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3 years ago

Could someone answer question 5

Mathematics

2 answers:

Savatey [412]3 years ago

5 0

P = 4s
P = (120)^1/2
4s = sqrt120
s = sqrt120/4

A = 2s
2*sqrt120/4 = sqrt120/2
Number 5 is the square root of 120 divided by 2

Rudiy273 years ago

3 0

P= sqrt120
P = 4s
4s= sqrt 120
4s = 2sqrt 30
s = (sqrt 30)/2
Area = s*s or s^2
A= ((sqrt 30)/2)^2
A= 30/4
A= 7 1/2 cm^2

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Step-by-step explanation:

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then the hypotenuse by Phitagors theory the triangle hypotenuse =

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2 years ago

In ΔNOP, OP = 7, PN = 13, and NO = 19. Which statement about the angles of ΔNOP must be true?

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The angles must be acute and all the measured angles must be different.

<h3>What is a triangle?</h3>

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If in ΔNOP, OP = 7, PN = 13, and NO = 19, then the measure of the angles must be different as well;

Hence the statement that must be true about the angles is that the angles must be acute and all the measured angles must be different. The given triangle must also be a scalene triangle

Learn more on triangles here: brainly.com/question/2938476

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2 years ago

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lozanna [386]

Answer:

$\displaystyle A = 300$

General Formulas and Concepts:

<u>Calculus</u>

Integrals

Definite Integrals
Area under the curve
Integration Constant C

Integration Rule [Reverse Power Rule]: $\displaystyle \int {x^n} \, dx = \frac{x^{n + 1}}{n + 1} + C$

Integration Rule [Fundamental Theorem of Calculus 1]: $\displaystyle \int\limits^b_a {f(x)} \, dx = F(b) - F(a)$

Integration Property [Multiplied Constant]: $\displaystyle \int {cf(x)} \, dx = c \int {f(x)} \, dx$

Integration Property [Addition/Subtraction]: $\displaystyle \int {[f(x) \pm g(x)]} \, dx = \int {f(x)} \, dx \pm \int {g(x)} \, dx$

Area of a Region Formula: $\displaystyle A = \int\limits^b_a {[f(x) - g(x)]} \, dx$

Step-by-step explanation:

<u>Step 1: Define</u>

<em>Identify</em>

f(x) = 6x + 19

Interval [12, 15]

Substitute in variables [Area of a Region Formula]: $\displaystyle A = \int\limits^{15}_{12} {(6x + 19)} \, dx$
[Integral] Rewrite [Integration Property - Addition/Subtraction]: $\displaystyle A = \int\limits^{15}_{12} {6x} \, dx + \int\limits^{15}_{12} {19} \, dx$
[Integrals] Rewrite [Integration Property - Multiplied Constant]: $\displaystyle A = 6\int\limits^{15}_{12} {x} \, dx + 19\int\limits^{15}_{12} {} \, dx$
[Integrals] Integrate [Integration Rule - Reverse Power Rule]: $\displaystyle A = 6(\frac{x^2}{2}) \bigg| \limits^{15}_{12} + 19(x) \bigg| \limits^{15}_{12}$
Evaluate [Integration Rule - Fundamental Theorem of Calculus 1]: $\displaystyle A = 6(\frac{81}{2}) + 19(3)$
Simplify: $\displaystyle A = 300$