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1 year ago

Then use f(x)= 5x+13 Choose a number between 1-10 and use that for your x value then SOLVE.

Mathematics

2 answers:

Sonja [21]1 year ago

5 0

We will choose x=2

Plug in x=2 as the input:
f(2)=5(2)+13

Use PEMDAS to simplify:
f(2)=10+13

f(2)=23

g100num [7]1 year ago

5 0

The answer to your question is. F(2)=23

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Please help me out!

seropon [69]

Answer:

Second option

Third option

Fourth option

Step-by-step explanation:

We have the following quadratic function

$f(x) =(x-1)(x+7)$

Use the distributive property to multiply the expression

$f(x) =x^2+7x-x-7$

$f(x) =x^2+6x-7$

For a function of the form $f (x) = ax ^ 2 + bx + c$ the x coordinate of the vertex is:

$x =-\frac{b}{2a}$

Then in this case the coordinate of the vertex is:

$x =-\frac{6}{2(1)}$

$x =-3$

To obtain the y coordinate of the vertex we evaluate the function at $x = -3$

$f(-3) =(-3)^2+6(-3)-7$

$f(-3) =9-18-7$

$f(-3) =-16$

Then the vertex is: (-3, -16)

We can see in the graph that the zeros of the function are x=1 and x=-7

Then the function is decreasing from -∞ to -3 and then it is increasing from -3 to ∞

The function is positive for $x$ and $x> 1$

The correct answers are:

Second option

Third option

Fourth option

7 0

2 years ago

For the inverse variation equation p = StartFraction 8 Over V EndFraction, what is the value of V when p = one-half?

likoan [24]

Answer:

V=16

Step-by-step explanation:

We are given that an inverse variation

$p=\frac{8}{V}$

We have to find the value of V when p=1/2

To find the value of V we will substitute the value of p

Substitute the value of p

Then, we get

$\frac{1}{2}=\frac{8}{V}$

$\frac{V}{2}=8$

Using multiplication property of equality

$V=8\times 2$

Using multiplication property of equality

$V=16$

Hence, the value of V=16

3 0

2 years ago

1.2(4x – 1) = 8x – 2

zysi [14]

Step-by-step explanation:

1.2(4x-1)=8x-2=6x

3 0

2 years ago

Solve 5x^2 − 3x + 17 = 9.

Vedmedyk [2.9K]

Answer:

The roots of the equation are x = $\frac{3+\sqrt{151}i}{10}$ and x = $\frac{3-\sqrt{151}i}{10}$

and there are no real roots of the equation given above

Step-by-step explanation:

To solve:

5x² − 3x + 17 = 9

⇒ 5x² − 3x + 17 - 9 = 0

⇒ 5x² − 3x + 8 = 0

Now,

the roots of the equation in the form ax² + bx + c = 0 is given as:

x = $\frac{-b\pm\sqrt{b^2-4ac}}{2a}$

in the above given equation

a = 5

b = -3

c = 8

therefore,

x = $\frac{-(-3)\pm\sqrt{(-3)^2-4\times5\times8}}{2\times5}$

x = $\frac{3\pm\sqrt{9-160}}{10}$

x = $\frac{3+\sqrt{-151}}{10}$ and x = $\frac{3-\sqrt{-151}}{10}$

x = $\frac{3+\sqrt{151}i}{10}$ and x = $\frac{3-\sqrt{151}i}{10}$

here i = √(-1)

Hence,

The roots of the equation are x = $\frac{3+\sqrt{151}i}{10}$ and x = $\frac{3-\sqrt{151}i}{10}$

and there are no real roots of the equation given above

7 0

2 years ago

Please and thank you

vampirchik [111]

Answer:

is this free points. cuz why you said please and thank you?

Step-by-step explanation:

7 0

2 years ago