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Finger [1]
2 years ago
6

Please help me i will mark as brainliest ​

Mathematics
1 answer:
Bingel [31]2 years ago
3 0
6. P = 2
Q = -5

7. C = -1

8. K = 3

9. A) y = 2x + 2
B) y = -3x + 5
C) y = 0x + -5

10. Y = 5x + -18

11. A)
i) (5, 0)
ii) (0, 5)

B) y = -1x + 0
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I need help with number 3
olya-2409 [2.1K]

Answer: He would earn $14

Step-by-step explanation: Since Sam had 11 games but only 9 of them were working just take 11 - 9 which = 2 then take 7 x 2 to get $14

Hope this helps

3 0
3 years ago
Read 2 more answers
10 points!!!!!?!!!! Tyrone wants to know how much wrapping paper is needed to cover a box with the dimensions shown
Leviafan [203]

Answer: B.54 Square inches

Step-by-step explanation:  Surface Area

= 2(lb + bh + lh)

= 2(6 x 3 + 3 x 1 + 6 x 1)

= 2(18 + 3 + 6)

= 2(27)

= 54 square inches

7 0
2 years ago
How do you prove each of the following theorems using either a two-column, paragraph, or flow chart proof?
lilavasa [31]

All the theorems are proved as follows.

<h3>What is a Triangle ?</h3>

A triangle is a polygon with three sides , three vertices and three angles.

1. The Triangle sum Theorem

According to the Triangle Sum Theorem, the sum of a triangle's angles equals 180 degrees.

To create a triangle ABC, starting at point A, move 180 degrees away from A to arrive at point B.

We turn 180 degrees from B to C and 180 degrees from C to return to A, giving a total turn of 360 degrees to arrive to A.

180° - ∠A + 180° - ∠B + 180° - ∠C = 360°

- ∠A - ∠B  - ∠C = 360° - (180°+ 180°+ 180°) = -180°

∠A + ∠B  + ∠C = 180°

(Hence Proved)

2. Isosceles Triangle Theorem

Considering an isosceles triangle ΔABC

with AB = AC, we have by sine rule;

\rm \dfrac{sinA}{BC} =  \dfrac{sinB}{AC} =  \dfrac{sinC}{AB}\\

as AB = AC

sin B = sin C

angle B = angle C

3.Converse of the Isosceles theorem

Consider an isosceles triangle ΔABC with ∠B= ∠C, we have by sine rule;

\rm \dfrac{sinA}{BC} =  \dfrac{sinB}{AC} =  \dfrac{sinC}{AB}\\

as  ∠B= ∠C ,

AB = AC

4. Midsegment of a triangle theorem

It states that the midsegment of two sides of a triangle is equal to (1/2)of the third side parallel to it.

Given triangle ABC with midsegment at D and F of AB and AC respectively, DF is parallel to BC

In ΔABC and ΔADF

∠A ≅ ∠A

BA = 2 × DA, BC = 2 × FA

Hence;

ΔABC ~ ΔADF (SAS similarity)

BA/DA = BC/FA = DF/AC = 2

Hence AC = 2×DF

5.Concurrency of Medians Theorem

A median of a triangle is a segment whose end points are on vertex of the triangle and the middle point of the side ,the medians of a triangle are concurrent and  the point of intersection is inside the triangle known as Centroid .

Consider a triangle ABC , X,Y and Z are the midpoints of the sides

Since the medians bisect the segment AB into AZ + ZB

BC into BX + XB

AC into AY + YC

Where:

AZ = ZB

BX = XB

AY = YC

AZ/ZB = BX/XB = AY/YC = 1

AZ/ZB × BX/XB × AY/YC = 1 and

the median segments AX, BY, and CZ are concurrent (meet at point within the triangle).

To know more about Triangle

brainly.com/question/2773823

#SPJ1

8 0
1 year ago
Can someone answer these please !!!!!!
katen-ka-za [31]
A. m^2 - 36 = 0
m^2 = 36
m = √36 = 6 or -6

b. 3x^2 + 15 = 0
3x^2 = -15
x^2 = -5
x = √(-5) not possible unless complex numbers is applied
-1 = i^2
x = √(-5) = √(5i^2) = √5i

c. 4d^2 + 16 = 16
4d^2 = 0
d^2 = 0
d = 0


Hope it helped!
4 0
2 years ago
What is the answer to this?
Svetach [21]

Answer: F: I only

Step-by-step explanation:

7 0
3 years ago
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