A coin is dropped in a 15.0 m deep well.

Divide using synthetic division (x^3+5x^2-x-5) divide (x+5)

Kamila [148]

Answer:

Quotient = x^ -1 and the remainder: 0

Step-by-step explanation:

Constant to the left with the opposite sign and coefficients to the right:

-5 | 1 5 -1 -5

Make an horizontal line

Bring down the first coefficient (1) under the line.

Multiply the constant -5 times 1=-5 and add -5 to 5 = 0

Keep doing the same with the coefficients.

8 0

3 years ago

Math homework please help

Tju [1.3M]

(1, 0) and (0, -2)
slope = (0 + 2)/(1 - 0) = 2/1 = 2

the slope is m = 2
the y intercept is b = -2

4 0

3 years ago

Read 2 more answers

Please help :)))) ( attachment )

Anastaziya [24]

Let,
f(x) = -2x+34
g(x) = (-x/3) - 10
h(x) = -|3x|
k(x) = (x-2)^2

This is a trial and error type of problem (aka "guess and check"). There are 24 combinations to try out for each problem, so it might take a while. It turns out that

g(h(k(f(15)))) = -6
f(k(g(h(8)))) = 2

So the order for part A should be: f, k, h, g
The order for part B should be: h, g, k f
note how I'm working from the right and moving left (working inside and moving out).

Here's proof of both claims

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

Proof of Claim 1:

f(x) = -2x+34
f(15) = -2(15)+34
f(15) = 4
-----------------
k(x) = (x-2)^2
k(f(15)) = (f(15)-2)^2
k(f(15)) = (4-2)^2
k(f(15)) = 4
-----------------
h(x) = -|3x|
h(k(f(15))) = -|3*k(f(15))|
h(k(f(15))) = -|3*4|
h(k(f(15))) = -12
-----------------
g(x) = (-x/3) - 10
g(h(k(f(15))) ) = (-h(k(f(15))) /3) - 10
g(h(k(f(15))) ) = (-(-12) /3) - 10
g(h(k(f(15))) ) = -6

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

Proof of Claim 2:

h(x) = -|3x|
h(8) = -|3*8|
h(8) = -24
---------------
g(x) = (-x/3) - 10
g(h(8)) = (-h(8)/3) - 10
g(h(8)) = (-(-24)/3) - 10
g(h(8)) = -2
---------------
k(x) = (x-2)^2
k(g(h(8))) = (g(h(8))-2)^2
k(g(h(8))) = (-2-2)^2
k(g(h(8))) = 16
---------------
f(x) = -2x+34
f(k(g(h(8))) ) = -2*(k(g(h(8))) )+34
f(k(g(h(8))) ) = -2*(16)+34
f(k(g(h(8))) ) = 2

5 0

4 years ago

The mass of a teapot and 4 cups is 1,280 grams. The mass of the same teapot and 10 cups is 2,720 grams. Each cup has the same ma

Aleksandr-060686 [28]

Answer: 320 grams

Step-by-step explanation:

Let the mas of the teapot be denoted by a

Let the mass of the cups be denoted by c

The question can be turned into an equation which will be:

a + 4c = 1280 ......... equation i

a + 10c = 2720 ........ equation ii

Subtract equation i from ii

6c = 1440

c = 1440/6

c = 240 grams

The mas of the cup is 240 grams

From equation i

a + 4c = 1280

a + 4(240) = 1280

a + 960 = 1280

a = 1280 - 960

a = 320 grams

The mass of the teapot is 320 grams

8 0

3 years ago

let sin(θ) =3/5 and tan(y) =12/5 both angels comes from 2 different right trianglesa)find the third side of the two tringles b)

statuscvo [17]

In a right triangle, we haev some trigonometric relationships between the sides and angles. Given an angle, the ratio between the opposite side to the angle by the hypotenuse is the sine of this angle, therefore, the following statement

$\sin (\theta)=\frac{3}{5}$

Describes the following triangle

To find the missing length x, we could use the Pythagorean Theorem. The sum of the squares of the legs is equal to the square of the hypotenuse. From this, we have the following equation

$x^2+3^2=5^2$

Solving for x, we have

$\begin{gathered} x^2+3^2=5^2 \\ x^2+9=25 \\ x^2=25-9 \\ x^2=16 \\ x=\sqrt[]{16} \\ x=4 \end{gathered}$

The missing length of the first triangle is equal to 4.

For the other triangle, instead of a sine we have a tangent relation. Given an angle in a right triangle, its tanget is equal to the ratio between the opposite side and adjacent side.The following expression

$\tan (y)=\frac{12}{5}$

Describes the following triangle

Using the Pythagorean Theorem again, we have

$5^2+12^2=h^2$

Solving for h, we have

$\begin{gathered} 5^2+12^2=h^2 \\ 25+144=h^2 \\ 169=h^2 \\ h=\sqrt[]{169} \\ h=13 \end{gathered}$

The missing side measure is equal to 13.

Now that we have all sides of both triangles, we can construct any trigonometric relation for those angles.

The sine is the ratio between the opposite side and the hypotenuse, and the cosine is the ratio between the adjacent side and the hypotenuse, therefore, we have the following relations for our angles

$\begin{gathered} \sin (\theta)=\frac{3}{5} \\ \cos (\theta)=\frac{4}{5} \\ \sin (y)=\frac{12}{13} \\ \cos (y)=\frac{5}{13} \end{gathered}$

To calculate the sine and cosine of the sum

$\begin{gathered} \sin (\theta+y) \\ \cos (\theta+y) \end{gathered}$

We can use the following identities

$\begin{gathered} \sin (A+B)=\sin A\cos B+\cos A\sin B \\ \cos (A+B)=\cos A\cos B-\sin A\sin B \end{gathered}$

Using those identities in our problem, we're going to have

$\begin{gathered} \sin (\theta+y)=\sin \theta\cos y+\cos \theta\sin y=\frac{3}{5}\cdot\frac{5}{13}+\frac{4}{5}\cdot\frac{12}{13}=\frac{63}{65} \\ \cos (\theta+y)=\cos \theta\cos y-\sin \theta\sin y=\frac{4}{5}\cdot\frac{5}{13}-\frac{3}{5}\cdot\frac{12}{13}=-\frac{16}{65} \end{gathered}$

4 0

1 year ago