F(x)=2(x)^2+5 square root (x+2) f(0) Posted on October 22, 2021 By mobete 9 Comments on F(x)=2(x)^2+5 square root (x+2) f(0) F(x)=2(x)^2+5 square root (x+2) f(0) Mathematics

[tex]f(x) = 2(x)^{2} +5 \sqrt{x+2}[/tex][tex]f(2) = 2(2)^{2} + 5 \sqrt{(2)+2}[/tex][tex]f(2) = 2(4) + 5\sqrt{4}[/tex][tex]f(2) = 8 + 5(2)[/tex][tex]f(2)= 8+10[/tex]f(2) = 18Reply

[tex]f(x)[/tex] is a fancy term for saying [tex]y[/tex]. It is mainly used in Trigonometry, Pre-Calculus, Calculus, and beyond.[tex]f(x) = 2x^2 + 5\sqrt{x+2}[/tex]plug in the [tex]2[/tex] into the [tex]x[/tex].[tex]f(2)= 2(2)^2 + 5\sqrt{2+2}[/tex]this gives:[tex]f(2) = 8 + 5\sqrt{4}[/tex]square the 4.[tex]f(2) = 8 + 5(2)[/tex]multiply [tex]5[/tex] and [tex]2[/tex].[tex]f(2) = 8 + 10[/tex]add them together.[tex]f(2) = 18[/tex]Glad to help. Reply

[tex]f(x)[/tex] is a fancy term for saying [tex]y[/tex]. It is mainly used in Trigonometry, Pre-Calculus, Calculus, and beyond.[tex]f(x) = 2x^2 + 5\sqrt{x+2}[/tex]plug in the [tex]2[/tex] into the [tex]x[/tex].[tex]f(2)= 2(2)^2 + 5\sqrt{2+2}[/tex]this gives:[tex]f(2) = 8 + 5\sqrt{4}[/tex]square the 4.[tex]f(2) = 8 + 5(2)[/tex]multiply [tex]5[/tex] and [tex]2[/tex].[tex]f(2) = 8 + 10[/tex]add them together.[tex]f(2) = 18[/tex]Glad to help. Reply

Hi Felicia![tex]\rm f(x)=2x^2+5\sqrt{x+2}[/tex]Remember that your square root function can only calculate positive values and zero. We can't put a negative under a square root.This means that the "stuff" under the root must be greater than or equal to zero (positive).[tex]\rm x+2\ge0[/tex]Solve for x in this equation, subtracting 2 from each side,[tex]\rm x\ge -2[/tex]This is our domain, all x values -2 or larger.Hope that helps! 🙂Reply

[tex]f(x)=2x^2+5\sqrt{x+2}\\\\f(0)\to substitute\ x=0:\\\\f(0)=2\cdot0^2+5\sqrt{0+2}=0+5\sqrt2=5\sqrt2\approx5\cdot1.41=7.05[/tex]Reply

[tex]f(x)=2x^2+5\sqrt{x+2}\\\\f(0)\to substitute\ x=0:\\\\f(0)=2\cdot0^2+5\sqrt{0+2}=0+5\sqrt2=5\sqrt2\approx5\cdot1.41=7.05[/tex]Reply

6.41Step-by-step explanation:[tex]If f(x)=2(x)^2+5 square root(x+2), complete the following statement f(0)=[/tex]Reply

[tex]f(x) = 2(x)^{2} +5 \sqrt{x+2}[/tex]

[tex]f(2) = 2(2)^{2} + 5 \sqrt{(2)+2}[/tex]

[tex]f(2) = 2(4) + 5\sqrt{4}[/tex]

[tex]f(2) = 8 + 5(2)[/tex]

[tex]f(2)= 8+10[/tex]

f(2) = 18

[tex]f(x)[/tex] is a fancy term for saying [tex]y[/tex]. It is mainly used in Trigonometry, Pre-Calculus, Calculus, and beyond.

[tex]f(x) = 2x^2 + 5\sqrt{x+2}[/tex]

plug in the [tex]2[/tex] into the [tex]x[/tex].

[tex]f(2)= 2(2)^2 + 5\sqrt{2+2}[/tex]

this gives:

[tex]f(2) = 8 + 5\sqrt{4}[/tex]

square the 4.

[tex]f(2) = 8 + 5(2)[/tex]

multiply [tex]5[/tex] and [tex]2[/tex].

[tex]f(2) = 8 + 10[/tex]

add them together.

[tex]f(2) = 18[/tex]

Glad to help.

[tex]f(x)[/tex] is a fancy term for saying [tex]y[/tex]. It is mainly used in Trigonometry, Pre-Calculus, Calculus, and beyond.

[tex]f(x) = 2x^2 + 5\sqrt{x+2}[/tex]

plug in the [tex]2[/tex] into the [tex]x[/tex].

[tex]f(2)= 2(2)^2 + 5\sqrt{2+2}[/tex]

this gives:

[tex]f(2) = 8 + 5\sqrt{4}[/tex]

square the 4.

[tex]f(2) = 8 + 5(2)[/tex]

multiply [tex]5[/tex] and [tex]2[/tex].

[tex]f(2) = 8 + 10[/tex]

add them together.

[tex]f(2) = 18[/tex]

Glad to help.

Hi Felicia!

[tex]\rm f(x)=2x^2+5\sqrt{x+2}[/tex]

Remember that your square root function can only calculate positive values and zero. We can't put a negative under a square root.

This means that the "stuff" under the root must be greater than or equal to zero (positive).

[tex]\rm x+2\ge0[/tex]

Solve for x in this equation, subtracting 2 from each side,

[tex]\rm x\ge -2[/tex]

This is our domain, all x values -2 or larger.

Hope that helps! 🙂

[tex]f(x)=2x^2+5\sqrt{x+2}\\\\f(0)\to substitute\ x=0:\\\\f(0)=2\cdot0^2+5\sqrt{0+2}=0+5\sqrt2=5\sqrt2\approx5\cdot1.41=7.05[/tex]

[tex]f(x)=2x^2+5\sqrt{x+2}\\\\f(0)\to substitute\ x=0:\\\\f(0)=2\cdot0^2+5\sqrt{0+2}=0+5\sqrt2=5\sqrt2\approx5\cdot1.41=7.05[/tex]

6.41

Step-by-step explanation:

[tex]If f(x)=2(x)^2+5 square root(x+2), complete the following statement f(0)=[/tex]

f(x) = 2 x^2 + 5 sqrt(x+2)

f(0)= 2 (0)^2 + 5 sqrt(2)

5* sqrt (0)= 7.07

hope that helps

n

Step-by-step explanation: