All Answers

#1

+2668

0

For the quadratic to be factorable, there must be 2 numbers that sum to $k$ and multiply to $23 \times -5 =-115$.

There are 4 pairs of numbers that multiply to $-115$: $(115, -1)$, $(23, -5)$, $(5, -23)$, $(1, -115)$.

This means that k can be the sum of these values or $\color{brown}\boxed{k = -114, -18, 18, 114}$

.

BuilderBoiApr 30, 2023

#1

+2668

0

What if x is 0? or negative?

BuilderBoiApr 30, 2023

#4

+2668

+1

Then the distance from $(0, 0)$ to $(1.8, -0.4)$ is $\sqrt{(1.8)^2 + (0.4)^2} = {\sqrt{85} \over 5} $ and the distance from $(0, 1) $ to $(1.8, -0.4)$ is $\sqrt{(1.8)^2 + (1.4)^2} = {\sqrt{130} \over 5}$.

Adding the two gives us ${\sqrt {85 }+ \sqrt {130} \over 5} \approx 4.12426$ but the correct answer was $\sqrt{17} \approx 4.12311$.

BuilderBoiApr 30, 2023

#2

+6

+1

I'm pretty sure 4/3 is incorrect, but can you show me your steps?

ImNotTheBestAtMathApr 30, 2023

#3

+15147

+2

the length of the shortest path

Hello BuilderBoy!

$y=2x-4\\\color{blue}P_1(0,0)\\ P(x,2x-4)\\ \color{blue}P_2(0,1)$

$L_{1,P}^2=x^2+y^2=x^2+(2x-4)^2\\ L_{P,2}^2=x^2+(1-y)^2=x^2+(1-2x+4)^2\\ L_{1,P}^2+L_{P,2}^2=x^2+(2x-4)^2+x^2+(5-2x)^2\\ $

$\frac{d(L_{1,P}^2+L_{P,2}^2)}{dx}=2x+2(2x-4)\cdot 2+2x+2(5-2x)\cdot (-2)=0\\ 4x+8x-16-20+8x=0\\ 20x=36\\ \color{blue}x_P=1.8\\ \color{blue}y_p=-0.4 \\ \color{blue}P(1.8,-0.4)$

You can calculate the distances between the points. The two-point equation is called:

$L=\sqrt{(y_2-y_1)^2+(x_2-x_1)^2}$

Have fun.

asinusApr 30, 2023

#11

+118725

+1

Sorry Cheesesteak, but that answer is NOT correct. I am very pleased that you have responded though.

Take a point that is 0.3 units from the cente, now rotate that point 180 degrees to get the second point.

The two points are 0.6 units and 180 degrees apart.

You said the 2 points had to be less than 60 degrees apart. You have only found a small portion of the solutions

MelodyApr 30, 2023

#2

+2668

+1

Sorry, but that's incorrect. The correct answer is actually $\sqrt{17}$, even though I don't know how to get it.

Thanks so much for your attempt, though!

BuilderBoiApr 30, 2023

#1

+2

The shortest path that starts at the origin, touches the line y = 2x - 4, and then ends at (0, 1) is a straight line segment. The length of this line segment is sqrt(2^2 + (-4 - 1)^2) = sqrt(21).

GuestApr 29, 2023

#1

0

The perimeter of rectangle OSTQ is 56.

GuestApr 29, 2023

#1

0

The perimeter of rectangle OSTQ is 40.

GuestApr 29, 2023

#1

0

sin PAQ = 2/3.

GuestApr 29, 2023

#2

0

Pearl writes down seven consecutive integers, and adds them up. The sum of the integers is equal to 21 times the largest of the seven integers. What is the smallest integer that Pearl wrote down?

Since the integers are consecutive,

each integer's value will be one larger

than the integer before it .

Call the first integer "x"

The seven integers are (x) + (x+1) + (x+2) + (x+3) + (x+4) + (x+5) + (x+6)

and their sum is 7x + 21

We want this to be 21 times the largest number

so 7x + 21 = 21 times (x+6)

7x + 21 = 21x + 126

Collect like terms 7x – 21x = 126 – 21

– 14x = 105

x = –7.5

Check answer by plugging into the original proposition

(x) + (x+1) + (x+2) + (x+3) + (x+4) + (x+5) + (x+6)

(–7.5) + (–7.5+1) + (–7.5+2) + (–7.5+3) + (–7.5+4) + (–7.5+5) + (–7.5+6)

(–7.5) + (–6.5) + (–5.5) + (–4.5) + (–3.5) + (–2.5) + (–1.5)

Add all those together and the sum is –31.5

Multiply 21 times –1.5 and the product is –31.5 so the answer checks out good

.

GuestApr 29, 2023

#1

0

Five times a number is divided by 7 more than that number. If the result is 0 then what was the original number?

5n

––––– = 0

n + 7

The only way this can happen is n = zero.

.

GuestApr 29, 2023

#2

+215

+11

Thanks so much!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!

xud33Apr 29, 2023

#1

+1

$e^{11 \pi i/2} = i$

.

GuestApr 29, 2023

#1

0

Let y = f(x) = arctan(x) + 1/2*arcsin(x). Then y = pi/2 - 1/2*arccos(1 - x^2). Since the range of arccos(1 - x^2) is [0,pi], the range of f(x) is [pi/4, 3*pi/4].

GuestApr 29, 2023

#1

0

$\sqrt{27}\cdot\sqrt{12}\cdot\sqrt{147}=\sqrt{27\cdot12\cdot147}=\sqrt{3^3\cdot2^2\cdot3\cdot3\cdot7^2}=\sqrt{2^2\cdot3^5\cdot7^2}=2\cdot7\cdot3^2\cdot\sqrt{3}=\boxed{126\sqrt{3}}$

.

GuestApr 29, 2023

#1

-1

To find the highest point of the parabolic arch, we need to find the value of $t$ that maximizes $y.$ Since the coefficient of $t^2$ is negative, the highest point occurs when the quadratic term is zero, that is, when $t = \frac{v \sin \theta}{g}.$ Substituting this value of $t$ into the equation for $y,$ we get:

$y_{\text{max}} = \frac{v^2 \sin^2 \theta}{2g}$

This expression gives the maximum height of the projectile for a given angle $\theta.$ Note that the height only depends on the square of the initial velocity and the sine of the angle, not on the actual value of the velocity or the angle itself.

To trace out the curve of maximum heights as $\theta$ varies, we need to find the maximum height for each angle. We can express this maximum height as a function of $\theta$:

$f(\theta) = \frac{v^2 \sin^2 \theta}{2g}$

The curve of maximum heights is then given by the graph of $f(\theta)$ over the interval $0^\circ \leq \theta \leq 180^\circ.$ Note that $f(\theta)$ is a symmetric function with respect to $\theta = 90^\circ,$ since $\sin^2 \theta = \sin^2 (180^\circ - \theta).$ Therefore, we can restrict our attention to the interval $0^\circ \leq \theta \leq 90^\circ.$

To find the area enclosed by the curve of maximum heights, we need to integrate $f(\theta)$ over the interval $0^\circ \leq \theta \leq 90^\circ.$

$\begin{aligned} \text{Area} &= \int_{0^\circ}^{90^\circ} f(\theta) d\theta \ &= \int_{0^\circ}^{90^\circ} \frac{v^2 \sin^2 \theta}{2g} d\theta \ &= \frac{v^2}{2g} \int_{0^\circ}^{90^\circ} \sin^2 \theta d\theta \ &= \frac{v^2}{4g} \int_{0^\circ}^{180^\circ} \sin^2 \theta d\theta \ &= \frac{v^2}{4g} \int_{0^\circ}^{180^\circ} \frac{1 - \cos 2\theta}{2} d\theta \ &= \frac{v^2}{8g} \int_{0^\circ}^{360^\circ} (1 - \cos 2\theta) d\theta \ &= \frac{v^2}{8g} \left[\theta - \frac{1}{2} \sin 2\theta \right]_{0^\circ}^{360^\circ} \ &= \frac{v^2}{8g} \cdot 360^\circ \ &= \frac{45}{\pi^2} \cdot \frac{v^4}{g^2} \end{aligned}$

where we have used the half-angle formula $\sin^2 \theta = \frac{1 - \cos 2\theta}{2}$ and integrated by parts to evaluate $\int \cos 2\theta d\theta.$

Therefore, $c = \frac{45}{\pi^2}$

.

GuestApr 29, 2023

#1

-3

We can use the Laplace expansion to evaluate the determinant of the given matrix. Expanding along the third column, we get:

| [[sin^2 A, cot A], [sin^2 B, cot B]] sin^2 C - [[sin^2 A, cot A], [sin^2 C, cot C]] sin^2 B + [[sin^2 B, cot B], [sin^2 C, cot C]] sin^2 A

where |M| denotes the determinant of matrix M. Note that the first term is the determinant of the 2x2 matrix [[sin^2 A, cot A], [sin^2 B, cot B]], which we can evaluate using the formula for the determinant of a 2x2 matrix:

[[sin^2 A, cot A], [sin^2 B, cot B]] = sin^2 A cot B - sin^2 B cot A

For the second and third terms, we can use the formula for the determinant of a 3x3 matrix:

[[a, b, c], [d, e, f], [g, h, i]] = a(ei - fh) - b(di - fg) + c(dh - eg)

Substituting the appropriate values, we get:

[[sin^2 A, cot A], [sin^2 C, cot C]] = sin^2 A cot C - sin^2 C cot A

[[sin^2 B, cot B], [sin^2 C, cot C]] = sin^2 B cot C - sin^2 C cot B

Substituting these values into our Laplace expansion, we get:

| [[sin^2 A, cot A], [sin^2 B, cot B], [sin^2 C, cot C]] = (sin^2 A cot B - sin^2 B cot A) sin^2 C

(sin^2 A cot C - sin^2 C cot A) sin^2 B

(sin^2 B cot C - sin^2 C cot B) sin^2 A

Factoring out sin^2 A, sin^2 B, and sin^2 C, respectively, we get:

| [[sin^2 A, cot A], [sin^2 B, cot B], [sin^2 C, cot C]] = sin^2 A (cot B sin^2 C - cot C sin^2 B)

sin^2 B (cot A sin^2 C - cot C sin^2 A)

sin^2 C (cot A sin^2 B - cot B sin^2 A)

Using the identity cot x = cos x / sin x, we can simplify the coefficients of cot A, cot B, and cot C:

cot B sin^2 C - cot C sin^2 B = cos B sin C - cos C sin B = sin(B - C) cot A sin^2 C - cot C sin^2 A = cos A sin C - cos C sin A = sin(A - C) cot A sin^2 B - cot B sin^2 A = cos A sin B - cos B sin A = sin(A - B)

Substituting these values, we get:

| [[sin^2 A, cot A], [sin^2 B, cot B], [sin^2 C, cot C]] = sin^2 A sin(B - C) - sin^2 B sin(A - C) + sin^2 C sin(A - B = 1.

Therefore, the determinant simplifies to 1.

GuestApr 29, 2023

#3

+215

+6

Thank you,That really helped!!!

xud33Apr 29, 2023

#2

+42

+4

Note that there is no way to get the same amount from two different combinations of coins. Thus, it suffices to find the nmber of ways to choose some nonzero number of coins from the coins Steve has where coins of the same value are considered identical.

We can have zero or one quarter, so there are two ways to choose how many quarters.

We can have zero, one, or two nickels, so there are three ways to choose how many nickels.

We can have any where from zero to three pennies, so there are four ways to choose the pennies.

However, note that we included the case when we choose no coins at all, which is not allowed, so we subtract $1$ from the product of $2,3,$and $4$, to get $2 \cdot 3 \cdot 4 - 1 = \boxed{23}.$

lpieleanuApr 29, 2023

#1

+215

+9

1 item for 1 penny

1 item for 2 pennies

1 item for 3 pennies

1 item for 5 cents

1 item for 6 cents

1 item for 7 cents

1 item for 8 cents

1 item for 10 cents

1 item for 11 cents

1 item for 12 cents

1 item for 13 cents

1 item for 25 cents

1 item for 26 cents

1 item for 27 cents

1 item for 28 cents

1 item for 30 cents

1 item for 31 cents

1 item for 32 cents

1 item for 33 cents

1 item for 35 cents

1 item for 36 cents

1 item for 37 cents

1 item for 38 cents.

----------------------

Total 23 different items for exact change.

xud33Apr 29, 2023

#1

+42

+2

1. By the area formula for a triangle, the area is $\frac{6 \cdot 8}{2} = 24$

2. Since the sum of the angles in any triangle is $180^\circ$, we have $\angle{BAC} = 180^\circ - \angle{ABC} - \angle{ACB} = 180^\circ - 135^\circ - 30^\circ = 15^\circ.$

lpieleanuApr 29, 2023

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