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3 years ago

15. A door wedge is shaped like a triangular prism. The wedge is filled with sand. How

Mathematics

2 answers:

kvasek [131]3 years ago

7 0

Answer:362

Step-by-step explanation:

You need to multiply

klemol [59]3 years ago

6 0

Answer:

126

Step-by-step explanation:

you add them all up and the sides that don't have the number there's already info about it it's like this

18 x 3 + 30 + 24 + 18 = 126

and you don't know how to do 18 x 3 lol it's 54

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X = 0.19

Luda [366]

Step-by-step explanation:

.19 = 19/100

.864 = 86.4/100

.5348 = 53.48 /100

I multiplied the decimal by 100 to get what it was before which is the same as moving two decimal places to the right and divided by 100

if any confusion ask

4 0

2 years ago

Read 2 more answers

Which of the following units for mass is an SI unit?

s2008m [1.1K]

Kilogram is an SI unit for mass.
Ounce and pound refer to American system for measuring things, and decakilogram is not the "basic" unit, but kilogram is.

5 0

3 years ago

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The directrix of a parabola is y=−4y=−4 . The focus of the parabola is (−2,−2)(−2,−2) .

salantis [7]

Refer to the diagram shown below.

The directrix is y = -4 and the focus is (-2, -2).
Therefore the vertex is at (-2, -3).

Consider an arbitrary point (x,y) on the parabola.
The square of distance from the focus to the point is
(x+2)² + (y+2)²
The square of the distance from the point to the directrix is
(y+4)²

Therefore
(y+4)² = (y+2)² + (x+2)²
y² + 8y + 16 = y² + 4y + 4 + (x+2)²
4y = (x+2)² - 12
y = (1/4)(x+2)² - 3

Answer: $y = \frac{1}{4} (x+2)^{2} - 3$

6 0

3 years ago

You know 1000 mg = 1g. Convert 2.5 grams into milligrams.

Nuetrik [128]

The answer is 2500 milligrams

7 0

3 years ago

Can someone please help me with this

Mama L [17]

Answer:

27.

Equation of 2 tangent lines at the given curve going through point P is:

y = -7x + 1

y = x + 1

28.

Part 1: 400 feet

Part 2: Velocity is +96 feet/second (or 96 feet/second UPWARD) & Speed is 96 feet per second

Part 3: acceleration at any time t is -32 feet/second squared

Part 4: t = 10 seconds

29.

Part 1: Average Rate of Change = -15

Part 2: The instantaneous rate of change at x = 2 is -8 & at x = 3 is -23

Step-by-step explanation:

27.

First of all, the equation of tangent line is given by:

$y-y_1=m(x-x_1)$

Where m is the slope, or the derivative of the function

Now,

If we take a point x, the corresponding y point would be $x^2-3x+5$ , so the point would be $(x,x^2-3x+5)$

Also, the derivative is:

$f(x)=x^2-3x+5\\f'(x)=2x-3$

Hence, we can equate the DERIVATIVE (slope) and the slope expression through the point given (0,1) and the point we found $(x,x^2-3x+5)$

The slope is $\frac{y_2-y_1}{x_2-x_1}$

So we have:

$\frac{x^2-3x+5-1}{x-0}\\=\frac{x^2-3x+4}{x}$

Now, we equate:

$2x-3=\frac{x^2-3x+4}{x}$

We need to solve this for x. Shown below:

$2x-3=\frac{x^2-3x+4}{x}\\x(2x-3)=x^2-3x+4\\2x^2-3x=x^2-3x+4\\x^2=4\\x=-2,2$

So, this is the x values of the point of tangency. We evaluate the derivative at these 2 points, respectively.

$f'(x)=2x-3\\f'(-2)=2(-2)-3=-7\\f'(2)=2(2)-3=1$

Now, we find 2 equations of tangent lines through the point (0,1) and with respective slopes of -7 and 1. Shown below:

$y-y_1=m(x-x_1)\\y-1=-7(x-0)\\y-1=-7x\\y=-7x+1$

and

$y-y_1=m(x-x_1)\\y-1=1(x-0)\\y-1=x\\y=x+1$

So equation of 2 tangent lines at the given curve going through point P is:

y = -7x + 1

y = x + 1

28.

Part 1:

The highest point is basically the maximum value of the position function. To get maximum, we differentiate and set it equal to 0. Let's do this:

$s(t)=160t-16t^2\\s'(t)=160-32t\\s'(t)=0\\160-32t=0\\32t=160\\t=\frac{160}{32}\\t=5$

So, at t = 5, it reaches max height. We plug in t = 5 into position equation to find max height:

$s(t)=160t-16t^2\\s(5)=160(5)-16(5^2)\\=400$

max height = 400 feet

Part 2:

Velocity is speed, but with direction.

We also know the position function differentiated, is the velocity function.

Let's first find time(s) when position is at 256 feet. So we set position function to 256 and find t:

$s(t)=160t-16t^2\\256=160t-16t^2\\16t^2-160t+256=0\\t^2-10t+16=0\\(t-2)(t-8)=0\\t=2,8$

At t = 2, the velocity is:

$s'(t)=v(t)=160-32t\\v(2)=160-32(2)\\v(2)=96$

It is going UPWARD at this point, so the velocity is +96 feet/second or 96 feet/second going UPWARD

The corresponding speed (without +, -, direction) is simply 96 feet/second

Part 3:

We know the acceleration is the differentiation of the velocity function. let's find it:

$v(t)=160-32t\\v'(t)=a(t)=-32$

hence, the acceleration at any time t is -32 feet/second squared

Part 4:

The rock hits the ground when the position is 0 (at ground). So we equate the position function, s(t), to 0 and find time when it hits the ground. Shown below:

$s(t)=160t-16t^2\\0=160t-16t^2\\16t^2-160t=0\\16t(t-10)=0\\t=0,10$