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kvv77 [185]
3 years ago
11

One hypotenuse and one leg of a right triangle have lengths 61 and 11. Find the length of the other leg

Mathematics
1 answer:
atroni [7]3 years ago
5 0
If the length is the same as one leg, than all the other legs should be the same exact dimensions, legs can not be different measurements by height. Hope that this help you!
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Examine the following table. Determine whether the rate of change is constant or variable. If it is constant, state the rate of
xenn [34]

Answer:

Variable rate of change.

Step-by-step explanation:

Plot the points and you'll see that the graph is not a straight line, therefore the rate of change is variable.

5 0
2 years ago
AB is tangent to the circle k(O) at B, and AD is a secant, which goes through center O. Point O is between A and D∈k(O). Find m∠
drek231 [11]

Answer:

∠BAD=20°20'

∠ADB=34°90'

Step-by-step explanation:

AB is tangent to the circle k(O), then AB⊥BO. If the measure of arc BD is 110°20', then central angle ∠BOD=110°20'.

Consider isosceles triangle BOD (BO=OD=radius of the circle). Angles adjacent to the base BD are equal, so ∠DBO=∠BDO. The sum of all triangle's angles is 180°, thus

∠BOD+∠BDO+∠DBO=180°

∠BDO+∠DBO=180°-110°20'=69°80'

∠BDO=∠DBO=34°90'

So ∠ADB=34°90'

Angles BOD and BOA are supplementary (add up to 180°), so

∠BOA=180°-110°20'=69°80'

In right triangle ABO,

∠ABO+∠BOA+∠OAB=180°

90°+69°80'+∠OAB=180°

∠OAB=180°-90°-69°80'

∠OAB=20°20'

So, ∠BAD=20°20'

5 0
3 years ago
Read 2 more answers
Mhm help please! quarantine got me
Karolina [17]
The answer would be 9.4
3 0
3 years ago
Read 2 more answers
A farmer has 5 times as many cows as horses.
Assoli18 [71]

Answer:

The correct answer is !40 cows and 28 Horses

Step-by-step explanation:

h+5h=168

6h=168

h=168/6

h=28 horses.

5*28=140 cows.

Proof:

28=140=168

168=168

7 0
3 years ago
Find an equation of the plane that contains the points p(5,−1,1),q(9,1,5),and r(8,−6,0)p(5,−1,1),q(9,1,5),and r(8,−6,0).
topjm [15]
Given plane passes through:
p(5,-1,1), q(9,1,5), r(8,-6,0)

We need to find a plane that is parallel to the plane through all three points, we form the vectors of any two sides of the triangle pqr:
pq=p-q=<5-9,-1-1,1-5>=<-4,-2,-4>
pr=p-r=<5-8,-1-6,1-0>=<-3,5,1>

The vector product pq x pr gives a vector perpendicular to both pq and pr.  This vector is the normal vector of a plane passing through all three points
pq x pr
=
  i   j   k
-4 -2 -4
-3  5  1
=<-2+20,12+4,-20-6>
=<18,16,-26>

Since the length of the normal vector does not change the direction, we simplify the normal vector as
N = <9,8,-13>

The required plane must pass through all three points.
We know that the normal vector is perpendicular to the plane through the three points, so we just need to make sure the plane passes through one of the three points, say q(9,1,5).

The equation of the required plane is therefore
Π :  9(x-9)+8(y-1)-13(z-5)=0
expand and simplify, we get the equation
Π  :  9x+8y-13z=24

Check to see that the plane passes through all three points:
at p: 9(5)+8(-1)-13(1)=45-8-13=24
at q: 9(9)+8(1)-13(5)=81+9-65=24
at r: 9(8)+8(-6)-13(0)=72-48-0=24
So plane passes through all three points, as required.

3 0
3 years ago
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