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3 years ago

Calculate the length of segment RS¯ with midpoint, M, if RM=5x and MS=x+12.

Mathematics

1 answer:

Colt1911 [192]3 years ago

8 0

Answer:

So M is the midpoint of RS lmbo

Step-by-step explanation:

What you have here are lengths of two equal segments, expressed as different algebraic expressions.

you just set those two expressions equal, and solve for x. Then plug that x into EITHER of the two expressions, to get the equal lengths of RM & MS.

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What is the Compound Inequality that is true for the missing side length? 3cm, 4cm, Xcm

Brrunno [24]

Answer:

$x< 3+4\: and\\\left | 3-4 \right |$

Step-by-step explanation:

1. Well, when it comes to triangles. Mainly Pythagorean Triple, right triangles whose sides are 3,4,5 or 5,11,12

2. Keeping in mind this a Pythagorean Triple, then x=5

3. Then we can write it as Compound Inequality by combining the Statements for an Inequality of Triangle:

We can write it as

$5$

Or we can do:

$4$

and:

$3$

4. Writing in terms of x:

$x$

3 0

3 years ago

Determine the interquartile range for the data. 16, 19, 25, 20, 22, 21, 17, 20

Roman55 [17]

Answer:

the answer is 15 subtract the highest number from the lowest

5 0

2 years ago

Solve <img src="https://tex.z-dn.net/?f=%5Csf%20%5Cdfrac%7B1%7D%7Bp%7D%20%2B%20%5Cdfrac%7B1%7D%7Bq%7D%20%2B%20%5Cdfrac%7B1%7D

Nostrana [21]

Answer:

$\displaystyle \begin{cases} \displaystyle {x} _{1} = - p \\ \displaystyle x _{2} = - q \end{cases}$

Step-by-step explanation:

we would like to solve the following equation for x:

$\displaystyle \frac{1}{p} + \frac{1}{q} + \frac{1}{x} = \frac{1}{p + q + x}$

to do so isolate $\frac{1}{x}$ to right hand side and change its sign which yields:

$\displaystyle \frac{1}{p} + \frac{1}{q} = \frac{1}{p + q + x} - \frac{1}{x}$

simplify Substraction:

$\displaystyle \frac{1}{p} + \frac{1}{q} = \frac{x - (q + p + x)}{x(p + q + x)}$

get rid of only x:

$\displaystyle \frac{1}{p} + \frac{1}{q} = \frac{ - (q + p )}{x(p + q + x)}$

simplify addition of the left hand side:

$\displaystyle \frac{q + p}{pq} = \frac{ - (q + p )}{x(p + q + x)}$

divide both sides by q+p Which yields:

$\displaystyle \frac{1}{pq} = \frac{ -1}{x(p + q + x)}$

cross multiplication:

$\displaystyle x(p + q + x) = - pq$

distribute:

$\displaystyle xp + xq + {x}^{2} = - pq$

isolate -pq to the left hand side and change its sign:

$\displaystyle xp + xq + {x}^{2} + pq = 0$

rearrange it to standard form:

$\displaystyle {x}^{2} + xp + xq + pq = 0$

now notice we end up with a quadratic equation therefore to solve so we can consider factoring method to use so

factor out x:

$\displaystyle x( {x}^{} + p ) + xq + pq = 0$

factor out q:

$\displaystyle x( {x}^{} + p ) +q (x + p)= 0$

group:

$\displaystyle ( {x}^{} + p ) (x + q)= 0$

by Zero product property we obtain:

$\displaystyle \begin{cases} \displaystyle {x}^{} + p = 0 \\ \displaystyle x + q= 0 \end{cases}$

cancel out p from the first equation and q from the second equation which yields:

$\displaystyle \begin{cases} \displaystyle {x}^{} = - p \\ \displaystyle x = - q \end{cases}$

and we are done!

3 0

2 years ago

2/9 of the students in a school are in sixth grade.

IRISSAK [1]

There are 20 students in sixth grade

Step-by-step explanation:

Step 1: Find the number of sixth-graders in the school.

Total number of students = 90

Number of sixth-graders = 2/9 of total students

⇒ Number of 6th graders = 2/9 × 90 = 20 students

5 0

2 years ago

Read 2 more answers

A village water tank measures 8m x5m x 3m. What is the capacity of the tank when full?

Bad White [126]

Answer:

Hence, the capacity of the tank when full is 120 cubic m.

Step-by-step explanation:

The measures of water tank is 8m x 5m x 3m.

The capacity of the tank when full is nothing but the capacity of the full tank which is the multiples of given dimensions.

The capacity of the tank when full is,

8m x 5m x 3m = 40 m^2 x 3m

= 120 m^3.

Hence, the capacity of the tank when full is 120 cubic m.

3 0

1 year ago