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3 years ago

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What is the domain of y equals 3 square root x

1 answer:

Akimi4 [234]3 years ago

5 0

The domain of the function $f(x) = 3 \sqrt{x}$ is x≥0 because negative values are not allowed inside "√ "

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Write the times of four trains as 12 hr. clock<br> a. 14:32<br> b. 16:30<br> c. 06:45<br> d. 10:29

romanna [79]

Answer:

a 2:32

b 4:30

c 6:45

d 10:29

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2 years ago

What is the square root of 11 times 8

Amanda [17]

i think the answer is 88 i hope i helped

7 0

3 years ago

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The point P(1,1/2) lies on the curve y=x/(1+x). (a) If Q is the point (x,x/(1+x)), find the slope of the secant line PQ correct

lukranit [14]

Answer:

See explanation

Step-by-step explanation:

You are given the equation of the curve

$y=\dfrac{x}{1+x}$

Point $P\left(1,\dfrac{1}{2}\right)$ lies on the curve.

Point $Q\left(x,\dfrac{x}{1+x}\right)$ is an arbitrary point on the curve.

The slope of the secant line PQ is

$\dfrac{y_2-y_1}{x_2-x_1}=\dfrac{\frac{x}{1+x}-\frac{1}{2}}{x-1}=\dfrac{\frac{2x-(1+x)}{2(x+1)}}{x-1}=\dfrac{\frac{2x-1-x}{2(x+1)}}{x-1}=\\ \\=\dfrac{\frac{x-1}{2(x+1)}}{x-1}=\dfrac{x-1}{2(x+1)}\cdot \dfrac{1}{x-1}=\dfrac{1}{2(x+1)}\ [\text{When}\ x\neq 1]$

1. If x=0.5, then the slope is

$\dfrac{1}{2(0.5+1)}=\dfrac{1}{3}\approx 0.3333$

2. If x=0.9, then the slope is

$\dfrac{1}{2(0.9+1)}=\dfrac{1}{3.8}\approx 0.2632$

3. If x=0.99, then the slope is

$\dfrac{1}{2(0.99+1)}=\dfrac{1}{3.98}\approx 0.2513$

4. If x=0.999, then the slope is

$\dfrac{1}{2(0.999+1)}=\dfrac{1}{3.998}\approx 0.2501$

5. If x=1.5, then the slope is

$\dfrac{1}{2(1.5+1)}=\dfrac{1}{5}\approx 0.2$

6. If x=1.1, then the slope is

$\dfrac{1}{2(1.1+1)}=\dfrac{1}{4.2}\approx 0.2381$

7. If x=1.01, then the slope is

$\dfrac{1}{2(1.01+1)}=\dfrac{1}{4.02}\approx 0.2488$

8. If x=1.001, then the slope is

$\dfrac{1}{2(1.001+1)}=\dfrac{1}{4.002}\approx 0.2499$

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3 years ago

Which of the following numbers is irrational

max2010maxim [7]

The square root of 5

6 0

3 years ago

Read 2 more answers

The article on the next page from the San Francisco Chronicle, January 27, 1996, describes very serious problems of subsidence,

Mekhanik [1.2K]

Answer:

246.3%

the complete question is found in the attached document

Step-by-step explanation:

1st step:

using 1936 data,

w1= 356% = 3.56(356/100) , H= 79 feet

specific gravity = 2.65

Sₓ= 100%= 1

initial void ratio(e₀)= (w1 x specific gravity)/Sₓ

=3.56 x 2.65/1 = 9.434

2nd step

using 1996 data

ΔH= 22ft

ΔH/H = Δe/(1 + e₀)

22/79 = Δe/(1+9.434)

0.278=Δe/10.434

Δe= 0.278 x 10.434

Δe= 2.905

Δe= e₀ - eₓ

eₓ= e₀-Δe

eₓ= 9.434 - 2.905

eₓ= 6.529

3rd step

calculating water content in 1996

eₓ =6.529, specific gravity= 2.65, Sₓ= 100%

W2 X 2.65 = 1 x 6.529

w2 = 6.529/2.65 = 2.463 = 246.3%

6 0

3 years ago