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3 years ago

A radius of a circle is 14 cm and a chord has 18 cm. What is the distance between the center of the circle and the chord

Mathematics

1 answer:

ryzh [129]3 years ago

6 0

Answer: $10.72\ cm$

Step-by-step explanation:

Given

Radius of circle $r=14\ cm$

Length of chord is $l=18\ cm$

Suppose the distance between the chord and center is x

From figure, we can write

$\Rightarrow r^2=x^2+9^2\\\Rightarrow 14^2=x^2+9^2\\\Rightarrow x^2=196-81\\\Rightarrow x^2=115\\\Rightarrow x=\sqrt{115}\\\Rightarrow x=10.72\ cm$

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Two trains leave a city at 3 P.M. One train goes east at a constant rate of 70 mph. The other goes west at a rate of 62 mph. How

andrey2020 [161]

Answer:

4 hours

Step-by-step explanation:

The general formula for distance, rate and time is:

d = rt, where distance is the product of rate multiplied by time

Given that one train is moving at a rate of 70 mph and one is moving at a rate of 62 mph in the opposite direction, the combined distance over time is equal to 528 miles. Using the variable 't' for time:

70t + 62t = 528

Combine like terms: 132t = 528

Divide both sides by 132: 132t/132 = 528/132 or t = 4 hours

8 0

3 years ago

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Ms. Carr's Class Is Selling Magazines To Raise Money For A Field Trip. The Students In The Class Device They Wanted To Make 5.50

FinnZ [79.3K]

Here you are giving only the amount they want to raise (namely profit times number of magazines sold), and here you are also giving Money they want to raise... So clarifying, the money they want to raise, should include the money they will spend on buying the magazines (there is no statement saying they found them, or were given the magazines, so a cost should be involved)

Now if they are only making the count of "Field trip costs X amount of money, and given we have to make a profit of $5.5, How many must we sell?" then the equation should be n=X/5.5

Should the story be, how much money must they raise to have a profit of 5.5 on each magazine and still have enough for the field trip, then you have a different equation which varies only in adding the cost of each magazine, either case, M should be defined not as money they need to raise (cause here they will be short on their goal) but Money they must earn. And again, you should rewrite your equation to be:

M=Amount they must raise
C=Cost per magazine
n=Number of magazines
p=profit $5.5 per magazine

C+p=M/n

And rewriting the previous they should make:

n(C+p)=M -----> n(C+5.5)=M m/n = 5.50 
m/n x n = 5.50 x n //// multiply each side by n 

m = 5.5n

6 0

3 years ago

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What is the value of x if 2x+4=3

kolezko [41]

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

# Bora 7#

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2 years ago

Find the volume of a square pyramid with base edge = 9 cm

maw [93]

Right square pyramid V=189cm³

3 0

3 years ago

−2x=x^2−6

Iteru [2.4K]

Step-by-step explanation:

Example 1

Solve the equation x3 − 3x2 – 2x + 4 = 0

We put the numbers that are factors of 4 into the equation to see if any of them are correct.

f(1) = 13 − 3×12 – 2×1 + 4 = 0 1 is a solution

f(−1) = (−1)3 − 3×(−1)2 – 2×(−1) + 4 = 2

f(2) = 23 − 3×22 – 2×2 + 4 = −4

f(−2) = (−2)3 − 3×(−2)2 – 2×(−2) + 4 = −12

f(4) = 43 − 3×42 – 2×4 + 4 = 12

f(−4) = (−4)3 − 3×(−4)2 – 2×(−4) + 4 = −100

The only integer solution is x = 1. When we have found one solution we don’t really need to test any other numbers because we can now solve the equation by dividing by (x − 1) and trying to solve the quadratic we get from the division.

Now we can factorise our expression as follows:

x3 − 3x2 – 2x + 4 = (x − 1)(x2 − 2x − 4) = 0

It now remains for us to solve the quadratic equation.

x2 − 2x − 4 = 0

We use the formula for quadratics with a = 1, b = −2 and c = −4.

We have now found all three solutions of the equation x3 − 3x2 – 2x + 4 = 0. They are: eftirfarandi:

x = 1

x = 1 + Ö5

x = 1 − Ö5

Example 2

We can easily use the same method to solve a fourth degree equation or equations of a still higher degree. Solve the equation f(x) = x4 − x3 − 5x2 + 3x + 2 = 0.

First we find the integer factors of the constant term, 2. The integer factors of 2 are ±1 and ±2.

f(1) = 14 − 13 − 5×12 + 3×1 + 2 = 0 1 is a solution

f(−1) = (−1)4 − (−1)3 − 5×(−1)2 + 3×(−1) + 2 = −4

f(2) = 24 − 23 − 5×22 + 3×2 + 2 = −4

f(−2) = (−2)4 − (−2)3 − 5×(−2)2 + 3×(−2) + 2 = 0 we have found a second solution.

The two solutions we have found 1 and −2 mean that we can divide by x − 1 and x + 2 and there will be no remainder. We’ll do this in two steps.

First divide by x + 2

Now divide the resulting cubic factor by x − 1.

We have now factorised

f(x) = x4 − x3 − 5x2 + 3x + 2 into

f(x) = (x + 2)(x − 1)(x2 − 2x − 1) and it only remains to solve the quadratic equation

x2 − 2x − 1 = 0. We use the formula with a = 1, b = −2 and c = −1.

Now we have found a total of four solutions. They are:

x = 1

x = −2

x = 1 +

x = 1 −

Sometimes we can solve a third degree equation by bracketing the terms two by two and finding a factor that they have in common.

6 0

3 years ago

Read 2 more answers