(8x 2 −15x)−(x 2 −27x)=ax 2 +bxleft parenthesis, 8, x, squared, minus, 15, x, right parenthesis, minus, left parenthesis, x, squ
quester [9]
Answer:
<h2>5</h2>
Step-by-step explanation:
Given the expression (8x² −15x)−(x² −27x) = ax² +bx, we are to determine the value of b-a. Before we determine the vwlue of b-a, we need to first calculate for the value of a and b from the given expression.
On expanding the left hand side of the expression we have;
= (8x² −15x)−(x² −27x)
Open the paranthesis
= 8x² −15x−x²+27x
collect the like terms
= 8x²−x²+27x −15x
= 7x²+12x
Comparing the resulting expression with ax²+bx
7x²+12x = ax²+bx
7x² = ax²
a = 7
Also;
12x = bx
b =12
The value of b - a = 12 - 7
b -a = 5
Hence the value of b-a is equivalent to 5
The answer is 10 because its in the tens place
Answer:
Given
Step-by-step explanation:
Given that: △RST ~ △VWX, TU is the altitude of △RST, and XY is the altitude of △VWX.
Comparing △RST and △VWX;
TU ~ XY (given altitudes of the triangles)
<TUS = <XYW (all right angles are congruent)
<UTS ≅ <YXW (angle property of similar triangles)
Thus;
ΔTUS ≅ ΔXYW (congruent property of similar triangles)
<UTS + <TUS + < UST = <YXW + <XTW + <XWY = (sum of angles in a triangle)
Therefore by Angle-Angle-Side (AAS), △RST ~ △VWX
So that:
= (corresponding side length proportion)