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quester [9]
3 years ago
6

Please help please help please help also so subscribe to mr beast

Mathematics
2 answers:
beks73 [17]3 years ago
7 0
Well Area is multiplying all the angles so just multiply the angles and your good hope this helps
vivado [14]3 years ago
5 0
I will subscribe to mr beast
You might be interested in
How do you find the measures of angles G,E and F? Help pleez
JulsSmile [24]

m∠HDG = 28°, m∠EFG = 50°, m∠DEG = 67°, m∠DGE = 65°

Solution:

Triangle sum property:

Sum of the angles of the triangle = 180°

In ΔDHG,

m∠HDG + 120° + 32° = 180°

m∠HDG + 152° = 180°

m∠HDG = 180° – 152°

m∠HDG = 28°

In ΔGEF,

m∠EFG + 17° + 113° = 180°

m∠EFG + 130° = 180°

m∠EFG = 180° – 130°

m∠EFG = 50°

Sum of the adjacent angles in a straight line is 180°

m∠DEG + m∠DEF = 180°

m∠DEG + 113° = 180°

m∠DEG = 180° – 113°

m∠DEG = 67°

In ΔDGE,

m∠DGE + 48° + 67° = 180°

m∠DGE + 115° = 180°

m∠DGE = 180° – 115°

m∠DGE = 65°

Hence m∠HDG = 28°, m∠EFG = 50°, m∠DEG = 67°, m∠DGE = 65°.

4 0
3 years ago
What is 12.9 written as a mixed number?
Elanso [62]

Answer:12 9/10

Step-by-step explanation:

7 0
3 years ago
Read 2 more answers
In a florist shop, the ratio of roses to tulips is 5 : 3. The shop has 72 tulips on Friday morning. If the florist uses up half
11Alexandr11 [23.1K]

Answer:

it would stay the same

Step-by-step explanation:

3 0
3 years ago
Which relation is a function?
gulaghasi [49]
For functions don’t repeat the x (x,y)
It is B
5 0
4 years ago
Α' β
yulyashka [42]

Answer:

8x² +36x -27=0

Step-by-step explanation:

2x²= 6x +3

Let's rewrite the equation into the form of ax²+bx+c= 0.

2x² -6x -3=0

Thus, a= 2

b= -6

c= -3

Sum of roots= -  \frac{b}{a}

Since your roots are p and q, sum of roots= p +q

p +q= - (\frac{ - 6}{2} )

p +q= 6 ÷2

p +q= 3

Product of roots= \frac{c}{a}

pq= -  \frac{3}{2}

<u>Quadratic equations</u><u>:</u>

x² -(sum of roots)x +(product of roots)= 0

Thus, we have to find the sum and the product of the new roots, p²q and pq².

p +q= 3

pq= -3/2

Product of new roots

= (p²q)(pq²)

= p³q³

= (pq)³

= ( -  \frac{3}{2} )^{3}  \\  =  -  \frac{27}{8}

sum of new roots

= p²q +pq²

= pq(p +q)

= (-3/2)(3)

= -9/2

Thus, the quadratic equation with roots p²q and pq² is

x² -(-9/2)x -27/8 = 0

x ^{2}  +  \frac{9}{2} x -  \frac{27}{8}  = 0

Multiply by 8 throughout:

8 {x}^{2}  + 36x - 27 = 0

3 0
3 years ago
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