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4 years ago

How many x-intercepts does the graph of y = 2x2 + 5x + 9 have?

Mathematics

1 answer:

Komok [63]4 years ago

4 0

Answer:

<h2>The graph of y = 2x² + 5x + 9 has zero x-intercepts.</h2>

Step-by-step explanation:

x-intercepts are for y = 0.

Put y = 0 to the equation y = 2x² + 5x + 9:

2x² + 5x + 9 = 0

Calculate the discriminant of quadratic equation ax² + bx + c = 0:

Δ = b² - 4ac

if Δ < 0, then an equation has no solution

if Δ = 0, then an equation has one solution

if Δ > 0, then an equation has two solution.

2x² + 5x + 9 = 0

a = 2, b = 5, c = 9

Δ = 5² - 4(2)(9) = 25 - 72 = -47 < 0 - no solution

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Read 2 more answers

4. Use the process outlined in the lesson to approximate the number 2√3. Use the approximation √3 ≈ 1.732 050 8.

jeka57 [31]

Answer:

sequence of five intervals

(1) 2 < $2^{\sqrt{3} }$ < $2^{2}$

(2) $2^{1.7}$ < $2^{\sqrt{3} }$ < $2^{1.8}$

(3) $2^{1.73}$ < $2^{\sqrt{3} }$ < $2^{1.74}$

(4) $2^{1.732}$ < $2^{\sqrt{3} }$ < $2^{1.733}$

(5) $2^{1.7320}$ < $2^{\sqrt{3} }$ < $2^{1.7321}$

Step-by-step explanation:

as per question given data

√3 ≈ 1.732 050 8

to find out

sequence of five intervals

solution

as we have given that √3 value that is here

√3 ≈ 1.732 050 8 ........................1

when we find $2^{\sqrt{3} }$ ................2

put here √3 value in equation number 2

we get $2^{\sqrt{3} }$ that is 3.322

sequence of five intervals

(1) 2 < $2^{\sqrt{3} }$ < $2^{2}$

(2) $2^{1.7}$ < $2^{\sqrt{3} }$ < $2^{1.8}$

(3) $2^{1.73}$ < $2^{\sqrt{3} }$ < $2^{1.74}$

(4) $2^{1.732}$ < $2^{\sqrt{3} }$ < $2^{1.733}$

(5) $2^{1.7320}$ < $2^{\sqrt{3} }$ < $2^{1.7321}$

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