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ch4aika [34]
3 years ago
10

Find out the value for brainiest

Mathematics
1 answer:
borishaifa [10]3 years ago
4 0

Answer:

126 degrees

Step-by-step explanation:

first we should find the measurements of these angles

so (I know these are the names of lines but there's no name for these angles, that's why I'm addressing them with the name of each one's nearest line name)

C= 90 degree

B= 65 degree

A= 79 (coz 180- 101= 79)

therefore D= ?

the total interior angle of a quadrilateral is 360

so 360- (90+ 65+ 79)

360- 234 = 126 degrees

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A little bit of geometry how many triangles do you see
Neko [114]
I would have to say That i see 8

6 0
3 years ago
Read 2 more answers
For what value is a is (a,9) a solution of the equation y=2x+1
yanalaym [24]

Answer:

a = 4

Step-by-step explanation:

y = 2x + 1 ......(1)

Going through y - y1 = m(x - x1)

m is slope and it is 2

y1 = 9 and x1 = a

y - 9 = 2(x - a)

y - 9 = 2x - 2a

y = 2x - 2a + 9 ........(2)

Equating (1) and (2)

2x + 1 = 2x - 2a + 9

Collecting like terms

2x - 2x + 2a = 9 - 1

2a = 8

a = 8/2

a = 4

7 0
3 years ago
Can anyone please explain? Need some help :)
DedPeter [7]

Answer:

93.5 square units

Step-by-step explanation:

Diameter of the Circle = 12 Units

Therefore:

Radius of the Circle = 12/2 =6 Units

Since the hexagon is regular, it consists of 6 equilateral triangles of side length 6 units.

Area of the Hexagon = 6 X Area of one equilateral triangle

Area of an equilateral triangle of side length s = \dfrac{\sqrt{3} }{4}s^2

Side Length, s=6 Units

\text{Therefore, the area of one equilateral triangle =}\dfrac{\sqrt{3} }{4}\times 6^2\\\\=9\sqrt{3} $ square units

Area of the Hexagon

= 6 X 9\sqrt{3} \\=93.5$ square units (to the nearest tenth)

7 0
3 years ago
The diagonals of a quadrilateral QRST intersect at P(-1,3). QRST has vertices at Q(3,6) and R(-4,5). What must be the coordinate
erica [24]

S = (-5,0)

T = (2,1)

Step-by-step explanation:

Step 1 :

Given

Q = (3,6) and R = (-4,5). P = (-1,3)

Let S be (a,b) and T be (c,d)

The diagonals of a parallelogram bisect each other. so in order to ensure that QRST is a parallelogram, P must be the mid point of the diagonals QS and RT.

Step 2 :

P is the midpoint  of QS

So we have (3+a) ÷ 2 =  -1  and (6 + b) ÷ 2 = 3

=> 3 + a = -2    and 6 + b = 6

=> a = -5  and b =0

So S should be (-5,0)

Step 3 :

P is the midpoint  of RT

So we have (-4+c) ÷ 2 =  -1  and (5 + d) ÷ 2 = 3

=> -4+ c = -2    and 5 + d = 6

=> c = 2  and d =1

So T should be (2,1)

Step 4 :

Answer :

S = (-5,0)

T = (2,1)

6 0
3 years ago
5)
Fudgin [204]

Answer:

C

Step-by-step explanation:

3 0
3 years ago
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