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elena55 [62]
3 years ago
9

Five times a number is 28 greater than the product of the number and -2. Find the number

Mathematics
1 answer:
Fynjy0 [20]3 years ago
6 0

Answer:

4

Step-by-step explanation:

Let the unknown number be x

Five times the number is expressed as 5x  

28 greater than the product of the number and -2 is expressed as;

28 + (-2x)

Equating both expressions

5x = 28 + (-2x)

5x = 28 - 2x

5x+2x = 28

7x = 28

x = 28/7

x = 4

Hence the number is 4

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(8w)^2 without parentheses
Nataly [62]

Answer:

I believe the answer would be 64w2.

Step-by-step explanation:

8 squared is 8 x 8, which equals 64, and then w squared is just w2, since we don't know what it is. Put them together, and the answer is 64w2.

5 0
4 years ago
X-y=5 and x^2y=5x+6​
sergeinik [125]

By applying algebraic handling on the two equations, we find the following three <em>solution</em> pairs: x₁ ≈ 5.693 ,y₁ ≈ 10.693; x₂ ≈ 1.430, y₂ ≈ 6.430; x₃ ≈ - 0.737, y₃ ≈ 4.263.

<h3>How to solve a system of equations</h3>

In this question we have a system formed by a <em>linear</em> equation and a <em>non-linear</em> equation, both with no <em>trascendent</em> elements and whose solution can be found easily by algebraic handling:

x - y = 5      (1)

x² · y = 5 · x + 6       (2)

By (1):

y = x + 5

By substituting on (2):

x² · (x + 5) = 5 · x + 6

x³ + 5 · x² - 5 · x - 6 = 0

(x + 5.693) · (x - 1.430) · (x + 0.737) = 0

There are three solutions: x₁ ≈ 5.693, x₂ ≈ 1.430, x₃ ≈ - 0.737

And the y-values are found by evaluating on (1):

y = x + 5

x₁ ≈ 5.693

y₁ ≈ 10.693

x₂ ≈ 1.430

y₂ ≈ 6.430

x₃ ≈ - 0.737

y₃ ≈ 4.263

By applying algebraic handling on the two equations, we find the following three <em>solution</em> pairs: x₁ ≈ 5.693 ,y₁ ≈ 10.693; x₂ ≈ 1.430, y₂ ≈ 6.430; x₃ ≈ - 0.737, y₃ ≈ 4.263.

To learn more on nonlinear equations: brainly.com/question/20242917

#SPJ1

8 0
2 years ago
Determine whether the improper integral converges or diverges, and find the value of each that converges.
Rus_ich [418]

Answer:

Diverges

Step-by-step explanation:

We can solve this by using integral by parts:

Let

u=lnx            dv=(1/x)dx

du=(1/x)dx    v=lnx

\int\limits {lnx/x} \, dx = (lnx)^2-\int\limits {lnx/x} \, dx

We can add

\int\limits {lnx/x} \, dx to both sides

\int\limits{lnx/x} \, dx = (1/2)\cdot{(lnx)^2}

We can evaluate the limit between 2 and infinity.

If x tends to infinity the limit will be infinity and therefore the integral diverges to ∞

6 0
3 years ago
PLEASE HELP QUICKLY how can I accurately solve this equation?
elena55 [62]

Answer:

multiply by the power of ten,

Step-by-step explanation:

3 0
3 years ago
Which statement is true about the discontinuities of the function f(x) = x-5/3x^2-17x-28
o-na [289]
1) function f(x)

              x - 5
f(x) = ----------------
          3x^2 - 17x - 28

2) factor the denominator:

3x^2 - 17x  - 28 = (3x + 4)(x - 7)

                        x - 5
=> f(x) = -----------------------
                (3x + 4) (x  - 7)

3) Find the limits when x → - 4/3 and when x → 7

Lim of f(x) when x → - 4/3 = +/- ∞

=> vertical assymptote x = - 4/3

Lim of f(x) when x → 7 = +/- ∞

=> vertical assymptote x = 7

Answer: there are assympotes at x = 7 and x = - 4/3
6 0
3 years ago
Read 2 more answers
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