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11Alexandr11 [23.1K]
2 years ago
13

Identify a pair of perpendicular lines in the given figure.

Mathematics
1 answer:
Georgia [21]2 years ago
8 0
S and t most likely
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Someone please help
VladimirAG [237]
The top answer is correct.
6 0
2 years ago
Please help figure out I will give brainliest!
noname [10]

Answer:

40 student 110 adult

this is using desmos

but to smash the equation you can go like basically we have to cancel out each variable to solve for the other

3x + 8y = 1000

x + y < =150

-3(x + y <= 150)

-3x -3y = -450

3x + 8y = 1000

5y = 550

y= 110

3x + 8 y = 1000

-8(x+y<=150) = -8x - 8y <= -1200

-5x = -200

\frac{ - 5x}{ - 5 }   = x

\frac{ - 200}{ - 5}  = 40

x= 40

40 kids 110 adults

7 0
2 years ago
4x+2y=-24<br> -4x+y = 12
allsm [11]

Answer:

Step-by-step explanation:

                                4x+2y = -24    -------------(i)

                              <u>  -4x+y =  12    -</u>------------(ii)

add (i) &(ii)                      3y = -12

                                         y = -12/3

                                         y = -4

Put y = (-4) in (i)

4x + 2*(-4) = -24

4x - 8 = -24

4x = -24 + 8

4x = -16

x = -16/4

x = -4

6 0
3 years ago
For which values of P and Q does the following equation have infinitely many solutions ?
solmaris [256]

Answer:

(P, Q) = (-75, 57)

Step-by-step explanation:

The equation will have infinitely many solutions when it is a tautology.

Subtract the right side from the equation:

Px +57 -(-75x +Q) = 0

x(P+75) +(57 -Q) = 0

This will be a tautology (0=0) when ...

P+75 = 0

P = -75

and

57-Q = 0

57 = Q

_____

These values in the original equation make it ...

-75x +57 = -75x +57 . . . . . a tautology, always true

4 0
3 years ago
In parallelogram DEFG, DH equals X +3, HF equals 3Y, GH equals 2X -5 and HE equals 5Y plus to find the values of X and Y
daser333 [38]

The values of X and Y are 30 and 11 respectively

<h3>How to determine the values of X and Y?</h3>

The figure that represents the complete question is added as an attachment

The given parameters are:

DH = X +3

HF  = 3Y

GH = 2X -5

HE = 5Y

From the attached parallelogram, we have:

DH = HF

GH = HE

Substitute the known values in the above equation

X + 3 = 3Y

2X - 5 = 5Y

Make X the subject in X + 3 = 3Y

X = 3Y - 3

Substitute X = 3Y - 3 in 2X - 5 = 5Y

2(3Y - 3) - 5 = 5Y

Expand

6Y - 6 - 5 = 5Y

Evaluate the like terms

Y = 11

Substitute Y = 11 in X = 3Y - 3

X = 3*11 - 3

Evaluate

X = 30

Hence, the values of X and Y are 30 and 11 respectively

Read more about parallelograms at:

brainly.com/question/3050890

#SPJ1

8 0
10 months ago
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