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3 years ago

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Jocelyn said that the triangle below is not a right triangle. Her work is shown below. Explain what she did wrong and show Jocel

yn the correct solution.

1 answer:

jekas [21]3 years ago

7 0

Jocelyn said that the triangle below is not a right triangle. Her work is shown below. Explain what she did wrong, and show Jocelyn the correct solution.

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In the Butterfly Museum, there are only pink and yellow butterflies. The ratio of the number of pink butterflies to the number o

RideAnS [48]

Answer: There are 48 butterflies altogether.

Step-by-step explanation:

Since we have given that

The ratio of number of pink butterflies to the number of yellow butterflies is 3:5

So,

Let the number of pink butterflies be 3x

Let the number of yellow butterflies be 5x

As we have given that there are 18 pink butterflies,

So,

$3x=18\\\\x=\frac{18}{3}\\\\x=6$

So, total number of butterflies is given by

$5x+3x=8x\\\\8x=8\times 6=48$

Hence, there are 48 butterflies altogether.

4 0

3 years ago

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Kaitaia has one more than 5 times the number or wristbands that shelly has. Rea has three more than twice the number that shelly

saul85 [17]

Answer: The required expression is 3x+4.

Step-by-step explanation:

Since we have given that

Let the number of wristbands that Shelly has be 'x'

Let the number of wristbands that Kaitaia has be $5x+1$

Let the number of wristbands that Rea has be three more than twice the number that shelly has be

$'2x+3'$

According to question, the expression that shows the number of wristbands more Katia has than Rea is given by

$5x+1-(2x-3)\\\\=5x+1-2x+3\\\\=3x+4$

Hence, the required expression is 3x+4.

3 0

3 years ago

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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:

<u>Pre-Algebra</u>

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

<u>Algebra I</u>

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

<u>Calculus</u>

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:

<u>Step 1: Define</u>

F: y = √(15 - x)

G: y = √(15 - 3x)

H: y = 0

<u>Step 2: Find Bounds of Integration</u>

<em>Solve each equation for the x-value for our bounds of integration.</em>

F

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

G

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

<u>Step 3: Find Area of Region</u>

<em>Integration Part 1</em>

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]: $\displaystyle A = \int\limits^{15}_0 {(15 - x)^{\frac{1}{2}}} \, dx - \int\limits^5_0 {(15 - 3x)^{\frac{1}{2}}} \, dx$

<u>Step 4: Identify Variables</u>

<em>Set variables for u-substitution for both integrals.</em>

Integral 1:

u = 15 - x

du = -dx

Integral 2:

z = 15 - 3x

dz = -3dx

<u>Step 5: Find Area of Region</u>

<em>Integration Part 2</em>

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

Topic: AP Calculus AB/BC (Calculus I/II)

Unit: Area Under the Curve - Area of a Region (Integration)

Book: College Calculus 10e

3 0

3 years ago

Njdcjkdckjcbnjbjkx bj. cbj.bjkcbjvkbjdvbjkdddc

andriy [413]

Answer:

lol

Step-by-step explanation:

can i have brainlyest pls?

7 0

3 years ago

Please help will give brainliest

Marat540 [252]

<h2>Answer:</h2>

This relation is a function because a function, in fact this is a linear function. We have that:

$\left[\begin{array}{cc}x & y\\2 & 3\\4 & 4\\6 & 5\\8 & 6\end{array}\right]$

As you can see below, all the points have been plotted an this is a linear function. Therefore, with two points we can get the equation, so:

$The \ equation \ of \ the \ line \ with \ slope \ m \\ passing \ through \ the \ point \ (x_{1},y_{1}) \ is:\\ \\ y-y_{1}=m(x-x_{1}) \\ \\ \\ y-3=\frac{4-3}{4-2}(x-2) \\ \\ \\ y-3=\frac{1}{2}(x-2) \\ \\ y=\frac{1}{2}x-1+3 \\ \\ y=\frac{1}{2}x+2 \\ \\ \\ Where: \\ \\ (x_{1},y_{1})=(2,3) \\ \\ (x_{2},y_{2})=(4,4)$

Finally, the equation is:

$\boxed{y=\frac{1}{2}x+2}$

5 0

3 years ago

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