Why isn't 0 at the far left of the scales?

Answer:

4cm - 4mm = 3.6 cm

So, B and C equals 3.6 cm

Answer:

64 Divided by 8

Step-by-step explanation:

Answer:

10+(n/2)=6

Step-by-step explanation:

Let the number be n

Quotient of n and 2 is (n/2)

Sum of this and 10 is

10+(n/2)=6

Answer:

4-5<-2 is false it should be 4-5>-2

Step-by-step explanation:

Answer:b

Step-by-step explanation:

Answer:

The pump should be added by 8PM

Step-by-step explanation:

An oil tanker can be emptied by the main pump in 4 hours.

An auxiliary pump can empty the tanker in 11 hours.

the main pump is started at 7 PM,and we have 10 PM - 7PM = 3 hours to empty the tank.

So we have to look for the combined rate of the both pumps .

Let's say the volume= x

Rate of first pump= x/4

Rate of second pump= x/11

The rate will be added together

Combined rate= x/11 + x/4

Combined rate= 15x/44

But they have three hours to complete their pump

15x/44 = 3

X=1.0227

The pump should be added by 8PM

Answer:

The answer could be any number above -7.

So, it could be...

(-8)

(-9)

(-10)

etc.

Step-by-step explanation:

The bigger the negative number, the lower the value.

Answer:

The standard equation of the sphere is [tex](x-5)^{2} + (y-5)^{2} + (z-4)^{2} = 53[/tex]

Step-by-step explanation:

From the question, the end point are (3,-2,4) and (7,12,4)

Since we know the end points of the diameter, we can determine the center (midpoint of the two end points) of the sphere.

The midpoint can be calculated thus

Midpoint = [tex](\frac{x_{1} + x_{2} }{2}, \frac{y_{1} + y_{2} }{2}, \frac{z_{1} + z_{2} }{2})[/tex]

Let the first endpoint be represented as [tex](x_{1}, y_{1}, z_{1})[/tex] and the second endpoint be [tex](x_{2}, y_{2}, z_{2})[/tex].

Hence,

Midpoint = [tex](\frac{x_{1} + x_{2} }{2}, \frac{y_{1} + y_{2} }{2}, \frac{z_{1} + z_{2} }{2})[/tex]

Midpoint = [tex](\frac{3 + 7 }{2}, \frac{-2+12 }{2}, \frac{4 + 4 }{2})[/tex]

Midpoint = [tex](\frac{10 }{2}, \frac{10}{2}, \frac{8 }{2})\\[/tex]

Midpoint = [tex](5, 5, 4)[/tex]

This is the center of the sphere.

Now, we will determine the distance (diameter) of the sphere

The distance is given by

[tex]d = \sqrt{(x_{2} - x_{1})^{2} +(y_{2} - y_{1})^{2} + (z_{2}- z_{1})^{2} }[/tex]

[tex]d = \sqrt{(7 - 3)^{2} +(12 - -2)^{2} + (4- 4)^{2}[/tex]

[tex]d = \sqrt{(4)^{2} +(14)^{2} + (0)^{2}[/tex]

[tex]d = \sqrt{16 +196 + 0[/tex]

[tex]d =\sqrt{212}[/tex]

[tex]d = 2\sqrt{53}[/tex]

This is the diameter

To find the radius, r

From [tex]Radius = \frac{Diameter}{2}[/tex]

[tex]Radius = \frac{2\sqrt{53} }{2}[/tex]

∴ [tex]Radius = \sqrt{53}[/tex]

r = [tex]\sqrt{53}[/tex]

Now, we can write the standard equation of the sphere since we know the center and the radius

Center of the sphere is [tex](5, 5, 4)[/tex]

Radius of the sphere is [tex]\sqrt{53}[/tex]

The equation of a sphere of radius r and center [tex](h,k,l)[/tex] is given by

[tex](x-h)^{2} + (y-k)^{2} + (z-l)^{2} = r^{2}[/tex]

Hence, the equation of the sphere of radius [tex]\sqrt{53}[/tex] and center [tex](5, 5, 4)[/tex] is

[tex](x-5)^{2} + (y-5)^{2} + (z-4)^{2} = \sqrt{(53} )^{2}[/tex]

[tex](x-5)^{2} + (y-5)^{2} + (z-4)^{2} = 53[/tex]

This is the standard equation of the sphere

Answer:

So in this sequence, we are adding by 9 in each term

So our pattern is: 7 , 7 +9 , 7 + 2(9) , 7 + 3(9) .....

hence, we can write it as: 7 + [tex]9^{n-1}[/tex] where n is the number of the term you want

Answer:

m=(y2-y1)/(x2-x1)

=(2-6)/(6-2)

=-4/4

=-1

Step-by-step explanation:

Answer:$12

Step-by-step explanation:

Answer:

months of game play (m) =$11

Step-by-step explanation:

The GCF of 45 and 75 is 15.

So, divide both numbers by 15.

45/15 = 3

7515 = 5

Therefore, the simplified version of the ratio is 3 to 5

Best of Luck!

Answer:

The parametric equations are

 x = - 2 t + 4 , y = - 5 t + 2 , z = - 3 t + 4

Step-by-step explanation:

Step(i):-

Given point P = ( 4,2,4)

Given vector is   -2 i - 5 j - 3 k

The vector equation of a line through the point a⁻ and parallel to b⁻ is

r⁻ = a⁻ + t b⁻  

Given r⁻ = 4 i + 2 j + 4 k

r⁻ = 4 i + 2 j + 4 k + t (-2 i - 5 j - 3 k)

Step(ii):-

The Cartesian form

[tex]\frac{x-a_{1} }{b_{1} } = \frac{y-a_{2} }{b_{2} } = \frac{z-a_{3} }{b_{3} } = t[/tex]

[tex]\frac{x-4 }{-2 } = \frac{y-2 }{-5 } = \frac{z-4 }{-3 } = t[/tex]

[tex]\frac{x-4 }{-2 } = t[/tex]

⇒ x - 4 = -2 t

⇒ x = - 2 t + 4

[tex]\frac{y-2 }{-5 } = t[/tex]

⇒ y - 2 = - 5 t

⇒ y = - 5 t + 2

[tex]\frac{z-4 }{-3 } = t[/tex]

z - 4 = -3 t

z = - 3 t + 4

Conclusion:-

The parametric equations are

 x = - 2 t + 4 , y = - 5 t + 2 , z = - 3 t + 4

Answer:

?

Step-by-step explanation:

no the frame will not fit

7 6/11 =7.545454

7.545454 is larger than 7.5

Answer:x=8

Step-by-step explanation:Reorder the terms:

10x + -2(6 + x) = 52

10x + (6 * -2 + x * -2) = 52

10x + (-12 + -2x) = 52

Reorder the terms:

-12 + 10x + -2x = 52

Combine like terms: 10x + -2x = 8x

-12 + 8x = 52

Solving

-12 + 8x = 52

Solving for variable 'x'.

Move all terms containing x to the left, all other terms to the right.

Add '12' to each side of the equation.

-12 + 12 + 8x = 52 + 12

Combine like terms: -12 + 12 = 0

0 + 8x = 52 + 12

8x = 52 + 12

Combine like terms: 52 + 12 = 64

8x = 64

Divide each side by '8'.

x = 8

Simplifying

x = 8

Answer:

x=5

Step-by-step explanation:

Answer:

9.9924 liters

Step-by-step explanation:

there are 3.785 liters in one gallon

multiply 3.785 liters times 2.64

that comes out to 9.9924 liters

Answer:

DQ = [tex]\frac{5}{(2-\sqrt{x})(2-\sqrt{x+h})(\sqrt{x}-\sqrt{x+h})}[/tex]

Step-by-step explanation:

Given function is f(x) = [tex]\frac{5}{2-\sqrt{x}}[/tex]

f(x + h) = [tex]\frac{5}{2-\sqrt{x+h} }[/tex]

Therefore, indicated difference quotient will be,

DQ = [tex]\frac{f(x+h)-f(x)}{h}[/tex]

Now we substitute the values in the difference quotient,

DQ = [tex]\frac{\frac{5}{2-\sqrt{x+h} }-\frac{5}{2-\sqrt{x}}}{h}[/tex]

      = [tex]\frac{5(2-\sqrt{x})-5(2-\sqrt{x+h})}{h(2-\sqrt{x})(2-\sqrt{x+h})}[/tex]

      = [tex]\frac{10-5\sqrt{x}-10+5\sqrt{x+h}}{h(2-\sqrt{x})(2-\sqrt{x+h})}[/tex]

      = [tex]\frac{-5\sqrt{x}+5\sqrt{x+h}}{h(2-\sqrt{x})(2-\sqrt{x+h})}[/tex]

      = [tex]\frac{5(-\sqrt{x}+\sqrt{x+h})}{h(2-\sqrt{x})(2-\sqrt{x+h})}[/tex]

      = [tex]\frac{5(-\sqrt{x}+\sqrt{x+h})(\sqrt{x}+\sqrt{x+h})}{h(2-\sqrt{x})(2-\sqrt{x+h})(\sqrt{x}+\sqrt{x+h})}[/tex]

      = [tex]\frac{5[(x+h)-x]}{h(2-\sqrt{x})(2-\sqrt{x+h})(\sqrt{x}+\sqrt{x+h})}[/tex]

      = [tex]\frac{5h}{h(2-\sqrt{x})(2-\sqrt{x+h})(\sqrt{x}+\sqrt{x+h})}[/tex]

      = [tex]\frac{5}{(2-\sqrt{x})(2-\sqrt{x+h})(\sqrt{x}+\sqrt{x+h})}[/tex]

Answer:

25 0r 26 F

Step-by-step explanation:

around .5 - 1 C

The correct answers are Planet A 16 hours - Inner planet and planet B 1,408 hours - Outer planet.

Large, circular celestial objects that are neither stars nor their remnants are known as planets. The inner planets are the ones that are much closer to the sun, thus they have much smaller orbits around it and it takes them much less time to circle around it. Also, these planets are much smaller, which enables them to spin around their axis much quicker too, so one day passes much quicker.

The outer planets are the total opposite. They are much further from the sun, so for them to be able to make a single circle around the sun it takes much more time, and also because they are much bigger, it takes them much more time to spin around their own axis, thus making one day much longer than the days of the inner planets.

The complete question is given below as:-

Which planet is an inner planet, and which one is an outer planet? Explain your answer. Planet A 16 hours planet B 1,408 hours

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Answer:

The answer is 18.

Step-by-step explanation:

You just need to plug in -7 to the equation 2x-4. You get -18, but absolute value is always positive, so its 18.

The answer is 18.

Sorry no explanation im soooo sorry

Answer: It’s the second one

Step-by-step explanation:

Proof by induction

Base case:

n=1: 1*2*3=6 is obviously divisible by six.

Assumption: For every n>1 n(n+1)(n+2) is divisible by 6.

For n+1:

(n+1)(n+2)(n+3)=

(n(n+1)(n+2)+3(n+1)(n+2))

We have assumed that n(n+1)(n+2) is divisble by 6.

We now only need to prove that 3(n+1)(n+2) is divisible by 6.

If 3(n+1)(n+2) is divisible by 6, then (n+1)(n+2) must be divisible by 2.

The "cool" part about this proof.

Since n is a natural number greater than 1 we can say the following:

If n is an odd number, then n+1 is even, then n+1 is divisible by 2 thus (n+1)(n+2) is divisible by 2,so we have proved what we wanted.

If n is an even number" then n+2 is even, then n+1 is divisible by 2 thus (n+1)(n+2) is divisible by 2,so we have proved what we wanted.

Therefore by using the method of mathematical induction we proved that for every natural number n, n(n+1)(n+2) is divisible by 6. QED.

The temperature in Fahrenheit, x, is converted into the temperature in Celsius with the formula . What transformations are used in transforming the parent graph y = x, to obtain the graph of blank. The answers would be:

To obtain the graph of c(x) = (5/9){ x - 32 } we need as horizontal transformation. The correct option is A.

A function is defined as the expression that set up the relationship between the dependent variable and independent variable. A function in mathematics from a set X to a set Y allocates exactly one element of Y to each element of X. The sets X and Y are collectively referred to as the function's domain and codomain, respectively.

Given function of temperature is c(x) = (5/9){ x - 32}. The two graphs are plotted on the x-y plane. It is observed that y = x is transformed to the horizontal direction.

The graph of the two functions y = x and c(x) = (5/9){ x - 32 } is attached with the answer below.

Therefore, to obtain the graph of c(x) = (5/9){ x - 32 } we need as horizontal transformation. The correct option is A.

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Answer:

A. 3-5= -2, which is not a counting number.

Step-by-step explanation:

Counting numbers are non negative numbers greater than 1 up to infinity. Counting numbers are also called natural numbers, and they do not include the negative numbers.

The statement states that "The difference of any two counting numbers will be a counting number"

From option A we can disprove that, since we can see that not all the difference between two counting numbers gives a counting number.

3 is a counting number

5 is a counting number

The difference between 3 and 5 gives -3, which is not a counting number.

This is a counterexample to the statement above.

Answer:

I think it would be 50 even numbers

Answer:

x = -7

Step-by-step explanation:

This isn't a quadratic.

Step 1: Write out equation

3x - 5 = -26

Step 2: Add 5 to both sides

3x = -21

Step 3: Divide both sides by 3

x = -7

Step-by-step explanation:

[tex]\sf \bf We \: have,[/tex]

[tex]\sf 3x - 5 = -26[/tex]

[tex]\sf : \implies \bf 3x - 5 \green{+5}= -26\green{+5}[/tex]

[tex]\sf : \implies \bf 3x =-21[/tex]

[tex]\sf : \implies \bf 3x× \green{\dfrac{1}{3}}=-21× \green{\dfrac{1}{3}}[/tex]

[tex]\sf : \implies \bf \underline{\boxed{\sf \bf x = -7}}[/tex]

This is basically saying [tex]1 \le x < 4[/tex]

The square bracket says "include this endpoint" while the curved parenthesis means "exclude this endpoint". We can tell what is included/excluded based on how the endpoint is drawn.

Open circle = exclude endpoint

Closed/filled in circle = include endpoint

Your teacher has also drawn a square bracket at the left endpoint, and a curved parenthesis at the right endpoint. This style of drawing the endpoints helps set up [1, 4). Though sometimes you might see the use of open/closed circles instead for the endpoints. I'm referring to the number line portion.

We want to express the shaded set of numbers on the x-axis with interval notation.

The interval notation is:

[1, 4)

Let's start by analyzing the graph.

We have two ends, one at x = 1, with a colored dot, which means that that value belongs to the set.

We have the other end at x = 4, with a white dot, which means that the value does not belong to the set.

Then the interval notation is:

[1, 4)

Where [] are used for colored dots.

() are used for white dots.

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Answer:

11

Step-by-step explanation:

sawarasenaaa kimi wa-