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2 years ago

What is the perimeter of a triangle with sides 11 inches, 5 inches, and 12 inches in length?

Mathematics

2 answers:

natulia [17]2 years ago

3 0

Answer: 28 in.

Step-by-step explanation:

Lorico [155]2 years ago

3 0

Answer:

Step-by-step explanation:

11+5+12=28

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Solve - pv + 40 < 65 for v<br> solve 7w - 3r = 15 for r

Katarina [22]

Answer:

v > -25/p

r = -5 +7/3 w

Step-by-step explanation:

- pv + 40 < 65

Subtract 40 from each side

- pv + 40-40 < 65-40

-pv < 25

Divide each side by -p (remember to flip the inequality since we are dividing by a negative)

-pv/-p > 25/-p

v > -25/p

7w - 3r = 15

Subtract 7w from each side

7w-7w - 3r = 15-7w

-3r = 15-7w

Divide by -3

-3r/-3 = (15-7w)/-3

r = -5 +7/3 w

8 0

2 years ago

7.<br>(-3x³)(-2x³y⁴z)(-3z²)÷(4x³z)(-3yz) (-3xyz)

serg [7]

Answer:

$-\frac{x^2y^2}{2}$

Step-by-step explanation:

We have an extremely large equation and are asked to divide it, so let's solve it step-by-step :

Remove the parenthesis to make it easier to read :

$\frac{-3x^3 *2x^3y^4z *3z^2}{4x^3z*3yz*3xyz}$

Multiply the numerators :

$\frac{-18x^6y^4z^3}{4*3*3x^3yyzzz}$

Multiply the denominators :

$\frac{-18x^6y^4z^3}{36x^4y^2z^3}$

Apply the negative rule :

$-\frac{18x^6y^4z^3}{36x^4y^2z^3}$

Cancel the common factor which is 18 :

$-\frac{x^6y^4z^3}{2x^4y^2z^3}$

Apply the addition exponent rule :

$\frac{y^4z^3x^{6-4} }{2y^2z^3}$

Subtract :

$\frac{x^2y^4z^3 }{2y^2z^3}$

Apply the rule for y :

$\frac{x^2y^{4-2} z^3 }{2z^3}$

Subtract :

$\frac{x^2y^2z^3 }{2z^3}$

Cancel the common factor of z^3 :

$-\frac{x^2y^2}{2}$

5 0

2 years ago

The ratio of girls to students at a middle school is 10 to 18. If there are 270 students at the school, how many are girls?

Stells [14]

A good way of doing this problem would be to use cross multiplication of ratios
<u /> $\frac{10 girls}{18 students} = \frac{X girls}{270 students}$

Then cross multiply

(10 girls)(270 students)=(X girls)(18 students)

2700=18x
150=x

4 0

2 years ago

Read 2 more answers

Solve: 7 x + 2 y = 16; -21 x-6 y = 24 Solve : 7 x + 2 y = 16 ; -21 x - 6 y = 24

Marat540 [252]

Answer:

No solution

Step-by-step explanation:

<h3>Solving system of linear equations:</h3>

7x + 2y = 16 ----------------(I)

-21x - 6y = 24 -----------------(II)

$\sf \dfrac{a_1}{a_2}=\dfrac{7}{-21}=\dfrac{-1}{3}\\$

$\sf \dfrac{b_1}{b_2}=\dfrac{2}{-6}=\dfrac{-1}{3}$

$\sf \dfrac{c_1}{c_2} =\dfrac{16}{24}=\dfrac{2}{3}\\\\ \dfrac{a_1}{a_2}=\dfrac{b_1}{b_2} \neq \dfrac{c_1}{c_2}$

So, the lines are parallel and has no solution

7 0

1 year ago

Find the expression that represents the length of the hypotenuse of a right triangle whose legs measure m^2-n^2 and 2mn

My name is Ann [436]

$\bf c^2=a^2+b^2\qquad 
\begin{cases}
c=hypotenuse\\
a=adjacent\\
b=opposite\\
------\\
a=m^2-n^2\\
b=2mn
\end{cases}\\\\\\ c^2=(m^2-n^2)^2+(\underline{2mn})^2
\\\\\\
c^2=m^4-2m^2n^2+n^4+\underline{2^2m^2n^2}
\\\\\\
c^2=m^4-2m^2n^2+n^4+{4m^2n^2}
\\\\\\
c^2=m^4+2m^2n^2+n^4\impliedby \textit{perfect square trinomial}
\\\\\\
c^2=(m^2+n^2)^2\implies c=\sqrt{(m^2+n^2)^2}\implies c=m^2+n^2$

7 0

3 years ago