Answer:
B) f(x) ≥ 3x + 4
g of x is less than or equal to negative one half times x minus 5
Step-by-step explanation:
From the above graph,
If point B is the solution of the system of inequalities, then point B must satisfies both inequalities
We can see from the graph that
Point B is located above (≥) the boundary line for f(x) and is also below (≤) the boundary line for g(x).
So we can say,
Point b satisfies the inequalities
Therefore,
The system is
f(x) ≥ (3x+4)
g(x) ≤ (-1/2x -5)
Hopes this Helps :)
X^2 - 13 x = -36
add 36 to both sides
x^2 - 13x +36 = 0
figure out the factor
(x-4)(x-9) = 0
so (x-4) or (x-9) equals 0
so solution is
x-4=0 add 4 to both sides and get x = 4
x-9=0 add 9 to both sides and get x=9
x={ 4, 9}
Answer:
x = 8
Step-by-step explanation:
Δ TSU and Δ TRV are similar , thus the ratios of corresponding sides are equal, that is
= 
, substitute values
= 
= 
( cross- multiply )
3x = x + 16 ( subtract x from both sides )
2x = 16 ( divide both sides by 2 )
x = 8
We will solve this by suing simultaneous equations,
⇒ 5s + 3j = 87
4s + 2j = 64
Multiply the first equation with 4 and the second one with 5, this is to get one of the values equal so that we can cancel them out,
⇒ (5s + 3j = 87) × 4
(4s + 2j = 64) × 5
∴ ⇒ 20s + 12j = 348
20s + 10j = 320
Subtract both the equations. This is how your result (after subtraction) should look like,
⇒ 2j = 28
∴ ⇒ j = $14
Now replace the value of 'j' in one of the original equations,
⇒ 4s + 2(14) = 64
⇒ 4s + 28 = 64
⇒ 4s = 64 - 28
⇒ 4s = 36
∴ ⇒ s = $9
Therefore, one pair of jeans cost $14 and a shirt costs $9
Hope you understood! Feel free to ask me if you didn't understand a step.
First term ,a=4 , common difference =4-7=-3, n =50
sum of first 50terms= (50/2)[2×4+(50-1)(-3)]
=25×[8+49]×-3
=25×57×-3
=25× -171
= -42925
derivation of the formula for the sum of n terms
Progression, S
S=a1+a2+a3+a4+...+an
S=a1+(a1+d)+(a1+2d)+(a1+3d)+...+[a1+(n−1)d] → Equation (1)
S=an+an−1+an−2+an−3+...+a1
S=an+(an−d)+(an−2d)+(an−3d)+...+[an−(n−1)d] → Equation (2)
Add Equations (1) and (2)
2S=(a1+an)+(a1+an)+(a1+an)+(a1+an)+...+(a1+an)
2S=n(a1+an)
S=n/2(a1+an)
Substitute an = a1 + (n - 1)d to the above equation, we have
S=n/2{a1+[a1+(n−1)d]}
S=n/2[2a1+(n−1)d]
