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3 years ago

6

Is it y=x ?? Pls help

1 answer:

DerKrebs [107]3 years ago

3 0

It's the first one. y can equal x as long as y doesn't equal x multiple times. We also know the last two are wrong because they're saying the same thing. A linear function is a straight line.

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Use linear approximation, i.e. the tangent line, to approximate √ 16.2 as follows: Let f ( x ) = √ x . Find the equation of the

Dvinal [7]

The linear approximation to $f(x)$ near $x=16$ is

$L(x)=f(16)+f'(16)(x-16)$

We have

$f(x)=\sqrt x\implies f(16)=4$

$f'(x)=\dfrac1{2\sqrt x}\implies f'(16)=\dfrac18$

$\implies L(x)=4+\dfrac18(x-16)=\dfrac x8+2$

Then

$f(16.2)\approx L(16.2)\iff\sqrt{16.2}\approx\dfrac{16.2}8+2$

$\implies\boxed{\sqrt{16.2}\approx4.025=\dfrac{161}{40}}$

5 0

3 years ago

The lengths of infants at birth in a certain hospital are normally distributed with a mean of 18 inches and a standard deviation

vaieri [72.5K]

Mean, m = 18 in
Standard deviation, SD = 2.2 in
Range: 16 ≤ X ≤ 21 in

Calculating Z value,
Z = (X-m)/SD

Then,
Z1 = (16-18)/2.2 ≈ -0.91
Z2 = (21-18)/2.2 ≈ 1.36
From Z table, and at Z1 = -0.91, and Z2 = 1.36;
P(16) = 0.1814
P(21) = 0.9131

Therefore,
P(16≤X≤21) = 0.9131 - 0.1814 = 0.7317
The probability that a child selected randomly measures between 16 and 21 in is 0.7317.

6 0

3 years ago

Susie has planned a trip to a city 60 miles away. She wishes to have an average speed of 60 miles/hour for the trip. Due to a tr

Nutka1998 [239]

Answer:

90 mi/h

Step-by-step explanation:

Given,

For first 30 miles, her speed is 30 miles per hour,

Let x be her speed in miles per hour for another 30 miles,

Since, here the distance are equal in each interval,

So, the average speed of the entire journey

$=\frac{\text{Average speed for first 30 miles + Average speed for another 30 miles}}{2}$

$=\frac{30+x}{2}$

According to the question,

$\frac{30+x}{2}=60$

$30+x=120$

$\implies x = 90$

Hence, she needs to go 90 miles per hour for remaining 30 miles.

7 0

4 years ago

Law of sines: StartFraction sine (uppercase A) Over a EndFraction = StartFraction sine (uppercase B) Over b EndFraction = StartF

Vsevolod [243]

First of all, this problem is properly done with the Law of Cosines, which tells us

$a^2 = b^2 + c^2 - 2 b c \cos A$

giving us a quadratic equation for b we can solve. But let's do it with the Law of Sines as asked.

$\dfrac{\sin A}{a} = \dfrac{\sin B}{b} = \dfrac{\sin C}{c}$

We have c,a,A so the Law of Sines gives us sin C

$\sin C = \dfrac{c \sin A}{a} = \dfrac{5.4 \sin 20^\circ}{3.3} = 0.5597$

There are two possible triangle angles with this sine, supplementary angles, one acute, one obtuse:

$C_a = \arcsin(.5597) = 34.033^\circ$

$C_o = 180^\circ - C_a = 145.967^\circ$

Both of these make a valid triangle with A=20°. They give respective B's:

$B_a = 180^\circ - A - C_a = 125.967^\circ$

$B_o = 180^\circ - A - C_o = 14.033^\circ$

So we get two possibilities for b:

$b = \dfrac{a \sin B}{\sin A}$

$b_a = \dfrac{3.3 \sin 125.967^\circ}{\sin 20^\circ} = 7.8$

$b_o = \dfrac{3.3 \sin 14.033^\circ}{\sin 20^\circ} = 2.3$

Answer: 2.3 units and 7.8 units

Let's check it with the Law of Cosines:

$a^2 = b^2 + c^2 - 2 b c \cos A$

$0 = b^2 - (2 c \cos A)b + (c^2-a^2)$

There's a shortcut for the quadratic formula when the middle term is 'even.'

$b = c \cos A \pm \sqrt{c^2 \cos^2 A - (c^2-a^2)}$

$b = c \cos A \pm \sqrt{c^2( \cos^2 A - 1) + a^2}$

$b = 5.4 \cos 20 \pm \sqrt{5.4^2(\cos^2 20 -1) + 3.3^2}$

$b = 2.33958 \textrm{ or } 7.80910 \quad\checkmark$

Looks good.

6 0

3 years ago

Read 2 more answers

Which of the following terms best fits this definition?

elena-14-01-66 [18.8K]

Step-by-step explanation:

hope you can understand

8 0

2 years ago