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3 years ago

Need actual help (please hurry)

Mathematics

1 answer:

mixer [17]3 years ago

3 0

A. n=5
20=4n
29/4=n
5=n

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What is 31/87 as a percentage? Give your answer rounded to one decimal place.

kicyunya [14]

Answer:

35.6%

Step-by-step explanation:

31/87(100)

3100/87

35.6%

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3 years ago

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Solve: (-6)(-7) Divide or multiple

Paul [167]

Answer: 42

Step-by-step explanation:

If they are in parentheses, you are going to multiply them, and when you multiply two negatives, you make a positive. which gives you 42.

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3 years ago

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Suppose θ is an angle in the standard position whose terminal side is in Quadrant IV and cot θ= -6/7 . Find the exact values of

Llana [10]

Answer:

Part 1) $csc(\theta)=-\frac{\sqrt{85}}{7}$

Part 2) $sin(\theta)=-\frac{7}{\sqrt{85}}$ or $sin(\theta)=-\frac{7\sqrt{85}}{85}$

Part 3) $tan(\theta)=-\frac{7}{6}$

Part 4) $cos(\theta)=\frac{6}{\sqrt{85}}$ or $cos(\theta)=\frac{6\sqrt{85}}{85}$

Part 5) $sec(\theta)=\frac{\sqrt{85}}{6}$

Step-by-step explanation:

we know that

The angle theta lie on the IV Quadrant

sin(θ) is negative

cos(θ) is positive

tan(θ) is negative

sec(θ) is positive

csc(θ) is negative

step 1

Find the value of csc(θ)

we know that

$1+cot^{2}(\theta)=csc^{2}(\theta)$

we have

$cot(\theta)=-\frac{6}{7}$

substitute

$1+(-\frac{6}{7})^{2}=csc^{2}(\theta)$

$1+\frac{36}{49}=csc^{2}(\theta)$

$\frac{85}{49}=csc^{2}(\theta)$ rewrite

$csc(\theta)=-\frac{\sqrt{85}}{7}$ ----> remember that is negative

step 2

Find the value of sin(θ)

we know that

$csc(\theta)=\frac{1}{sin(\theta)}$

we have

$csc(\theta)=-\frac{\sqrt{85}}{7}$

therefore

$sin(\theta)=-\frac{7}{\sqrt{85}}$

$sin(\theta)=-\frac{7\sqrt{85}}{85}$

step 3

Find the value of tan(θ)

we know that

$tan(\theta)=\frac{1}{cot(\theta)}$

we have

$cot(\theta)=-\frac{6}{7}$

therefore

$tan(\theta)=-\frac{7}{6}$

step 4

Find the value of cos(θ)

we know that

$sin^{2}(\theta)+cos^{2}(\theta)=1$

we have

$sin(\theta)=-\frac{7}{\sqrt{85}}$

substitute

$(-\frac{7}{\sqrt{85}})^{2}+cos^{2}(\theta)=1$

$\frac{49}{85}+cos^{2}(\theta)=1$

$cos^{2}(\theta)=1-\frac{49}{85}$

$cos^{2}(\theta)=\frac{36}{85}$

$cos(\theta)=\frac{6}{\sqrt{85}}$ ------> the cosine is positive

$cos(\theta)=\frac{6\sqrt{85}}{85}$

step 5

Find the value of sec(θ)

we know that

$sec(\theta)=\frac{1}{cos(\theta)}$

we have

$cos(\theta)=\frac{6}{\sqrt{85}}$

therefore

$sec(\theta)=\frac{\sqrt{85}}{6}$ ----> is positive

8 0

4 years ago

The graph of f(x)= 3/1+x^2 is shown in the figure to the right. Use the second derivative of f to find the intervals on which f

GenaCL600 [577]

Answer:

Concave Up Interval: $(- \infty,\frac{-\sqrt{3} }{3} )U(\frac{\sqrt{3} }{3} , \infty)$

Concave Down Interval: $(\frac{-\sqrt{3} }{3}, \frac{\sqrt{3} }{3} )$

General Formulas and Concepts:

Calculus

Derivative of a Constant is 0.

Basic Power Rule:

f(x) = cxⁿ
f’(x) = c·nxⁿ⁻¹

Quotient Rule: $\frac{d}{dx} [\frac{f(x)}{g(x)} ]=\frac{g(x)f'(x)-g'(x)f(x)}{g^2(x)}$

Chain Rule: $\frac{d}{dx}[f(g(x))] =f'(g(x)) \cdot g'(x)$

Second Derivative Test:

Possible Points of Inflection (P.P.I) - Tells us the possible x-values where the graph f(x) may change concavity. Occurs when f"(x) = 0 or undefined
Points of Inflection (P.I) - Actual x-values when the graph f(x) changes concavity
Number Line Test - Helps us determine whether a P.P.I is a P.I

Step-by-step explanation:

Step 1: Define

$f(x)=\frac{3}{1+x^2}$

Step 2: Find 2nd Derivative

1st Derivative [Quotient/Chain/Basic]: $f'(x)=\frac{0(1+x^2)-2x \cdot 3}{(1+x^2)^2}$
Simplify 1st Derivative: $f'(x)=\frac{-6x}{(1+x^2)^2}$
2nd Derivative [Quotient/Chain/Basic]: $f"(x)=\frac{-6(1+x^2)^2-2(1+x^2) \cdot 2x \cdot -6x}{((1+x^2)^2)^2}$
Simplify 2nd Derivative: $f"(x)=\frac{6(3x^2-1)}{(1+x^2)^3}$

Step 3: Find P.P.I

Set f"(x) equal to zero: $0=\frac{6(3x^2-1)}{(1+x^2)^3}$

Case 1: f" is 0

Solve Numerator: $0=6(3x^2-1)$
Divide 6: $0=3x^2-1$
Add 1: $1=3x^2$
Divide 3: $\frac{1}{3} =x^2$
Square root: $\pm \sqrt{\frac{1}{3}} =x$
Simplify: $\pm \frac{\sqrt{3}}{3} =x$
Rewrite: $x= \pm \frac{\sqrt{3}}{3}$

Case 2: f" is undefined

Solve Denominator: $0=(1+x^2)^3$
Cube root: $0=1+x^2$
Subtract 1: $-1=x^2$

We don't go into imaginary numbers when dealing with the 2nd Derivative Test, so our P.P.I is $x= \pm \frac{\sqrt{3}}{3}$ (x ≈ ±0.57735).

Step 4: Number Line Test

See Attachment.

We plug in the test points into the 2nd Derivative and see if the P.P.I is a P.I.

x = -1

Substitute: $f"(x)=\frac{6(3(-1)^2-1)}{(1+(-1)^2)^3}$
Exponents: $f"(x)=\frac{6(3(1)-1)}{(1+1)^3}$
Multiply: $f"(x)=\frac{6(3-1)}{(1+1)^3}$
Subtract/Add: $f"(x)=\frac{6(2)}{(2)^3}$
Exponents: $f"(x)=\frac{6(2)}{8}$
Multiply: $f"(x)=\frac{12}{8}$
Simplify: $f"(x)=\frac{3}{2}$

This means that the graph f(x) is concave up before $x=\frac{-\sqrt{3}}{3}$ .

x = 0

Substitute: $f"(x)=\frac{6(3(0)^2-1)}{(1+(0)^2)^3}$
Exponents: $f"(x)=\frac{6(3(0)-1)}{(1+0)^3}$
Multiply: $f"(x)=\frac{6(0-1)}{(1+0)^3}$
Subtract/Add: $f"(x)=\frac{6(-1)}{(1)^3}$
Exponents: $f"(x)=\frac{6(-1)}{1}$
Multiply: $f"(x)=\frac{-6}{1}$
Divide: $f"(x)=-6$

This means that the graph f(x) is concave down between and .

x = 1

Substitute: $f"(x)=\frac{6(3(1)^2-1)}{(1+(1)^2)^3}$
Exponents: $f"(x)=\frac{6(3(1)-1)}{(1+1)^3}$
Multiply: $f"(x)=\frac{6(3-1)}{(1+1)^3}$
Subtract/Add: $f"(x)=\frac{6(2)}{(2)^3}$
Exponents: $f"(x)=\frac{6(2)}{8}$
Multiply: $f"(x)=\frac{12}{8}$
Simplify: $f"(x)=\frac{3}{2}$

This means that the graph f(x) is concave up after $x=\frac{\sqrt{3}}{3}$ .

Step 5: Identify

Since f"(x) changes concavity from positive to negative at $x=\frac{-\sqrt{3}}{3}$ and changes from negative to positive at $x=\frac{\sqrt{3}}{3}$ , then we know that the P.P.I's $x= \pm \frac{\sqrt{3}}{3}$ are actually P.I's.

Let's find what actual point on f(x) when the concavity changes.

$x=\frac{-\sqrt{3}}{3}$

Substitute in P.I into f(x): $f(\frac{-\sqrt{3}}{3} )=\frac{3}{1+(\frac{-\sqrt{3} }{3} )^2}$
Evaluate Exponents: $f(\frac{-\sqrt{3}}{3} )=\frac{3}{1+\frac{1}{3} }$
Add: $f(\frac{-\sqrt{3}}{3} )=\frac{3}{\frac{4}{3} }$
Divide: $f(\frac{-\sqrt{3}}{3} )=\frac{9}{4}$

$x=\frac{\sqrt{3}}{3}$

Substitute in P.I into f(x): $f(\frac{\sqrt{3}}{3} )=\frac{3}{1+(\frac{\sqrt{3} }{3} )^2}$
Evaluate Exponents: $f(\frac{\sqrt{3}}{3} )=\frac{3}{1+\frac{1}{3} }$
Add: $f(\frac{\sqrt{3}}{3} )=\frac{3}{\frac{4}{3} }$
Divide: $f(\frac{\sqrt{3}}{3} )=\frac{9}{4}$

Step 6: Define Intervals

We know that before f(x) reaches $x=\frac{-\sqrt{3}}{3}$ , the graph is concave up. We used the 2nd Derivative Test to confirm this.

We know that after f(x) passes $x=\frac{\sqrt{3}}{3}$ , the graph is concave up. We used the 2nd Derivative Test to confirm this.

Concave Up Interval: $(- \infty,\frac{-\sqrt{3} }{3} )U(\frac{\sqrt{3} }{3} , \infty)$

We know that after f(x) passes $x=\frac{-\sqrt{3}}{3}$ , the graph is concave up until $x=\frac{\sqrt{3}}{3}$ . We used the 2nd Derivative Test to confirm this.

Concave Down Interval: $(\frac{-\sqrt{3} }{3}, \frac{\sqrt{3} }{3} )$

6 0

3 years ago

Identify the factors of the function y = 3x2 - 7x – 10

NikAS [45]

Answer:

(x + 1)(3x - 10)

Step-by-step explanation:

Given

3x² - 7x - 10

Consider the factors of the product of the x² term and the constant which sum to give the coefficient of the x- term.

product = 3 × - 10 = - 30 and sum = - 7

The factors are + 3 and - 10

Use these factors to split the x- term

3x² + 3x - 10x - 10 ( factor the first/second and third/fourth terms )

3x(x + 1) - 10(x + 1) ← factor out (x + 1) from each term

(x + 1)(3x - 10), thus

y = (x + 1)(3x - 10) ← in factored form

8 0

3 years ago