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3 years ago

an 8 in. by 10 in. photo is reduced so that the width (the shorter dimension) is 3 in. what is the length of the reduced photo

Mathematics

1 answer:

Sindrei [870]3 years ago

6 0

Answer:

Step-by-step explanation:

The length can be found by setting up a proportion. If the width is reduced by some ratio, then the length will share the same ratio.

3 / 8 = x / 10 Multiply both sides by 10.

3 * 10 / 8 = x Multiply first

30/8 = x Divide next. You don't need to do it in this order.

3 6/8 Simplify

3 3/4 Give another choice

3.75

The length = 3.75 inches

If none of these are correct, please leave a note with your choices.

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The indefinite integral can be found in more than one way. First use the substitution method to find the indefinite integral. Th

Fantom [35]

Answer:

∫ $6x^5(x^6-2)\,dx$ = $\frac{1}{2}(x^6-2)^2+C$

Step-by-step explanation:

To find:

∫ $6x^5(x^6-2)\,dx$

Solution:

Method of substitution:

Let $x^6-2=t$

Differentiate both sides with respect to $t$

$6x^5\,dx=dt$

[use $(x^n)'=nx^{n-1}$ ]

So,

∫ $6x^5(x^6-2)\,dx$ = ∫ $t\,dt$ = $\frac{t^2}{2}+C_1$ where $C_1$ is a variable.

(Use ∫ $t^n\,dt=\frac{t^{n+1} }{n+1}$ )

Put $t=x^6-2$

∫ $6x^5(x^6-2)\,dx$ = $\frac{1}{2}(x^6-2)^2+C_1$

Use $(a-b)^2=a^2+b^2-2ab$

So,

∫ $6x^5(x^6-2)\,dx$ = $\frac{1}{2}(x^6-2)^2+C_1=\frac{1}{2}(x^{12}+4-4x^6)+C_1=\frac{x^{12} }{2}-2x^6+2+C_1=\frac{x^{12} }{2}-2x^6+C$

where $C=2+C_1$

Without using substitution:

∫ $6x^5(x^6-2)\,dx$ = ∫ $6x^{11}-12x^5\,dx$ = $\frac{6x^{12} }{12}-\frac{12x^6}{6}+C=\frac{x^{12} }{2}-2x^6+C$

So, same answer is obtained in both the cases.

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3 years ago

The graphs of two polynomial equations do not intersect. Kelly concludes that the system has no solution. Which statement best e

ololo11 [35]

Answer:

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Step-by-step explanation:

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3 years ago

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To two decimal places, find the value of k that will make the function f(x) continuous everywhere. f of x equals the quantity 3x

Fiesta28 [93]

Given function is

$f(x)=\left\{\begin{matrix}3x+k & x\leq -4 \\ kx^2-5 & x> -4\end{matrix}\right.$

now we need to find the value of k such that function f(x) continuous everywhere.

We know that any function f(x) is continuous at point x=a if left hand limit and right hand limits at the point x=a are equal.

So we just need to find both left and right hand limits then set equal to each other to find the value of k

To find the left hand limit (LHD) we plug x=-4 into 3x+k

so LHD= 3(-4)+k

To find the Right hand limit (RHD) we plug x=-4 into

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so RHD= $k(-4)^2-5$

Now set both equal

$k(-4)^2-5=3(-4)+k$

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$16k-k=-12+5$

$15k=-7$

$k=-\frac{7}{15}$

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<u>Hence final answer is -0.47.</u>

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Rzqust [24]

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