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4 years ago

the number of words x tina can type per minute is at least 50. write and graph an inequality to describe this situation

Mathematics

2 answers:

IgorLugansk [536]4 years ago

5 0

Answer:-x≥50

Explanation:-

In algebra, an inequality is a relation that stack between two values when they are different. Notations for inequality are ≠,<,>,≤ and≥.

Given: The number of words Tine type per minute be x.

We know that the notation for 'at least' is ≥, thus

Inequality for the given situation is x≥50,which means she can write more than or equal to 50 words per minute.

In the graph we can see shaded part which represents number of words type more than or equal to 50.

marshall27 [118]4 years ago

3 0

I can only write a inequality here but here goes...
since she can atleast type 50 per minute that means the equation would be x >50 (you would also put the line under the > because it can equal it) (x is the amount she can type)

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The ideal width of a safety belt strap for a certain automobile is 6 cm. The actual width can vary by at most 0.45 cm. Write an

OlgaM077 [116]

Answer:

$|x-6|\le 0.45$

$x\in [5.55,6.45]$

Step-by-step explanation:

Absolute Value Inequality

Assume the actual width of a safety belt strap for a certain automobile is x. We know the ideal width of the strap is 6 cm. This means the variation from the ideal width is x-6.

Note if x is less than 6, then the variation is negative. We usually don't care about the sign of the variation, just the number. That is why we need to use the absolute value function.

The variation (unsigned) from the ideal width is:

$|x-6|$

The question requires that the variation is at most 0.45 cm. That poses the inequality:

$|x-6|\le 0.45$

That is the range of acceptable widths. Let's now solve the inequality.

To solve an inequality for an absolute value less than a positive number N, we write:

$-0.45\le x-6 \le 0.45$

This is a double inequality than can be easily solved by adding 6 to all the sides.

$-0.45+6\le x \le 0.45+6$

Operating:

$5.55\le x \le 6.45$

That is the solution in inequality form. Expressing in interval form:

$\boxed{x\in [5.55,6.45]}$

3 0

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