lambda=2+3\lambda=-2\implies \lambda=-\dfrac43

and plugging this into the first three equations, you find a critical point at (x,y,z)=\left(\dfrac{14}3,\dfrac23,-\dfrac{13}3\right).

The squared distance is then d\left(\dfrac{14}3,\dfrac23,-\dfrac{13}3\right)^2=\dfrac43, which means the shortest distance must be \sqrt{\dfrac43}=\dfrac2{\sqrt3}.
7 0
3 years ago
The length of a room is 4 feet less than twice its width. The perimeter of
AlladinOne [14]

L=2W-4

PERIMETER=2L+2W

58=2(2W-4)+2W

58=4W-8+2W

58=6W-8

6W=58+8

6W=66

W=66/6

W=11 ANS. FOR THE WIDTH.

L=2*11-4

L=22-4

L=18 ANS. FOR THE LENGTH.

PROOF:

58=2*18=2*11

58=36+22

58=58

5 0
3 years ago
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