Answer:
HELPPPPPP ); WILL GIVE BRAINLIST.
1 answer:
<h3>Answer:</h3>
- f(1) = 2
- No. The remainder was not 0.
Synthetic division is quick and not difficult to learn. The number in the upper left box is the value of x you're evaluating the function for (1). The remaining numbers across the top are the coefficients of the polynomial in decreasing order by power (the way they are written in standard form). The number at lower left is the same as the number immediately above it—the leading coefficient of the polynomial.
Each number in the middle row is the product of the x-value (the number at upper left) and the number in the bottom row just to its left. The number in the bottom row is the sum of the two numbers above it.
So, the number below -4 is the product of x (1) and 1 (the leading coefficient). That 1 is added to -4 to give -3 on the bottom row. Then that is multiplied by 1 (x, at upper left) and written in the next column of the middle row. This proceeds until you run out of numbers.
The last number, at lower right, is the "remainder", also the value of f(x). Here, it is 2 (not 0) for x=1, so f(1) = 2.
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Answer:
MC = 4.5cm
Step-by-step explanation:
Question:
Let the isosceles triangle ABC with AB = AC = 3 cm. if the mediator of the sides AC intersects with the side BC in M and the perimeter of the triangle AMC = 12 cm. Calculate MC.
Solution:
Find attached the diagram used in solving the question.
Given:
∆ABC is an isosceles triangle (two sides and angles are equal)
AB = BC = 3cm
Perimeter of ∆AMC = 12cm
From the diagram, M cuts AC at the the middle.
AD = CD = AC/2 = 3/2
Perimeter of Right angled ∆AMD = AM + AD + MD
= 3/2 + AM +MD
Perimeter of Right angled ∆CMD =CM + CD + MD
= 3/2 + CM +MD
Right angled ∆AMD = Right angled ∆CMD
CM = AM
Therefore ∆AMC is an isosceles triangle
CM = AM (two sides of an isosceles triangle are equal)
Let CM = AM = x
Perimeter of ∆AMC = AM + CM + AC
12 = x + x + 3
12 = 2x + 3
2x = 12-3
2x = 9
x = 9/2 = 4.5
CM = AM = 4.5cm
MC = CM = 4.5cm
The bus stop method is just like division.
Ex
137÷5=
1.) 5 goes into one zero times so there is a zero on top.
2.) 5 goes into 13 two times, so after the zero there is a 2.
3.) 5 goes into 37 seven times so there is a 7 after the 2.
4.) There is a 2 left and five can't go into school therefore that is a remainder.
027R2
5/137
<u>- 10</u>
37
<u>-35</u>
2
Hoped I helped:D
Ex
137÷5=
1.) 5 goes into one zero times so there is a zero on top.
2.) 5 goes into 13 two times, so after the zero there is a 2.
3.) 5 goes into 37 seven times so there is a 7 after the 2.
4.) There is a 2 left and five can't go into school therefore that is a remainder.
027R2
5/137
<u>- 10</u>
37
<u>-35</u>
2
Hoped I helped:D