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2 years ago

10

Question 1 (1 point)

1 answer:

alisha [4.7K]2 years ago

5 0

Answer:

See below

Step-by-step explanation:

1) x=2-1

x=1

2) (x+1)(x+1)=2(x+1)

3) 10x-20 -11 = 4+3x

10x-3x=20+11+4

7x=35

x=5

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3A + 4N - 5A + 3 - 8N + 11 =

Anit [1.1K]

Answer:

-2a-4n+14

Step-by-step explanation:

7 0

3 years ago

1252 divided by 48 equals 26 r 4

Ksenya-84 [330]

I assume that in this question, you are asked to evaluate the value(s) of r. To answer that question, dividing 1252 by 48 gives us 26.0833 or 26. Mathematically expressing the solution,
26 = 26r^4
Then, divide the equation by 26, giving us 1 = r^4. Evaluate r. This will give us answers of +-1.

7 0

3 years ago

4. Find the vertex, focus, and directrix of the parabola. Sketch its graph, showing the focus and the directrix.

valentina_108 [34]

Answer:

vertex - (0,0)

focus - (0, 3/4)

directrix - 3/4

parabola- x^2=-3y

graph of equation:

8 0

2 years ago

Find the length of the curve. R(t) = cos(8t) i + sin(8t) j + 8 ln cos t k, 0 ≤ t ≤ π/4

arsen [322]

we are given

$R(t)=cos(8t)i+sin(8t)j+8ln(cos(t))k$

now, we can find x , y and z components

$x=cos(8t),y=sin(8t),z=8ln(cos(t))$

Arc length calculation:

we can use formula

$L=\int\limits^a_b {\sqrt{(x')^2+(y')^2+(z')^2} } \, dt$

$x'=-8sin(8t),y=8cos(8t),z=-8tan(t)$

now, we can plug these values

$L=\int _0^{\frac{\pi }{4}}\sqrt{(-8sin(8t))^2+(8cos(8t))^2+(-8tan(t))^2} dt$

now, we can simplify it

$L=\int _0^{\frac{\pi }{4}}\sqrt{64+64tan^2(t)} dt$

$L=\int _0^{\frac{\pi }{4}}8\sqrt{1+tan^2(t)} dt$

$L=\int _0^{\frac{\pi }{4}}8\sqrt{sec^2(t)} dt$

$L=\int _0^{\frac{\pi }{4}}8sec(t) dt$

now, we can solve integral

$\int \:8\sec \left(t\right)dt$

$=8\ln \left|\tan \left(t\right)+\sec \left(t\right)\right|$

now, we can plug bounds

and we get

$=8\ln \left(\sqrt{2}+1\right)-0$

so,

$L=8\ln \left(1+\sqrt{2}\right)$ ..............Answer

5 0

2 years ago

The weights of newborn baby boys born at a local hospital are believed to have a normal distribution with an average weight of 5

Lisa [10]

Answer: 0.4987

Step-by-step explanation:

Given : The weights of newborn baby boys born at a local hospital are believed to have a normal distribution with

Mean : $\mu=5\text { lbs}$

Standard deviation : $\sigma= 2\text{ lbs}$

Let X be the random variable that represents the weight of randomly selected student .

Z score : $z=\dfrac{x-\mu}{\sigma}$

For x = 5 lbs

$z=\dfrac{5-5}{2}=0$

For x = 11 lbs

$z=\dfrac{11-5}{2}=3$

By using the standard normal distribution table , the probability that the weight of the newborn baby boy will be between 5 lbs and 11 lbs :-

$P(5$

Hence, the probability that the weight of the newborn baby boy will be between 5 lbs and 11 lbs =0.4987

5 0

3 years ago