X is greater than 7 or less than 5

DochEvi [55]

Use Ga_thMath(u) (brainly doesn't allow me to type it) To use the app u need to take a pic of the problem and then it will process it and you'll get ur answer ASAP(most of the time). Many questions have been asked before so search it on brainly.

8 0

2 years ago

I will mark branliest if correct

marishachu [46]

Answer:

I believe it would be 1

Step-by-step explanation:

6 0

2 years ago

Which graph represents y = StartRoot x EndRootWhat is the solution of StartFraction 1 Over c minus 3 EndFraction minus StartFrac

romanna [79]

The equation is undefined for singularity points at 0, 3, then c ≠ 0, c ≠ 3

<h3>What is the graph of the parent function (y)?</h3>

The set of all coordinates (x, y) in the plane that satisfy the equation y = f(x) is the graph of the function. Suppose a function is only specified for a small set of input values, the graph of the function will only have a small number of points, in which each point's x-coordinate represents an input number and its y-coordinate represents an output number.

From the given information,

The domain for the $\mathbf{y=\sqrt{x}}$ is at x ≥ 0,
The range is the set of values that the dependent variable for which the function is defined. f(x) ≥ 0.

In the second question:

$\mathbf{\dfrac{1}{c-3}-\dfrac{1}{c}= \dfrac{3}{c(c-3)}}$

Multiply by LCM

c - (c - 3) = 3

Solve c - (c - 3) = 3: True for all c

c ≠ 0, c ≠ 3

Therefore, we can conclude that since the equation is undefined for singularity points at 0, 3, then c ≠ 0, c ≠ 3

Learn more about the graph of a function here:

brainly.com/question/3939432

#SPJ1

5 0

1 year ago

Expand the following X(5x-1)

RSB [31]

Answer:

5x^2-x

Step-by-step explanation:

pls mark as brainliest

7 0

2 years ago

Find the linear approximating polynomial for the following function centered at the given point a.point a. b. Find the quadratic

ZanzabumX [31]

Answer:

$(a)L(x)=-\frac{\sqrt{2} }{2}+\frac{\sqrt{2} }{2}(x+\frac{\pi}{4})\\(b)Q(x)=-\frac{\sqrt{2} }{2}+\frac{\sqrt{2} }{2}(x+\frac{\pi}{4})+\frac{\sqrt{2} }{4}(x+\frac{\pi}{4})^2\\(c)L(-0.23\pi)=-0.6626\\Q(-0.23\pi)=-0.6613$

Step-by-step explanation:

Given the function:

$f(x)=sin x, a=-\frac{\pi}{4}$

(a)Linear approximating polynomial

$L(x)=f(a)+f'(a)(x-a)\\f(a)=sin(-\frac{\pi}{4})=-\frac{\sqrt{2} }{2}\\f'(x)=cos x, a=-\frac{\pi}{4}\\f'(a)=cos (-\frac{\pi}{4})=\frac{\sqrt{2} }{2} \\Therefore:\\L(x)=-\frac{\sqrt{2} }{2}+\frac{\sqrt{2} }{2}(x-(-\frac{\pi}{4}))\\L(x)=-\frac{\sqrt{2} }{2}+\frac{\sqrt{2} }{2}(x+\frac{\pi}{4})$

(b)Quadratic approximating polynomial

$Q(x)=L(x)+\frac{1}{2}f''(a)(x-a)^2\\f''(x)=-sin(x), \\f''(a)=-sin(-\frac{\pi}{4})=\frac{\sqrt{2} }{2}\\Q(x)=-\frac{\sqrt{2} }{2}+\frac{\sqrt{2} }{2}(x+\frac{\pi}{4})+\frac{\sqrt{2} }{4}(x+\frac{\pi}{4})^2$

(c)When $x=-0.23\pi$

Using Linear Approximation polynomial

$L(x)=-\frac{\sqrt{2} }{2}+\frac{\sqrt{2} }{2}(x+\frac{\pi}{4})\\L(-0.23\pi)=-\frac{\sqrt{2} }{2}+\frac{\sqrt{2} }{2}(-0.23\pi+\frac{\pi}{4})\\L(-0.23\pi)=-0.6626$

Using the Quadratic approximating polynomial

$Q(x)=-\frac{\sqrt{2} }{2}+\frac{\sqrt{2} }{2}(x+\frac{\pi}{4})+\frac{\sqrt{2} }{4}(x+\frac{\pi}{4})^2\\Q(-0.23\pi)=-\frac{\sqrt{2} }{2}+\frac{\sqrt{2} }{2}(-0.23\pi+\frac{\pi}{4})+\frac{\sqrt{2} }{4}(-0.23\pi+\frac{\pi}{4})^2=-0.6613$

4 0

2 years ago