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3 years ago

If MN=NQ and MQ=QR=RP, calculate for x

Mathematics

1 answer:

Akimi4 [234]3 years ago

7 0

Answer:

C. 18°

Step-by-step explanation:

Refer to attached diagram

<h3>Given </h3>

MN = NQ
MQ = QR = RP
NMR = 3x
QMR = x

Value of x

<h3>Solution</h3>

MN = NQ ⇒ ∠MRN = ∠QMN = 3x+x = 4x
MQ = QR ⇒ ∠MRQ = ∠QMR = x

RQP is exterior angle of ΔMQR ⇒

∠RQP = x + x = 2x

QR = RP ⇒

∠RPQ = ∠RQP = 2x

∠QRN is exterior angle of ΔQRP ⇒

∠QRN = 2x + 2x = 4x

∠MRN = ∠QRN - ∠QRM = 4x - x = 3x

∠MRN = ∠NMR = 3x ⇒

NR = MQ

NR = MQ ⇒

∠NQR = ∠QRN = 4x

We now have a straight angle MQP:

∠MQP = ∠MQN + ∠RQN + ∠RQP
180° = 4x + 4x + 2x
10x = 180°
x = 18°

Correct choice is C

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Answer:

From the calculation for both the accounts, it is clear that both the account double in the same time period of 14 years 26 days .

Step-by-step explanation:

Given as :

The principal for the first account = p = $100

The rate of interest = r = 5% compounded annually

The account gets double , so, Amount = A = $200

Let the time after which account gets double = t years

So, From Compound Interest method

Amount = Principal × $(1+\dfrac{\trxtrm rate}{100})^{\textrm time}$

As amount is double its principal

So, A = 2 × $100 = $200

Or, A = p × $(1+\dfrac{\trxtrm r}{100})^{\textrm t}$

Or, $200 = $100 × $(1+\dfrac{\trxtrm 5}{100})^{\textrm t}$

Or, $\dfrac{200}{100}$ = $(1.05)^{\textrm t}$

Or, 2 = $(1.05)^{\textrm t}$

Taking Log both side

$Log_{10}2$ = $Log_{10}(1.05)^{t}$

Or, 0.3010 = t $Log_{10}1.05$

Or, 0.3010 = t × 0.0211

∴ t = $\dfrac{0.3010}{0.0211}$

I.e t = 14.26

So, The time period to get account double is 14 years 26 days

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Amount = Principal × $(1+\dfrac{\trxtrm rate}{100})^{\textrm time}$

Or, A = p × $(1+\dfrac{\trxtrm r}{100})^{\textrm t}$

As amount is double its principal

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Or, $2000 = $1000 × $(1+\dfrac{\trxtrm 5}{100})^{\textrm t}$

Or, $\dfrac{2000}{1000}$ = $(1.05)^{\textrm t}$

Or, 2 = $(1.05)^{\textrm t}$

Taking Log both side

$Log_{10}2$ = $Log_{10}(1.05)^{t}$

Or, 0.3010 = t $Log_{10}1.05$

Or, 0.3010 = t × 0.0211

∴ t = $\dfrac{0.3010}{0.0211}$

I.e t = 14.26

So, The time period to get account double is 14 years 26 days

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Step-by-step explanation:

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4 years ago

Find the area of the region enclosed by the graphs of these equations. (CALCULUS HELP)

sergiy2304 [10]

Answer:

$\displaystyle A = \frac{20\sqrt{15}}{3}$

General Formulas and Concepts:

Pre-Algebra

Order of Operations: BPEMDAS

Brackets
Parenthesis
Exponents
Multiplication
Division
Addition
Subtraction

Left to Right

Equality Properties

Multiplication Property of Equality
Division Property of Equality
Addition Property of Equality
Subtraction Property of Equality

Algebra I

Terms/Coefficients
Graphing
Exponential Rule [Root Rewrite]: $\displaystyle \sqrt[n]{x} = x^{\frac{1}{n}}$

Calculus

Derivatives

Derivative Notation

Derivative of a constant is 0

Basic Power Rule:

f(x) = cxⁿ
f’(x) = c·nxⁿ⁻¹

Area - Integrals

U-Substitution

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

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

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

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

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

Step-by-step explanation:

Step 1: Define

F: y = √(15 - x)

G: y = √(15 - 3x)

H: y = 0

Step 2: Find Bounds of Integration

Solve each equation for the x-value for our bounds of integration.

Set y = 0: 0 = √(15 - x)
[Equality Property] Square both sides: 0 = 15 - x
[Subtraction Property of Equality] Isolate x term: -x = -15
[Division Property of Equality] Isolate x: x = 15

Set y = 0: 0 = √(15 - 3x)
[Equality Property] Square both sides: 0 = 15 - 3x
[Subtraction Property of Equality] Isolate x term: -3x = -15
[Division Property of Equality] Isolate x: x = 5

This tells us that our bounds of integration for function F is from 0 to 15 and our bounds of integration for function G is 0 to 5.

We see that we need to subtract function G from function F to get our area of the region (See attachment graph for visual).

Step 3: Find Area of Region

Integration Part 1

Rewrite Area of Region Formula [Integration Property - Subtraction]: $\displaystyle A = \int\limits^b_a {f(x)} \, dx - \int\limits^d_c {g(x)} \, dx$
[Integral] Substitute in variables and limits [Area of Region Formula]: $\displaystyle A = \int\limits^{15}_0 {\sqrt{15 - x}} \, dx - \int\limits^5_0 {\sqrt{15 - 3x}} \, dx$
[Area] [Integral] Rewrite [Exponential Rule - Root Rewrite]: $\displaystyle A = \int\limits^{15}_0 {(15 - x)^{\frac{1}{2}}} \, dx - \int\limits^5_0 {(15 - 3x)^{\frac{1}{2}}} \, dx$

Step 4: Identify Variables

Set variables for u-substitution for both integrals.

Integral 1:

u = 15 - x

du = -dx

Integral 2:

z = 15 - 3x

dz = -3dx

Step 5: Find Area of Region

Integration Part 2

[Area] Rewrite [Integration Property - Multiplied Constant]: $\displaystyle A = -\int\limits^{15}_0 {-(15 - x)^{\frac{1}{2}}} \, dx + \frac{1}{3}\int\limits^5_0 {-3(15 - 3x)^{\frac{1}{2}}} \, dx$
[Area] U-Substitution: $\displaystyle A = -\int\limits^0_{15} {u^{\frac{1}{2}}} \, du + \frac{1}{3}\int\limits^0_{15} {z^{\frac{1}{2}}} \, dz$
[Area] Reverse Power Rule: $\displaystyle A = -(\frac{2u^{\frac{3}{2}}}{3}) \bigg|\limit^0_{15} + \frac{1}{3}(\frac{2z^{\frac{3}{2}}}{3}) \bigg|\limit^0_{15}$
[Area] Evaluate [Integration Rule - FTC 1]: $\displaystyle A = -(-10\sqrt{15}) + \frac{1}{3}(-10\sqrt{15})$
[Area] Multiply: $\displaystyle A = 10\sqrt{15} + \frac{-10\sqrt{15}}{3}$
[Area] Add: $\displaystyle A = \frac{20\sqrt{15}}{3}$