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kicyunya [14]
3 years ago
10

If line t is perpendicular to both line l and line m, then ∠1 and ∠2 are both ____ angles. Question acute right obtuse complemen

tary

Mathematics
2 answers:
ella [17]3 years ago
8 0
They are both right angles. 
Masja [62]3 years ago
8 0

Answer:

∠1 and ∠2 both are right angles.

Step-by-step explanation:

Given that if line t is perpendicular to both line l and line m, then

we have to find about the angle ∠1 and ∠2

As given line l is perpendicular to both line l and line m,

Therefore the point at which the ;line intersect the two lines l and m at 90° i.e

∠1 and ∠2 both angles are formed at the intersection points of line t with l and m.

Hence, ∠1 and ∠2 both are right angles.

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Evaluate the expression<br><br> 5c - 2 = when c = 3<br><br> HELPPP quick
Ugo [173]

Answer:

13

Step-by-step explanation:

Substitute 3 into c --> 5(3)-2 = 15-2 = 13

8 0
3 years ago
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Will give brainliest lots of points A is the midpoint of LB,Lb =6x-17 and AB=2x+3
NARA [144]
A is the mid point ==> AL = AB = 2AB
6x-17=2(2x+3) => 6x -17 = 4x +6 => 2x = 23 =>x=23/2
then just plug x into LB and AB
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I WILL GIVE BRAINLISET!! no links por favor <br><br> 8) Simplify.
Marina CMI [18]

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√ 15  + √ 10

Step-by-step explanation:

The answer is above                                                                            

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3 years ago
What is 4p+5=33 ? For the equation
faltersainse [42]

Answer:

p=7

Step-by-step explanation:

4p+5=33

33-5=28

4p=28

28/4=7

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Suppose you solved a second-order equation by rewriting it as a system and found two scalar solutions: y = e^5x and z = e^2x. Th
xenn [34]

Answer:

The solutions are linearly independent because the Wronskian is not equal to 0 for all x.

The value of the Wronskian is \bold{W=-3e^{7x}}

Step-by-step explanation:

We can calculate the Wronskian using the fundamental solutions that we are provided and their corresponding the derivatives, since the Wroskian is defined as the following determinant.

W = \left|\begin{array}{cc}y&z\\y'&z'\end{array}\right|

Thus replacing the functions of the exercise we get:

W = \left|\begin{array}{cc}e^{5x}&e^{2x}\\5e^{5x}&2e^{2x}\end{array}\right|

Working with the determinant we get

W = 2e^{7x}-5e^{7x}\\W=-3e^{7x}

Thus we have found that the Wronskian is not 0, so the solutions are linearly independent.

3 0
3 years ago
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