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lapo4ka [179]
3 years ago
8

Question 1 (1 point)

Mathematics
1 answer:
Iteru [2.4K]3 years ago
4 0

Answer:

The answer would be (B) .

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Taxi A charges $0.20 per mile and an initial fee of $4. Taxi B charges $0.40 per mile and an initial fee of $2. Write an inequal
makkiz [27]

The expression for the cost of Taxi B is 0.40<em>x</em> + 2.<em>  </em>The expression for the cost of Taxi A is 0.20<em>x</em> + 4.  The inequality asks for when the cost of Taxi B will be greater than Taxi A, so you write it as 0.40<em>x </em>+ 2 > 0.20<em>x</em> + 4,  or  0.20<em>x</em> + 4 < 0.40<em>x </em>+ 2.

The answer is B) 0.20<em>x</em> + 4 < 0.40<em>x </em>+ 2.

8 0
3 years ago
Read 2 more answers
Combine like terms to create an equivalent expression -1/2(-3y+10)
marysya [2.9K]

Answer:

3/2y - 5

Step-by-step explanation:

-1/2(-3y+10)

Expand the brackets.

-1/2(-3y) -1/2(10)

Multiply.

3/2y - 5

3 0
3 years ago
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Mrs. Kennard has a circular garden with radius of 4.5 ft. she wants to put up a wood fence around half of garden and a steel fen
sveticcg [70]

Answer:108.33, assuming that the wood fence goes around the garden and doesn’t cut across it as well

Step-by-step explanation: 4.5 times 2 times pi to find circumference. Divide by 2 because it is only half of garden. Times by 12 to convert to inches. Divide by 36. Multiply by 23.

6 0
3 years ago
The perimeter of a square is 32cm. what is the lenght of each side​
pshichka [43]

Answer:

<h2>s = 8 cm</h2>

Step-by-step explanation:

For a square, P = 4s, where s is the length of one side.

Thus,            32 cm = 4s, and s = 8 cm

8 0
3 years ago
Use​ l'Hôpital's Rule to find the following limit. ModifyingBelow lim With x right arrow 0StartFraction 3 sine (x )minus 3 x Ove
steposvetlana [31]

Answer:

\lim_{x \to 0} \frac{3sinx-3x}{7x^3}=-\frac{1}{14}

Step-by-step explanation:

The limit is:

\lim_{x \to 0} \frac{3sinx-3x}{7x^3}=\frac{0}{0}

so, you have an indeterminate result. By using the l'Hôpital's rule you have:

\lim_{x \to 0} \frac{a(x)}{b(x)}= \lim_{x \to 0} \frac{a'(x)}{b'(x)}

by replacing, and applying repeatedly you obtain:

\lim_{x \to 0} \frac{3sinx-3x}{7x^3}= \lim_{x \to 0}\frac{3cosx-3}{21x^2}= \lim_{x \to 0}\frac{-3sinx}{42x}= \lim_{x \to 0}\frac{-3cosx}{42}\\\\ \lim_{x \to 0} \frac{3sinx-3x}{7x^3}=\frac{-3cos0}{42}=-\frac{1}{14}

hence, the limit of the function is -1/14

8 0
3 years ago
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