solve th variables for x,y,z​

Answer:

The kinetic energy of the car is 43,200 Joules.

Step-by-step explanation:

KE = (1/2)mv^2

KE = (1/2)(600 kg)(12 m/s)^2

KE = (1/2)(600 kg)(144 m^2/s^2)

KE = 43,200 kg*m^2/s^2 = 43,200 Joules

Answer:

Step-by-step explanation:

KE= 1/2mv^2

KE = 1/2(600) 12^2

KE = 300 * 144 = 43200J

Answer:

[tex]\left \{ {{y=x; \ \ x\le 0} \atop {y=4+ \frac12x;\ \ x>0}} \right.[/tex]

Step-by-step explanation:

If you look at the graph you see that:

before 0, the graph has same y as it has x, or y=x.

after 0, the graph starts at 4, and increases by 1 every 2 steps horizontally, or has a slope of 1/2.

Finally, the 0 has to be included in the blue part of the graph based on where the solid dot is.

Answer:

d & e

Step-by-step explanation:

I'm assuming  by "same side interior angle" they mean consecutive interior angles. I'll attach a document explaining what that is.

Answer:

Step-by-step explanation:

a. [tex]\left(5\times \:10^3\right)\left(9\times \:10^7\right)[/tex]

[tex]5\times 10^3 = 5000\\\\5000\times \:9\times \:10^7[/tex]

[tex]\mathrm{Multiply\:the\:numbers:}\:5000\times \:9=45000[/tex]

[tex]=45000\times \:10^7[/tex]

b. [tex]\left(7\times 10^5\right)\div \left(2\times 10^2\right)[/tex]

[tex]\left(7\times 10^5\right)\div \left(2\times 10^2\right)=\frac{7\times \:10^5}{2\times \:10^2}[/tex]

[tex]\frac{10^5}{10^2} = 10^3[/tex]

[tex]\frac{7\times \:10^3}{2}[/tex]

[tex]\mathrm{Factor}\:\: 10^3 : 2^3\times5^3[/tex]

[tex]\frac{7\times \:2^3\times \:5^3}{2}[/tex]

[tex]\frac{2^3}{2}=2^2[/tex]

[tex]7\times \:2^2\times \:5^3=3500[/tex]

Answer:

M’ = (2, -7)

N’ = (1, -5)

O’ = (7, - 2)

P’ = (8,- 4)

Step-by-step explanation:

Swap the x and y coordinates and negate the y coordinate.

M(-7, -2) swap = (- - 2, -7) = (2,-7)

N(-5, - 1) swap = (- - 1, - 5) = (1, - 5)

O(-2,-7) swap = (- -7, -2) = (7, - 2)

P(-4,-8) swap = (- - 8, - 4) = (8, - 4)

Answer:

104

Step-by-step explanation:

Answer:

24

Step-by-step explanation:

m=8x4 n=3x4 so that is the answer

Answer:

(by using the first principle of differentiation)

We have the "Limit definition of Derivatives":

[tex]\boxed{\mathsf{f'(x)= \lim_{h \to 0} \{\frac{f(x+h)-f(x)}{h} \} ....(i)}}[/tex]

Here, f(x) = sec x, f(x+h) = sec (x+h)

[tex]\implies \mathsf{f'(x)= \lim_{h \to 0} \{\frac{sec(x+h)-sec(x)}{h} \} }[/tex]

[tex]\implies \mathsf{f'(x)= \lim_{h \to 0} \frac{1}{h} \{\frac{1}{cos(x+h)} -\frac{1}{cos(x)} \} }[/tex]

[tex]\implies \mathsf{f'(x)= \lim_{h \to 0} \frac{1}{h} \{\frac{cos(x)-cos(x+h)}{cos(x)cos(x+h)} \} }[/tex]

[tex]\boxed{\mathsf{cos\:A-cos\:B=-2sin(\frac{A+B}{2} )sin(\frac{A-B}{2} )}}[/tex]

=> cos(x) - cos(x+h) = -2sin{(x+x+h)/2}sin{(x-x-h)/2}

[tex]\implies \mathsf{f'(x)= \lim_{h \to 0} \frac{2sin(x+\frac{h}{2} )}{cos(x+h)cos(x)}\:.\: \lim_{h \to 0} \frac{sin(\frac{h}{2} )}{h} }[/tex]

[tex]\implies \mathsf{f'(x)= \lim_{h \to 0} \frac{sin(x+\frac{h}{2} )}{cos(x+h)cos(x)}\:.\: \lim_{h \to 0} \frac{2sin(\frac{h}{2} )}{h} }[/tex]

[tex]\implies \mathsf{f'(x)= \lim_{h \to 0} \frac{sin(x+\frac{h}{2} )}{cos(x+h)cos(x)}\:.\: \lim_{h \to 0} \frac{1\times sin(\frac{h}{2} )}{\frac{h}{2} } }[/tex]

[tex]\implies \mathsf{f'(x)= \lim_{h \to 0} \frac{sin(x+\frac{h}{2} )}{cos(x+h)cos(x)}\:.\: 1 }[/tex]

Substituting 0 for h and h/ 2

[tex]\implies \mathsf{f'(x)= \lim_{h \to 0} \frac{sin(x+0)}{cos(x+0)cos(x)} }[/tex]

[tex]\implies \mathsf{f'(x)= \lim_{h \to 0} \frac{sin(x)}{cos(x)cos(x)} }[/tex]

[tex]\implies \mathsf{f'(x)= \lim_{h \to 0} \frac{sin(x)}{cos(x)}\times \frac{1}{cos x} }[/tex]

[tex]\implies \mathsf{f'(x)= tan(x)\times sec(x) }[/tex]

Hence, we got

[tex]\underline{\mathsf{\overline{\frac{d}{dx} (sec(x))=sec(x)tan(x)}}}[/tex]

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

(by using other standard derivatives)

[tex] \boxed{ \mathsf{ \frac{d}{dx} ( \sec \: x) = \sec x \tan x }}[/tex]

We have a standard derivative for variables in x raised to an exponent:

[tex] \boxed{ \mathsf{ \frac{d}{dx}(x)^{n} = n(x)^{n - 1} }}[/tex]

Therefore,

[tex] \mathsf{ \frac{d}{dx}( \cos x)^{ - 1} = - 1( \cos \: x) ^{( - 1 - 1} } \\ \implies \mathsf{\ - 1( \cos \: x) ^{- 2 }}[/tex]

[tex] \implies \: \mathsf{ - \frac{1}{ (\cos x) {}^{2} } }[/tex]

The process of differentiating doesn't just end here. It follows chain mechanism, I.e.,

while calculating the derivative of a function that itself contains a function, the derivatives of all the inner functions are multiplied to that of the exterior to get to the final result.

[tex] \implies \mathsf{ - \frac{1}{ (\cos x )^{2} } \times ( - \sin x) }[/tex]

[tex] \implies \mathsf{ \frac{ \sin x }{ \cos x } \times ( \frac{1}{cos x} ) }[/tex]

[tex] \implies \mathsf{ \tan x \times \sec x }[/tex]

= sec x. tan x

Answer:

30

Step-by-step explanation:

Answer:

the value of x is 12

......

Answer:

101

HOPE THIS HELPS

- Todo ❤️

Step-by-step explanation:

Suplemetary angles

180-79=101

Answer:

here the answer is 24

Step-by-step explanation:

answerrrrrerrr issss hereeeeeee

Answer:

0

Step-by-step explanation:

4 multiply by x(3) = 12

6 multiply by y(2) = 12

12 -12 =0

Answer:

4x - 12 = 84

Step-by-step explanation:

The last answer choice is correct because when you cross-multiply:

you get 4x - 12 = 84.

Therefore, the last option is correct.

Answer:

D would be the answer (4x-12=84)

Step-by-step explanation:

4/7=12/x-3

=>1/7=3/x-3

=>x-3=21

Multiplying both sides by 4

4(x-3)=4x21

=>4x-12=84

Hope this helped :)

Answer:

No

Step-by-step explanation:

The answer is to find the sum of each number, because factors are pulling out from total numbers, but when multiplying you don't need to pull out anything so it would be number

Yep!!! Your correct

Answer:

x = -.1715 ≈ - .172    or x = -5.83

Step-by-step explanation:

x² + 6x + 1 = 0

x² + 6x = -1

Complete the square   Add to both sides (1/2 of the x-term, then square it.)

x² + 6x + 9  = -1 + 9

(x + 3)(x + 3) = 8

(x + 3)² = 8

[tex]\sqrt{(x + 3)^{2}[/tex]  = [tex]\sqrt{8}[/tex]

 x + 3 = ±  [tex]\sqrt{8}[/tex]

  x = -3 ± [tex]\sqrt{8}[/tex]

   x = -3 + [tex]\sqrt{8}[/tex]   or x = -3 - [tex]\sqrt{8}[/tex]

   x = -.1715 ≈ - .172    or x = -5.83

Answer:

1.) x = 2

2.) x = 5

3.) x = 9

4.) x = 0

Step-by-step explanation:

1.) -3x - 8 = -14

-3x = -6

x = 2

2.) 4x - 6 = 14

4x = 20

x = 5

3.) -3 - 3x = -30

-3x = -27

x = 9

4.) -5x + 5 = 5

-5x = 0

x = 0

Step-by-step explanation:

f(x) = -3x - 8

f(-3) = -3(-3) - 8

f(-3) = 9 - 8

f(-3) = 1

Answer:

3, 4, 5 and 5, 12, 13

Step-by-step explanation:

The square of the largest side is equal to the sum of the squares of the other 2 sides.

5² = 3² + 4²

13² = 5² + 12²

The 2 triplets are (3, 4, 5 ) and (5, 12, 13)

Answer:

i think it is 49:9

Step-by-step explanation:

because number going in 49:9 by which multiple it is 7x7 is 49 3x3 is 9 so the answer is 49:9

Answer:

28:12

Step-by-step explanation:

Multiply both 7 and 3 by 4 and you get the ratio 28:12.

Step-by-step explanation:

5x - 6 = 3x + 2

5x - 3x = 2 + 6

2x = 8

x = 8/2

x = 4

Answer:

a

Step-by-step explanation:

Answer:

20400

Step-by-step explanation:

its 2.03x10x10x10x10 so first ten: 20.4 and ten: 204 3rd ten: 2040 last ten :20400

Answer:

Step-by-step explanation:

Here we need to find a 3rd degree polynomial that has only two distinct zeros:  {-3, 8}.  

Focusing on the zero x = 8 and assuming that this 8 has a multiplicity of 2, we come up with the following polynomial in which the factor x - 8 shows up twice:

f(x) = a(x - 8)(x - 8)(x + 3), or f(x) = a(x - 8)^2(x + 3)

One such polynomial is thus

f(x) = a(x - 8)^2(x + 3), where 'a' is a constant coefficient.  This polynomial has a double zero at x = 8 and a third zero at x = -3.

Step-by-step explanation:

The polynomial has degree 3 and two zeros: - 3 and 8.

Since it has degree 3 it should have 3 zero's.

Two possible scenarios

1. -3 has multiplicity of two:

2. 8 has multiplicity of two:

Answer:

first put 2 in the numerator for the first blank and 2/9 in the second blank

Step-by-step explanation:

1/3 equals 3/9 and 3/9-1/9=2/9

Answer:

[tex]\frac{3}{9}[/tex]

Step-by-step explanation:

Step-by-step explanation:

very simple thought experiment :

he is going 90 km/h (90 kilometers per hour), so, he moving 90 km in 1 hour.

how far is he going in just 2 seconds ?

so, we get two ratios.

90km/1 hour = 90,000 meter / 1 hour

x meter / 2 seconds

we need to bring both ratios (they are fractions) to the same dimension of the denominators, so that we can then bring them to the same denominator.

the first is dealing with an hour.

the second one with (2) seconds.

so, how many seconds are in 1 hour ?

60 seconds per minute, 60 minutes in the hour.

60×60 = 3600 seconds.

so, now we know, the first ratio is also

90,000 meter / 3600 seconds

now we need the factor to to multiply numerator and denominator with to bring 3600 down to 2.

3600 × f = 2

f = 2/3600 = 1/1800

now we get

90,000 / 3600 × (1/1800) / (1/1800) =

(90000 / 1800) / (3600 / 1800) = 50 / 2

so, 90km/h is the same speed as 50 meter / 2 seconds.

and therefore we know, the building was 50 meters long.

Answer:

Move it to the right by 3 spaces.

Step-by-step explanation:

10^3 is 1000

On a number line the negative numbers are on the left and the positive numbers are on the right. So since the exponent is a positive 3 we move it to the right by 3 spaces to get 140

Juan llenó 3 cajas grandes y 4 cajas pequeñas.

En esta pregunta debemos derivar ecuaciones algebraicas a partir de la información dada en el enunciado:

Diferencia entre cajas grandes y cajas pequeñas

[tex]m - n = 1[/tex] (1)

Donde:

Diferencia entre la cantidad de bombones en una caja grande y la cantidad de bombones en una caja pequeña

[tex]x-y = 5[/tex] (2)

Donde:

Total de bombones en cajas grandes

[tex]x\cdot m = 60[/tex] (3)

Total de bombones en cajas pequeñas

[tex]y\cdot n = 60[/tex] (4)

By (3) and (4) in (2):

[tex]\frac{60}{m} - \frac{60}{n} = 5[/tex]

[tex]60\cdot n - 60\cdot m = 5\cdot m\cdot n[/tex] (5)

By (1):

[tex]m = 1 + n[/tex]

(1) in (5):

[tex]60\cdot n - 60\cdot (1+n) = 5\cdot (1+n)\cdot n[/tex]

[tex]60 = 5\cdot n +5\cdot n^{2}[/tex]

[tex]5\cdot n^{2}+5\cdot n -60 = 0[/tex]

[tex]n^{2}+n-12 = 0[/tex]

[tex](n+4)\cdot (n-3) = 0[/tex]

La única raíz que encaja con el enunciado es [tex]n = 3[/tex]. Por (1) tenemos que [tex]m = 4[/tex].

Juan llenó 3 cajas grandes y 4 cajas pequeñas.

Invitamos cordialmente a ver este problema sobre sistemas de ecuaciones: https://brainly.com/question/12149981