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3 years ago

9

What do all experiments have in common?

1 answer:

Ksivusya [100]3 years ago

5 0

C. there are at least two different ways in which participants are treated

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Why are there two solutions for the equation |6 + y| = 2? Explain.

GaryK [48]

$Solution, \left|6+y\right|=2\quad :\quad y=-8\quad \mathrm{or}\quad \:y=-4$

$Steps:$

$|f\left(y\right)|=a\quad \Rightarrow \:f\left(y\right)=-a\quad \mathrm{or}\quad \:f\left(y\right)=a, 6+y=-2\quad \quad \mathrm{or}\quad \:\quad \:6+y=2$

$6+y=-2\quad :\quad y=-8,\\6+y=-2,\\\mathrm{Subtract\:}6\mathrm{\:from\:both\:sides}, 6+y-6=-2-6,\\\mathrm{Simplify}, y=-8$

$6+y=2\quad :\quad y=-4,\\6+y=2,\\\mathrm{Subtract\:}6\mathrm{\:from\:both\:sides},6+y-6=2-6\\\mathrm{Simplify},y=-4$

$\mathrm{Combine\:the\:ranges}, y=-8\quad \mathrm{or}\quad \:y=-4$

$\mathrm{The\:Correct\:Answer\:is\:Because\:of\:the\:absolute\:value,\:It\:could\:be\:Positive\:or\:negative.}$

$\mathrm{Hope\:This\:Helps!!!}$

$\mathrm{-Austint1414}$

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4 years ago

9x-3.3=6.8+8x what is x

trasher [3.6K]

Answer:

<h2>x = 10.1</h2>

Step-by-step explanation:

$9x-3.3=6.8+8x\qquad\text{add 3.3 to both sides}\\\\9x-3.3+3.3=6.8+3.3+8x\\\\9x=10.1+8x\qquad\text{subtract 8x from both sides}\\\\9x-8x=10.1+8x-8x\\\\x=10.1$

7 0

4 years ago

What number must you add to complete the square?<br> x^2 + 12x = 16

dmitriy555 [2]

Answer: B. 36

The complete square is (x+6)^2. It's missing the third term, which is 36 because 6 x 6 is 36.

7 0

4 years ago

Read 2 more answers

Limit as x approaches 0 of (sin^2x)/x

melisa1 [442]

Answer:

<h3>0</h3>

Step-by-step explanation:

Given the expression

$\lim_{x \to \ 0} \frac{sin^2x}{x}$

Substitute the value of x in the function

$= \frac{sin ^2(0)}{0}\\= 0/0 (indeterminate) \\$

Apply l'hospital rule

$\lim_{x \to \ 0} \frac{d/dx(sin^2x)}{d/dx(x)} \\= \lim_{x \to \ 0} \frac{(2sinxcosx)}{1} \\$

Substitute the value of x

= 2 sin(0)cos(0)

= 2 * 0 * 1

= 0

Hence the limit of the function is 0

5 0

3 years ago

Describe the region enclosed by the circle x^2 +y^2 = 11x in polar coordinates.

vivado [14]

Answer:

the region of the circle in polar coordinate can be expressed as:

$\mathbf{- \dfrac{\pi}{2} \leq \theta \leq \dfrac{\pi}{2} , 0 \leq r \leq 11 cos \theta}$

Step-by-step explanation:

Given that:

$x^2 +y^2 = 11x$

This implies that :

$(x-5.5)^2 + y^2 = 5.5^2$

center (5.5, 0) and radius = 5.5

The relation between the cartesian coordinates (x,y) and the polar coordinates (r, θ) is as follows:

x = rcosθ

y = rsinθ

Thus, $x^2 + y^2 = r^2$

$x^2 + y^2 = 11x$

$(rcos \theta)^2 + (r sin \theta)^2 = 11(rcos \theta)$

$r^2 = 11r cos \theta$

Thus, the angle θ runs between $- \dfrac{\pi}{2}$ to $\dfrac{\pi}{2}$

Therefore, the region of the circle in polar coordinate can be expressed as:

$\mathbf{- \dfrac{\pi}{2} \leq \theta \leq \dfrac{\pi}{2} , 0 \leq r \leq 11 cos \theta}$

7 0

4 years ago