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LUCKY_DIMON [66]
3 years ago
9

Plz plz answer this i'll give 11 points

Mathematics
1 answer:
shutvik [7]3 years ago
4 0

Answer:

A =bh

Step-by-step explanation:

2 1/2 times 5 1/4 = 13.125

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Ax + c = R (solve for x)
sp2606 [1]
Ax + c = R     (subtract c from each side)

Ax + c - c = R - c

Ax = R - c     (divide A from each side)

Ax/A = (R-c)/A

x = \frac{R-c}{A}  <-answer

4 0
3 years ago
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4 and one-third divided by 5 and StartFraction 1 over 6 EndFraction
tigry1 [53]

Answer:

31/6

Step-by-step explanation:

I got it right on my quiz

3 0
2 years ago
Find an equation of the line l that passes through the point (x, y) = (−9, 2) and satisfies the condition. the x-intercept of l
spayn [35]
-9(2)+4=-16 you do -9 times 2 which is -18 then you subtract that to 4 which you get -16. hope this helps!

3 0
3 years ago
Please help!! best answer gets brainless ​
katen-ka-za [31]

Answer:

  33, 39, 45, 51

Step-by-step explanation:

The general term of an arithmetic sequence is given by the formula ...

  an = a1 + d(n-1) . . . where a1 is the first term and d is the common difference

Comparing this formula to the one you are given, you see that ...

  a1 = 33, d = 6

This means the first term is 33, and each successive term is 6 more than the previous one. The first 4 terms are ...

  33, 39, 45, 51

6 0
2 years ago
−2x=x^2−6
Iteru [2.4K]

Step-by-step explanation:

Example 1

Solve the equation x3 − 3x2 – 2x + 4 = 0

We put the numbers that are factors of 4 into the equation to see if any of them are correct.

f(1) = 13 − 3×12 – 2×1 + 4 = 0 1 is a solution

f(−1) = (−1)3 − 3×(−1)2 – 2×(−1) + 4 = 2

f(2) = 23 − 3×22 – 2×2 + 4 = −4

f(−2) = (−2)3 − 3×(−2)2 – 2×(−2) + 4 = −12

f(4) = 43 − 3×42 – 2×4 + 4 = 12

f(−4) = (−4)3 − 3×(−4)2 – 2×(−4) + 4 = −100

The only integer solution is x = 1. When we have found one solution we don’t really need to test any other numbers because we can now solve the equation by dividing by (x − 1) and trying to solve the quadratic we get from the division.

Now we can factorise our expression as follows:

x3 − 3x2 – 2x + 4 = (x − 1)(x2 − 2x − 4) = 0

It now remains for us to solve the quadratic equation.

x2 − 2x − 4 = 0

We use the formula for quadratics with a = 1, b = −2 and c = −4.

We have now found all three solutions of the equation x3 − 3x2 – 2x + 4 = 0. They are: eftirfarandi:

x = 1

x = 1 + Ö5

x = 1 − Ö5

Example 2

We can easily use the same method to solve a fourth degree equation or equations of a still higher degree. Solve the equation f(x) = x4 − x3 − 5x2 + 3x + 2 = 0.

First we find the integer factors of the constant term, 2. The integer factors of 2 are ±1 and ±2.

f(1) = 14 − 13 − 5×12 + 3×1 + 2 = 0 1 is a solution

f(−1) = (−1)4 − (−1)3 − 5×(−1)2 + 3×(−1) + 2 = −4

f(2) = 24 − 23 − 5×22 + 3×2 + 2 = −4

f(−2) = (−2)4 − (−2)3 − 5×(−2)2 + 3×(−2) + 2 = 0 we have found a second solution.

The two solutions we have found 1 and −2 mean that we can divide by x − 1 and x + 2 and there will be no remainder. We’ll do this in two steps.

First divide by x + 2

Now divide the resulting cubic factor by x − 1.

We have now factorised

f(x) = x4 − x3 − 5x2 + 3x + 2 into

f(x) = (x + 2)(x − 1)(x2 − 2x − 1) and it only remains to solve the quadratic equation

x2 − 2x − 1 = 0. We use the formula with a = 1, b = −2 and c = −1.

Now we have found a total of four solutions. They are:

x = 1

x = −2

x = 1 +

x = 1 −

Sometimes we can solve a third degree equation by bracketing the terms two by two and finding a factor that they have in common.

6 0
3 years ago
Read 2 more answers
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