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3 years ago

Solve: 3(-x+2) < 15 A X>7 B x < 3 C x<-> D x>-3

Mathematics

1 answer:

sukhopar [10]3 years ago

7 0

Answer:

D x>-3

Step-by-step explanation:

3(-x+2) < 15

-3x + 6< 15 (Subtract 6 from both sides)

-3x < 9 (Divide 3 on both sides)

x>-3 (Flip the inequality because the variable was negative)

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Find the equation of the line that contains the given point and has the given slope-

mylen [45]

Answer:

y-5 = -3x

y = -3x+5

Step-by-step explanation:

We can use the point slope form of the equation

y-y1 = m(x-x1) where m is the slope and (x1,y1) is a point on the line

y-5 = -3(x-0)

y-5 = -3x

We can change it to slope intercept form y= mx+b by adding 5 to each side

y-5+5 = -3x+5

y = -3x+5

3 0

3 years ago

Read 2 more answers

Verify that the roots of 5x²- 6x -2 = 0 are <img src="https://tex.z-dn.net/?f=%5Cfrac%7B3%20%2B%20%5Csqrt%7B19%7D%20%7D%7B5%7D%2

Mice21 [21]

Answer:

Proof below.

Step-by-step explanation:

Quadratic Formula

$x=\dfrac{-b \pm \sqrt{b^2-4ac} }{2a}\quad\textsf{when }\:ax^2+bx+c=0$

Given quadratic equation:

$5x^2-6x-2=0$

Define the variables:

a = 5
b = -6
c = -2

Substitute the defined variables into the quadratic formula and solve for x:

$\implies x=\dfrac{-(-6) \pm \sqrt{(-6)^2-4(5)(-2)}}{2(5)}$

$\implies x=\dfrac{6 \pm \sqrt{36+40}}{10}$

$\implies x=\dfrac{6 \pm \sqrt{76}}{10}$

$\implies x=\dfrac{6 \pm \sqrt{4 \cdot 19}}{10}$

$\implies x=\dfrac{6 \pm \sqrt{4}\sqrt{19}}{10}$

$\implies x=\dfrac{6 \pm2\sqrt{19}}{10}$

$\implies x=\dfrac{3 \pm \sqrt{19}}{5}$

Therefore, the exact solutions to the given quadratic equation are:

$x=\dfrac{3 + \sqrt{19}}{5} \:\textsf{ and }\:x=\dfrac{3 - \sqrt{19}}{5}$

Learn more about the quadratic formula here:

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3 0

2 years ago

Read 2 more answers

Let f(x) = log(x). Find values of a such that f(kaa) = kf(a).

meriva

Answer:

a = $k^{\frac{1}{k-2}}$

Step-by-step explanation:

Given:

f(x) = log(x)

and,

f(kaa) = kf(a)

now applying the given function, we get

⇒ log(kaa) = k × log(a)

⇒ log(ka²) = k × log(a)

Now, we know the property of the log function that

log(AB) = log(A) + log(B)

and,

log(Aᵇ) = b × log(A)

Thus,

⇒ log(k) + log(a²) = k × log(a) (using log(AB) = log(A) + log(B) )

⇒ log(k) + 2log(a) = k × log(a) (using log(Aᵇ) = b × log(A) )

⇒ k × log(a) - 2log(a) = log(k)

⇒ log(a) × (k - 2) = log(k)

⇒ log(a) = (k - 2)⁻¹ × log(k)

⇒ log(a) = $\log(k^{\frac{1}{k-2}})$ (using log(Aᵇ) = b × log(A) )

taking anti-log both sides

⇒ a = $k^{\frac{1}{k-2}}$

3 0

3 years ago

A help would be greatly appreciated!

dalvyx [7]

The tree was approximately 40.8ft tall.

You use the equation, a^2+b^2=c^2.
12^2 + 39^2 = x^2, add the results of 12^2 and 39^2 to make 1665 = x^2. You can take the positive root of both sides, 3√185, and it equals 40.80441152, which is rounded to 40.8.

7 0

4 years ago

Graph the following three equations on your own device and answer the questions below.

Alina [70]

Due to length restrictions, we kindly invite to read the explanation of this question for further details on functions.

<h3>What are the characteristics of each of the three functions?</h3>

a) In this part we must evaluate each of the three functions at the given value of x:

Case 1

f(10) = 5 · 10 + 7

f(10) = 57

Case 2

f(10) = 10² + 6

f(10) = 106

Case 3

f(10) = 2¹⁰ + 3

f(10) = 1027

b) In this part we must evaluate each of the three functions at the given value of x:

Case 1

f(100) = 5 · 100 + 7

f(100) = 507

Case 2

f(100) = 100² + 6

f(100) = 10006

Case 3

f(100) = 2¹⁰⁰ + 3

f(100) = 1.267 × 10³⁰ + 3

c) In this part we must evaluate each of the three functions at the given value of x:

Case 1

f(1000) = 5 · 1000 + 7

f(1000) = 5007

Case 2

f(1000) = 1000² + 6

f(1000) = 1000006

Case 3

f(1000) = 2¹⁰⁰⁰ + 3

f(1000) = (1.268 × 10³⁰)¹⁰ + 3

f(1000) = 10.744 × 10³⁰⁰ + 3

f(1000) = 1.074 × 10³⁰¹ + 3

e) The third function increases the fastest.

f) In this part we need to compare the third function with respect to the first and second functions:

5 · x + 7 = 2ˣ + 3

2ˣ - 5 · x = 4

The solutions of the equation are x = - 0.675 and x = 4.81. The function will exceed the other first function at x = 4.81.

x² + 6 = 2ˣ + 3

2ˣ - x² = 3

The solution of the equation is x = 4.588. The function will exceed the other second function at x = 4.588.

g) Yes, exponential functions with bases greater than 1 will surpass polynomic function at any point x such that x > 0.

h) The domain represents the set of x-values of a function and the range represents the set of y-values of a function. Then, the domain and range of each function is:

Case 1

Domain - All real numbers.

Range - All real numbers

Case 2

Domain - All real numbers.

Range - [6, +∞)

Case 3

Domain - All real numbers.

Range - (3, +∞)

To learn more on functions: brainly.com/question/12431044

#SPJ1

5 0

2 years ago