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Art [367]
2 years ago
8

The distance between two towns on a map varies directly with the actual distance between the towns.

Mathematics
1 answer:
Hitman42 [59]2 years ago
6 0

Answer:  220 miles

Step-by-step explanation:

(150 miles)/(7.5 inches) = 20 miles/inch

(20 miles/inch)*(11 inches) = 220 miles

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Write the inequality!! Connie’s dog Fido weighs 35 pounds. Her vet placed Fido on a diet. What inequality can you write to find
Butoxors [25]

Answer:   \frac{35-28}{x} \leq 6

Step-by-step explanation:

Let x be the average number of pounds Fido must loss.

Since, the initial weight of Fido is 35 pounds.

And, After losing the weight, the new weight of Fido in pounds = 28 pounds.

Then the time taken for losing the weight

= \frac{\text{ The weight it losses}}{\text{ Average of losing weight per month}}

= \frac{35-28}{x}

According to the question,  it must lose weight within 6 months,

Thus,   \frac{35-28}{x}\leq 6

Which is the required inequality to find the average number of pounds per month.

By solving it we, get,    x\geq \frac{7}{6}

6 0
2 years ago
What equation can represent 3 increased by a numbwr x is 9
Ksivusya [100]

Answer:

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Step-by-step explanation:

4 0
3 years ago
2 5/6 - 11/4=?ggffcgyujjh
ElenaW [278]
1/12 I think I’m sorry if it isn’t
6 0
3 years ago
If f(x)=x^3-x+2, then (f^-1)'(2)
yawa3891 [41]

Note that f(x) as given is <em>not</em> invertible. By definition of inverse function,

f\left(f^{-1}(x)\right) = x

\implies f^{-1}(x)^3 - f^{-1}(x) + 2 = x

which is a cubic polynomial in f^{-1}(x) with three distinct roots, so we could have three possible inverses, each valid over a subset of the domain of f(x).

Choose one of these inverses by restricting the domain of f(x) accordingly. Since a polynomial is monotonic between its extrema, we can determine where f(x) has its critical/turning points, then split the real line at these points.

f'(x) = 3x² - 1 = 0   ⇒   x = ±1/√3

So, we have three subsets over which f(x) can be considered invertible.

• (-∞, -1/√3)

• (-1/√3, 1/√3)

• (1/√3, ∞)

By the inverse function theorem,

\left(f^{-1}\right)'(b) = \dfrac1{f'(a)}

where f(a) = b.

Solve f(x) = 2 for x :

x³ - x + 2 = 2

x³ - x = 0

x (x² - 1) = 0

x (x - 1) (x + 1) = 0

x = 0   or   x = 1   or   x = -1

Then \left(f^{-1}\right)'(2) can be one of

• 1/f'(-1) = 1/2, if we restrict to (-∞, -1/√3);

• 1/f'(0) = -1, if we restrict to (-1/√3, 1/√3); or

• 1/f'(1) = 1/2, if we restrict to (1/√3, ∞)

6 0
2 years ago
Algebra 1. Please show work.
MAVERICK [17]
Try this:
Given:
\left \{ {{x-y=1} \ [1] \atop {x+y=3} \ [2]} \right.
1) [1]+[2]: 2x=2 ⇒ x=2
2) [2]-[1]: 2y=1 ⇒ y=1.
3) answer: (2;1)
7 0
2 years ago
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