Here in an idea:
lets say his total is repreented with the variable t.
lets say the number of bags of apples is represented by a.
t = a + 2
unless there is missing info.
hope this helps!
The answer is 29 and 31 because by the definition of a prime number it is a natural number that is greater than 1 and has no positive divisors other than 1 and itself.
Answer:
a = A/πb
Step-by-step explanation:
To solve this subject of the formulae given that A = πab.
<u>solution</u>
A = πab
the next step is to look for a unique way to get rid of the variables disturbing "a" from standing alone. this variables are π and b, we need to detach dem from a
A = πab
divide both sides by πb
A/πb = πab/πb
A/πb = a
a = A/πb
therefore the value of a in the fomular A = πab is evaluated to be a = A/πb
*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.