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sladkih [1.3K]
3 years ago
12

PQR is right-angled triangle.calculate the soze of the angle marked correct.

Mathematics
1 answer:
OleMash [197]3 years ago
7 0
The correct answer is 40.3 degrees
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In January, Jeff bought three rabbits. By February, the rabbit population
Murrr4er [49]

Answer:

Jeff will have 48 rabbits by May

5 0
3 years ago
I will Mark you brainliest if you explain your work
Vlad1618 [11]

Answer:

(-1/2, 1)

Step-by-step explanation:

I just used the midpoint formula, and plugged in the points. I hope this helps!

=(x^2+x^1 /2 , y^2+y^1 /2)

=(-2+1 /2 , -5+3 /2)

=(-1/2, 1)

8 0
3 years ago
Read 2 more answers
the perimeter of a rectangle is 60 meters. if the length of the rectangle is 14 meters. what is the width of the rectangle.,
Ray Of Light [21]
Perimeter is 2width+2length
Therefore since you already know that the perimeter is 60 and the length is 14 you can set up the equation; 
60= 2w+14
with this you then need to find w so;
60=2w+14
-14     -14
46=2w
(46/2)=(2w/2)
13=w
The width is 13
8 0
3 years ago
Read 2 more answers
Pls show work! questions on pic
Lena [83]

Answer:

a.(10,23)

b.(7,30)

e.(20,20)

Step-by-step explanation:

We are given that

x=Width of garden

y=Length of garden

According to question

x\geq 5

y\geq 20

Perimeter of rectangle=2(x+y)

Where x=Width  of rectangle

y=Length of rectangle

Fencing used =2(x+y)

2(x+y)\leq 80

First we change inequality  into equality

x=5...(1)

y=20...(2)

2(x+y)=80

x+y=40...(3)

Substitute x=0 in equation (3)

y=40

Substitute y=0  in equation (3)

x=40

Substitute x=0 and y=0 in equation (3)

2(x+y)\leq 80

0\leq 80

It is true. Therefore, the  shaded region below the line.

0\geq 5

0\geq 20

These are false equation .Therefore, the shaded region above the line.

Substitute x=5 in equation (3)

5+y=40

y=40-5=35

Substitute y=20 in equation (3)

x+20=40

x=40-20=20

Intersection point of equation (1) and (3) is (5,35) and intersect point of equation (2) and (3) is (20,20).

Now,

a.(10,23)

Substitute in

2(x+y)\leq 80

2(10+23)\leq 80

66\geq 80

10>5 and 23>20

it satisfied all inequality.

Hence, it is a solution.

b.(7,30)

7>5

30>20

2(7+30)=74

it satisfied all inequality.

Hence, it is a solution.

c.(18,25)

18>5

25>20

2(18+25)=86>80

It does not satisfied third inequality

Hence, it is not a solution.

d.(8,35)

8>5

35>20

2(8+35)=86>80

It does not satisfied third inequality

Hence, it is not a solution.

e.(20,20)

2(20+20)=80

20>5

20=20

it satisfied all inequality.

Hence, it is a solution.

4 0
3 years ago
Find the number of the different ways, for 3 students to sit on 7 seats in one row.
qwelly [4]

Answer:

210

Step-by-step explanation:

Here comes the problem from Combination.

We are being asked to find the number of ways out in which 3 students may sit on 7 seats in a row. Please see that in this case the even can not be repeated.

Let us start with the student one. For him all the 7 seats are available to sit. Hence number of ways for him to sit = 7

Let us see the student second. For him there are only 6 seats available to sit as one seat has already been occupied. Hence number of ways for him to sit = 6

Let us see the student third. For him there are only 5 seats available to sit as two seat has already been occupied. Hence number of ways for him to sit = 5

Hence the total number of ways for three students to be seated will be

7 x 6 x 5

=210

3 0
3 years ago
Read 2 more answers
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