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3 years ago

I need help on this question as well.

Mathematics

2 answers:

Y_Kistochka [10]3 years ago

7 0

8 is the answer, because 4 x 8 = 32

klio [65]3 years ago

7 0

Answer:

Step-by-step explanation:

Move 4 to the other hand side by division 32/4 ..x=8

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Can someone pls help me with this math The function shown crosses the y-axis at....

Leona [35]

Answer:

(0,5)

Step-by-step explanation:

Hope this helps. Pls give brainliest.

4 0

3 years ago

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Use place value pattern to complete the table

svp [43]

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3 years ago

Find the vertex and axis of symmetry for f(x)=2x^2-1

vfiekz [6]

<h3>Axis of symmetry is 0 and vertex is (0, -1)</h3>

Solution:

Given is:

$f(x) = 2x^2 - 1$

We have to find the vertex and axis of symmetry

The general equation is given as:

$f(x) = ax^2 + bx + c$

Comparing with given equation,

a = 2

b = 0

c = -1

The axis of symmetry is given as:

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

$x = \frac{0}{2(2)}\\\\x = 0$

Thus axis of symmetry is 0

The x coordinate of the vertex is the same

x coordinate of the vertex = 0

h = 0

The y coordinate of the vertex is:

k = f(h)

k = f(0)

$f(0) = 2(0) - 1\\\\f(0) = 0 - 1\\\\f(0) = -1$

Thus, y coordinate of the vertex is -1

Therefore, vertex is (0, -1)

3 0

3 years ago

Three hundred eighty-five defective alcohol swabs were produced in a day. If 4% of the alcohol swabs produced were defective, ho

dusya [7]

Answer:

9625

Step-by-step explanation:

If 385 alcohol swabs were defective, then 385 is 4% of all produced alcohol swabs.

385*25 = 9625

9625 alcohol swabs were produced

6 0

2 years ago

Simplify each rational expression to lowest terms, specifying the values of xx that must be excluded to avoid division

k0ka [10]

Answer:

(a) $\frac{x^2-6x+5}{x^2-3x-10}=\frac{x-1}{x+2}$ . The domain of this function is all real numbers not equal to -2 or 5.

(b) $\frac{x^3+3x^2+3x+1}{x^3+2x^2-x}=1+\frac{x^2+4x+1}{x^3+2x^2-x}$ . The domain of this function is all real numbers not equal to 0, $-1+\sqrt{2}$ or $-1+\sqrt{2}$ .

(c) $\frac{x^2-16}{x^2+2x-8}=\frac{x-4}{x-2}$ .The domain of this function is all real numbers not equal to 2 or -4.

(d) $\frac{x^2-3x-10}{x^3+6x^2+12x+8}=\frac{x-5}{\left(x+2\right)^2}$ . The domain of this function is all real numbers not equal to -2.

(e) $\frac{x^3+1}{x^2+1}=x+\frac{-x+1}{x^2+1}$ . The domain of this function is all real numbers.

Step-by-step explanation:

To reduce each rational expression to lowest terms you must:

(a) For $\frac{x^2-6x+5}{x^2-3x-10}$

$\mathrm{Factor}\:x^2-6x+5\\\\x^2-6x+5=\left(x^2-x\right)+\left(-5x+5\right)\\x^2-6x+5=x\left(x-1\right)-5\left(x-1\right)\\\\\mathrm{Factor\:out\:common\:term\:}x-1\\x^2-6x+5=\left(x-1\right)\left(x-5\right)$

$\mathrm{Factor}\:x^2-3x-10\\\\x^2-3x-10=\left(x^2+2x\right)+\left(-5x-10\right)\\x^2-3x-10=x\left(x+2\right)-5\left(x+2\right)\\\\\mathrm{Factor\:out\:common\:term\:}x+2\\x^2-3x-10=\left(x+2\right)\left(x-5\right)$

$\frac{x^2-6x+5}{x^2-3x-10}=\frac{\left(x-1\right)\left(x-5\right)}{\left(x+2\right)\left(x-5\right)}$

$\mathrm{Cancel\:the\:common\:factor:}\:x-5\\\\\frac{x^2-6x+5}{x^2-3x-10}=\frac{x-1}{x+2}$

The denominator in a fraction cannot be zero because division by zero is undefined. So we need to figure out what values of the variable(s) in the expression would make the denominator equal zero.

To find any values for x that would make the denominator = 0 you need to set the denominator = 0 and solving the equation.

$x^2-3x-10=\left(x+2\right)\left(x-5\right)=0$

Using the Zero Factor Theorem: = 0 if and only if = 0 or = 0

$x+2=0\\x=-2\\\\x-5=0\\x=5$

The domain is the set of all possible inputs of a function which allow the function to work. Therefore the domain of this function is all real numbers not equal to -2 or 5.

(b) For $\frac{x^3+3x^2+3x+1}{x^3+2x^2-x}$

$\mathrm{Divide\:the\:leading\:coefficients\:of\:the\:numerator\:}x^3+3x^2+3x+1\mathrm{\:and\:the\:divisor\:}x^3+2x^2-x\mathrm{\::\:}\frac{x^3}{x^3}=1$

Quotient = 1

$\mathrm{Multiply\:}x^3+2x^2-x\mathrm{\:by\:}1:\:x^3+2x^2-x$

$\mathrm{Subtract\:}x^3+2x^2-x\mathrm{\:from\:}x^3+3x^2+3x+1\mathrm{\:to\:get\:new\:remainder}$

Remainder = $x^2+4x+1}$

$\frac{x^3+3x^2+3x+1}{x^3+2x^2-x}=1+\frac{x^2+4x+1}{x^3+2x^2-x}$

The domain of this function is all real numbers not equal to 0, $-1+\sqrt{2}$ or $-1+\sqrt{2}$ .

$x^3+2x^2-x=0\\\\x^3+2x^2-x=x\left(x^2+2x-1\right)=0\\\\\mathrm{Solve\:}\:x^2+2x-1=0:\quad x=-1+\sqrt{2},\:x=-1-\sqrt{2}$

(c) For $\frac{x^2-16}{x^2+2x-8}$

$x^2-16=\left(x+4\right)\left(x-4\right)$

$x^2+2x-8= \left(x-2\right)\left(x+4\right)$

$\frac{x^2-16}{x^2+2x-8}=\frac{\left(x+4\right)\left(x-4\right)}{\left(x-2\right)\left(x+4\right)}\\\\\frac{x^2-16}{x^2+2x-8}=\frac{x-4}{x-2}$

The domain of this function is all real numbers not equal to 2 or -4.

$x^2+2x-8=0\\\\x^2+2x-8=\left(x-2\right)\left(x+4\right)=0$

(d) For $\frac{x^2-3x-10}{x^3+6x^2+12x+8}$

$\mathrm{Factor}\:x^2-3x-10\\\left(x^2+2x\right)+\left(-5x-10\right)\\x\left(x+2\right)-5\left(x+2\right)$

$\mathrm{Apply\:cube\:of\:sum\:rule:\:}a^3+3a^2b+3ab^2+b^3=\left(a+b\right)^3\\\\a=x,\:\:b=2\\\\x^3+6x^2+12x+8=\left(x+2\right)^3$

$\frac{x^2-3x-10}{x^3+6x^2+12x+8}=\frac{\left(x+2\right)\left(x-5\right)}{\left(x+2\right)^3}\\\\\frac{x^2-3x-10}{x^3+6x^2+12x+8}=\frac{x-5}{\left(x+2\right)^2}$

The domain of this function is all real numbers not equal to -2

$x^3+6x^2+12x+8=0\\\\x^3+6x^2+12x+8=\left(x+2\right)^3=0\\x=-2$

(e) For $\frac{x^3+1}{x^2+1}$

$\frac{x^3+1}{x^2+1}=x+\frac{-x+1}{x^2+1}$

The domain of this function is all real numbers.

$x^2+1=0\\x^2=-1\\\\\mathrm{For\:}x^2=f\left(a\right)\mathrm{\:the\:solutions\:are\:}x=\sqrt{f\left(a\right)},\:\:-\sqrt{f\left(a\right)}\\\\x=\sqrt{-1},\:x=-\sqrt{-1}$

4 0

3 years ago