2/3 of books is what

Khan shennanigans ;(

Snezhnost [94]

Answer:

15

Step-by-step explanation:

6 0

2 years ago

· Amanda has $210 in her school lunch account. She spends $35 each week on school lunches. The

tigry1 [53]

Answer:

(See step-by-step)

Step-by-step explanation:

Let’s start by making A-(X) and W-(Y). So, when substituting the variables you get x=210-35y. When x=0, the answer to the equation is 6. So, the y-intercept (when x=0) is 6 or (0,6). When y=0, the answer to the equation/x-intercept(y=0) is 210 or (210,0). The graph itself looks like the image included.

The domain, realistically, is x>=0 however on the graph itself, the domain is x= All Real Numbers. The range realistically is y>=0, but on the graph it is also All Real Numbers.

The x-intercept is the account when no weeks have passed and no money is lost (210) and the y-intercept is the maximum amount of weeks before there is no more money in the account (6).

3 0

3 years ago

What is the equation of the line that is perpendicular to y = one-fifth x + 4 and that passes through (5,–4)?

Brilliant_brown [7]

Answer:

y = -5x + 21

Step-by-step explanation:

perpendicular lines have slopes that are negative reciprocals.

The given line has a slope of 1/5, so the perpendicular line will have a slope of -5.

y = -5x + B

-4 = -5(5) + B

B = -4 + 25 = 21

y = -5x + 21

8 0

3 years ago

Read 2 more answers

Prove these identities <br>(i) secx cosecx - cotx = tanx<br>

brilliants [131]

Answer:

$(A)\frac 1 {cosx} \frac 1 {sin x} - \frac {cosx} {sin x} = \frac {sin x} {cosx} \\\\(B)\frac{1-cos^2 x }{cos x sin x} = \frac {sin x} {cosx} \\\\(C) \frac{(sin^2x + cos^2x)-cos^2 x }{cos x sin x} = \frac {sin x} {cosx} \\\\(D) \frac{sin^2x }{cos x sin x} = \frac {sin x} {cosx}\\\\(E)\frac {sin x}{cos x} = \frac {sin x}{cos x}$

In step A, you replace all trigonometric functions with their definitions in terms of sine and cosine.

In step B, you add the two together as you add fractions.

In step C, you replace 1 with the sum of the squares of sine and cosine.
In step D, you add the cosine terms together so they disappear.

In step E, finally, you divide numerator and denominator by $sin x$ .

3 0

2 years ago

The position function of a particle in rectilinear motion is given by s(t) = 2t3 – 21t2 + 60t + 3 for t ≥ 0 with t measured in s

Norma-Jean [14]

The positions when the particle reverses direction are:

$s(t_1)=55ft\\\\s(t_2)=28ft$

The acceleraton of the paticle when reverses direction is:

$a(t_1)=-18\frac{ft}{s^{2}}\\ \\a(t_2)=a(5s)=18\frac{ft}{s^{2}}$

Why?

To solve the problem, we need to remember that if we derivate the position function, we will get the velocity function, and if we derivate the velocity function, we will get the acceleration function. So, we will need to derivate two times.

Also, when the particle reverses its direction, the velocity is equal to 0.

We are given the following function:

$s(t)=2t^{3}-21t^{2}+60t+3$

So,

- Derivating to get the velocity function, we have:

$v(t)=\frac{ds}{dt}=(2t^{3}-21t^{2}+60t+3)\\\\v(t)=3*2t^{2}-2*21t+60*1+0\\\\v(t)=6t^{2}-42t+60$

Now, making the function equal to 0, to find the times when the particle reversed its direction, we have:

$v(t)=6t^{2}-42t+60\\\\0=6t^{2}-42t+60\\\\0=t^{2}-7t+10\\(t-5)*(t-2)=0\\\\t_{1}=5s\\t_{2}=2s$

We know that the particle reversed its direction two times.

- Derivating the velocity function to find the acceleration function, we have:

$a(t)=\frac{dv}{dt}=6t^{2}-42t+60\\\\a(t)=12t-42$

Now, substituting the times to calculate the accelerations, we have:

$a(t_1)=a(2s)=12*2-42=-18\frac{ft}{s^{2}}\\ \\a(t_2)=a(5s)=12*5-42=18\frac{ft}{s^{2}}$

Now, substitutitng the times to calculate the positions, we have:

$s(t_1)=2*(2)^{3}-21*(2)^{2}+60*2+3=16-84+120+3=55ft\\\\s(t_2)=2*(5)^{3}-21*(5)^{2}+60*5+3=250-525+300+3=28ft$

Have a nice day!

3 0

3 years ago