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3 years ago

What's the common denominator between 210 and 770

Mathematics

1 answer:

Kipish [7]3 years ago

7 0

Suppose you want to combine the fractions 13/210 and 29/770. What's the LCD?

Factor both 210 and 770: 3*7*10 and 7*11*10. Form the LCD using the common factors 7 and 10 and the unique factors 3 and 11. The LCD is thus 3(7)(10)(11), or 2310.

Then 13/210 and 29/770 become 143/2310 and 87/2310. We can add these because their denominators are the same.

So the desired LCD is 2310.

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What is the solution to 4x + 2 = 6(-2x - 5) ?

exis [7]

Answer:

x= -2

Step-by-step explanation:

4x + 2 = 6(-2x - 5)

Distribute

4x+2 = -12x-30

Add 12x to each side

4x+2+12x = -12x-30+12x

16x+2 = -30

Subtract 2 from each side

16x+2-2=-30-2

16x = -32

Divide by 16

16x/16x = -32/16

x = -2

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2 years ago

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1 year ago

Solve 11x+10y=32..... Will mark the brainliest!

erica [24]

The equation 11x + 10y = 32 is a linear equation, and the solution to the linear equation 11x + 10y = 32 is $y = \frac{32 - 11x}{10}$

<h3>How to solve the equation?</h3>

The equation is given as:

11x + 10y = 32

Subtract 11x from both sides

10y = 32 - 11x

Divide both sides by 10

$y = \frac{32 - 11x}{10}$

Hence, the solution to the linear equation 11x + 10y = 32 is $y = \frac{32 - 11x}{10}$

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brainly.com/question/14323743

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4 0

1 year ago

Fill in the table with the appropriate value to describe the amount of hot dogs eaten at a hot dog eating contest.

Ivan

20 hot dogs was eaten by the 10 people who attended the hot dog eating contest.

<h3>What is an equation?</h3>

An equation is an expression that shows the relationship between two or more numbers and variables.

Let us assume that there were 10 people at the hot dog eating contest. If each person eats 2 hot dog, hence:

Number of hot dogs eaten = 2 hot dog per person * 10 people = 20 hot dogs.

20 hot dogs was eaten by the 10 people who attended the hot dog eating contest.

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7 0

1 year 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:

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