Which math expression means "an unknown number divided by 180"?

Which equation has solutions of 6 and -6? x2 – 12x + 36 = 0 x2 + 12x – 36 = 0 x2 + 36 = 0 x2 – 36 = 0

Svetlanka [38]

Answer:

The equation that has solutions 6 and -6 is $x^2 - 36 = 0$

Solution:

We have to find which equation has the solutions 6 and -6.

We have been given three equations.

$x^{2}-12 x+36=0$ --- eqn 1

$x^{2}+12 x-36=0$ -- eqn 2

$x^{2}-36=0$ ---- eqn 3

The 6 and -6 to satisfy any of these equations they have to be the roots of the equation.

This means that when we substitute 6 and -6 in any of the equations and then solve them the answer on simplification should be 0.

This condition should individually be satisfied by both 6 and -6 for any one of the equations.

Now let us try and substitute 6 and -6 in eq1.

Now, substituting 6 in eq1.

62-12×6+36=0

Now we simply the equation to check is the LHS is equal to the RHS of the equation.

LHS:

72-72=0

RHS: 0

Since LHS=RHS it is the root of the equation.

Now we check if -6 satisfies eq1.

-62-12×-6+36=0

LHS:

72+72=144

RHS: 0

Hence LHS is not equal to RHS, -6 is not the root of eq1.

Similarly we check for eq2

Checking for 6 and -6 we get

LHS is not equal to RHS hence this does not satisfy eq2.

Now in the same way we check for eq3

LHS=RHS for both 6 and -6 hence they are the solutions for eq3.

Hence the equation that has solutions 6 and -6 is $x^2 - 36 = 0$

3 0

3 years ago

Read 2 more answers

After computing a confidence interval, the user believes the results are meaningless because the width of the interval is too la

navik [9.2K]

Answer:

The best recommendation is to: Increase the sample size.

Step-by-step explanation:

The width of a confidence interval $W=2\times CV\times \frac{SD}{\sqrt{n}}$ ,depends on three things,

Sample size (n)
Standard deviation
Confidence level

To decrease the width the following options can be used:

Increase the sample size.

Since the sample is inversely proportional to the width of the confidence interval, increasing the value of n will decrease the width.

Decrease the standard deviation.

The standard deviation is directly proportional to the width of the confidence interval. On decreasing the standard deviation value the width of the interval will also decrease.

Decrease the confidence level.

The critical value of the distribution is based on the confidence level. Higher the confidence level, higher will be critical value.

So, on deceasing the confidence level the critical value will decrease, hence decreasing the width of the interval.

Thus, the best recommendation is to: Increase the sample size.

3 0

3 years ago

the perimeter of a basketball court is 108 meters and the length is 6 meters longer than twice the width. what are the lengthy a

geniusboy [140]

The length is 38 meters and the width is 16 meters.

3 0

4 years ago

The box plots compare the number of calories in each snack pack of crackers and cookies.

Nikitich [7]

Answer:

4th statement is true.

Step-by-step explanation:

We have been two box plots, which represents the number of calories in each snack pack of crackers and cookies. We are asked to find the correct statement about our given box plots.

1. More packets of crackers have 80 calories than any other number of calories.

We can see that median of box plot representing calories of cookies is 80. This means that half of the packets of crackers have less than 80 calories and half of the packets have more than 80 calories, therefore, 1st statement is false.

2. The value 70 is an outlier for the number of calories in the cookie pack.

Since an outlier is 1.5 times the interquartile range.

$IQR=Q_3-Q_1$

$\text{IQR of cookie packs}=105-90$

$\text{IQR of cookie packs}=15$

$\text{Lower outlier}=Q_1-(1.5*IQR)$

$\text{Lower outlier}=90-(1.5*15)$

$\text{Lower outlier}=90-22.5$

$\text{Lower outlier}=67.5$

Since any number less than 67.5 will be an outlier and 70 is grater than 67.5, therefore, 70 is not an outlier in number of calories in cookie packs and 2nd statement is false.

3. The upper quartile of the cookie data is equivalent to the maximum in the cracker data.

We can see that upper quartile of cookie data is 105 and the maximum in cracker data is 100. Since 105 is greater than 100, therefore, 3rd statement is false.

4. The number of calories in each pack of cookies has a greater variation than the number of calories in each pack of crackers.

Since range and IQR are good measures of variation of box-plots, so we will find the range and IQR of our both box-plots.

We have already seen that IQR of cookie packs is 15.

$\text{IQR of cracker packs}=85-75$

$\text{IQR of cracker packs}=10$

$\text{Range}=\text{Maximum value - Minimum value}$

$\text{Range of calories in cracker packs}=100-70$

$\text{Range of calories in cracker packs}=30$

$\text{Range of calories in cookie packs}=115-70$

$\text{Range of calories in cookie packs}=45$

We can see that the range of calories in cookie packs (45) is greater than range of calories in cracker packs (30) and IQR of calories in cookie packs (15) is greater than IQR of calories in cracker packs (10), therefore, 4th statement is true.

3 0

4 years ago

Read 2 more answers

In triangle OPQ, p = 38 cm, q = 29 cm and Angle O=39º. Find the length of o, to the nearest

Lisa [10]

Answer:

The length of o is 24 centimeters

Step-by-step explanation:

In Δ OPQ

∵ Side p is opposite to ∠P

∵ Side q is opposite to ∠Q

∵ Side o is opposite to ∠O

→ To find side o we must use the cosine rule

∴ o = $\sqrt{p^{2}+q^{2}-2(p)(q).cos(O) }$

∵ p = 38 cm

∵ q = 29 cm

∵ m∠O = 39°

→ Substitute them in the rule to find o

∴ o = $\sqrt{(38)^{2}+(29)^{2}-2(38)(29).cos(39)}$

∴ o = 23.92008154

→ Round it to the nearest centimeter (whole number)

∴ o = 24 centimeters

∴ The length of o is 24 centimeters

6 0

3 years ago