I not sure if you mean the gradient of the line or the equation, so here is both.
Gradient (=m) :
( 4, 4 ) ( 6, 5 )
( x1 , y1) ( x2, y2 )
m = y2-y1 / x2-x1
= -5-4 / 6 - 4
= -9 / 2 (as a fraction)
= -4.5 (as a decimal)
Equation of the line :
m = -9/2
( 4, 4 )
( a, b )
y - b =m (x - a)
y - 4= -9/2 (x - 4)
(Multiply both sides by 2 to get rid of fraction)
2y - 4 = -9 (x - 4)
2y - 4 = -9x + 36
2y = -9x + 36 + 4
2y = -9x + 40
2y + 9x = 40
Please feel free to ask if you have any questions
Answer:
2
Step-by-step explanation:
2
Answer:
20
Step-by-step explanation:
Do the opposite to get the answer
8*2.5=20
Then to check it
20/8=2.5
*the diagram of the Russian stringed instrument is attached below.
Answer/Step-by-step explanation:
To show that the traingular parts of the two balalaikas instruments are congruent, substitute x = 6, to find the missing measurements that is given in both ∆s.
Parts of the first ∆:
WY = (2x - 2) in = 2(6) - 2 = 12 - 2 = 10 in
m<Y = 9x = 9(6) = 54°.
XY = 12 in
Parts of the second ∆:
m<F = 72°
HG = (x + 6) in = 6 + 6 = 12 in
HF = 10 in
m<G = 54°
m<H = 180 - (72° + 54°)
m<H = 180 - 126
m<H = 54°
From the information we have, let's match the parts that are congruent to each other in both ∆s:
WY ≅ FH (both are 10 in)
XY ≅ GH (both are 12 in)
<Y ≅ <G (both are 54°)
Thus, since two sides (WY and XY) and an included angle (<Y) of ∆WXY is congruent to two corresponding sides (FH and GH) and an included angle (<G) in ∆FGH, therefore, ∆WXY ≅ ∆FGH by the Side-Angle-Side (SAS) Congruence Theorem.
This is enough proof to show that the triangular parts of the two balalaikas are congruent for x = 6.
Answer:
24.
Step-by-step explanation:
First, use the Order of Operations:
PEMDAS.
First, divide, then multiply to get your solution.
21 / 7 = 3.
3 · 8 = 24.
Therefore, your answer is 24.