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2 years ago

Factorize x²+17x+6 =????

Mathematics

1 answer:

meriva2 years ago

3 0

Answer:

x²+17x+6=xx+17x+6=(17+x)x +6=17+xx+6= 17+x²+6=23+x²

=>x²+17x+6=23+x²

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A baby gains 11 pounds in its first year of life. The baby gained 4 1/4 pounds during the first four months and 3 1/2 pounds in

Ede4ka [16]

Answer:

3.25 pounds

Step-by-step explanation:

x + 4 (1/2) + 3 (1/2) = 11

x + 4.25 + 3.5 = 11

x + 7.75 = 11

x = 11 - 7.75

x = 3.25 pounds

8 0

2 years ago

Change each of the following points from rectangular coordinates to spherical coordinates and to cylindrical coordinates.

FromTheMoon [43]

Answer and Step-by-step explanation: Spherical coordinate describes a location of a point in space: one distance (ρ) and two angles (Ф,θ).To transform cartesian coordinates into spherical coordinates:

$\rho = \sqrt{x^{2}+y^{2}+z^{2}}$

$\phi = cos^{-1}\frac{z}{\rho}$

For angle θ:

If x > 0 and y > 0: $\theta = tan^{-1}\frac{y}{x}$ ;

If x < 0: $\theta = \pi + tan^{-1}\frac{y}{x}$ ;
If x > 0 and y < 0: $\theta = 2\pi + tan^{-1}\frac{y}{x}$ ;

Calculating:

a) (4,2,-4)

$\rho = \sqrt{4^{2}+2^{2}+(-4)^{2}}$ = 6

$\phi = cos^{-1}(\frac{-4}{6})$

$\phi = cos^{-1}(\frac{-2}{3})$

For θ, choose 1st option:

$\theta = tan^{-1}(\frac{2}{4})$

$\theta = tan^{-1}(\frac{1}{2})$

b) (0,8,15)

$\rho = \sqrt{0^{2}+8^{2}+(15)^{2}}$ = 17

$\phi = cos^{-1}(\frac{15}{17})$

$\theta = tan^{-1}\frac{y}{x}$

The angle θ gives a tangent that doesn't exist. Analysing table of sine, cosine and tangent: θ = $\frac{\pi}{2}$

c) (√2,1,1)

$\rho = \sqrt{(\sqrt{2} )^{2}+1^{2}+1^{2}}$ = 2

$\phi = cos^{-1}(\frac{1}{2})$

$\phi$ = $\frac{\pi}{3}$

$\theta = tan^{-1}\frac{1}{\sqrt{2} }$

d) (−2√3,−2,3)

$\rho = \sqrt{(-2\sqrt{3} )^{2}+(-2)^{2}+3^{2}}$ = 5

$\phi = cos^{-1}(\frac{3}{5})$

Since x < 0, use 2nd option:

$\theta = \pi + tan^{-1}\frac{1}{\sqrt{3} }$

$\theta = \pi + \frac{\pi}{6}$

$\theta = \frac{7\pi}{6}$

Cilindrical coordinate describes a 3 dimension space: 2 distances (r and z) and 1 angle (θ). To express cartesian coordinates into cilindrical:

$r=\sqrt{x^{2}+y^{2}}$

Angle θ is the same as spherical coordinate;

z = z

Calculating:

a) (4,2,-4)

$r=\sqrt{4^{2}+2^{2}}$ = $\sqrt{20}$

$\theta = tan^{-1}\frac{1}{2}$

z = -4

b) (0, 8, 15)

$r=\sqrt{0^{2}+8^{2}}$ = 8

$\theta = \frac{\pi}{2}$

z = 15

c) (√2,1,1)

$r=\sqrt{(\sqrt{2} )^{2}+1^{2}}$ = $\sqrt{3}$

$\theta = \frac{\pi}{3}$

z = 1

d) (−2√3,−2,3)

$r=\sqrt{(-2\sqrt{3} )^{2}+(-2)^{2}}$ = 4

$\theta = \frac{7\pi}{6}$

z = 3

5 0

3 years ago

In f(x) = 2x2 − 8x − 10, the y-intercept is ? at ? and the x-intercepts are (-1, 0) and ? .

IgorLugansk [536]

<h2>1. y-intercept</h2>

$\boxed{(0,-10)}$

The quadratic function $f(x)=2x^22-8x-10$ represents a parabola. In fact, the graph of a quadratic function is a special type of U-shaped curve called a parabola. To find the y intercept, we set $x=0$ as follows:

$f(x)=2x^2-8x-10 \\ \\ If \ x=0 \rightarrow f(0)=-10 \\ \\ Then \ y-intercept: \\ \\ (0,-10)$

<h2>2. x-intercepts</h2>

$\boxed{(-1,0) \ and \ (5,0)}$

To find the other x-intercept, we must set $y=0$ as follows:

$f(x)=2x^2-8x-10 \\ \\ If \ y=0 \rightarrow 2x^2-8x-10=0 \\ \\ Using \ the \ quadratic \ formula: \\ \\ x=\frac{-b \pm \sqrt{b^2-4ac}}{2a} \\ \\ \therefore x=\frac{-(-8) \pm \sqrt{(-b)^2-4(2)(-10)}}{2(2)} \\ \\ \therefore x_{1}=-1 \ and \ x_{2}=5$