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maks197457 [2]
3 years ago
11

A road is 4 cm long on the map. Find the actual length of the road.

Mathematics
1 answer:
Usimov [2.4K]3 years ago
7 0

The distance between two cities is

75

k

m

s

.

Explanation:

As a road of

4

c

m

.

on road map indicates a road of length equal to

30

k

m

.

Each

c

m

.

on map indicates

30

4

=

7.5

k

m

s

.

Hence a distance of

10

c

m

.

between two cities on the map indicates

7.5

×

10

=

75

k

m

s

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Morgan made a mistake when subtracting the rational expressions below 3t^2-4t+1/t+3-t^2+2t+2/t+3=2t^2-2t+3/t+3. What was Morgan'
marusya05 [52]
<h3>Answer: Choice D. </h3>

Morgan forgot to distribute the negative sign to two of the terms in the second expression.

=============================================================

Explanation:

Focus on the numerators.

We have (3t^2-4t+1) as the first numerator and we subtract off (t^2+2t+2) as the second numerator.

Morgan needs to simplify (3t^2-4t+1)-(t^2+2t+2) for the numerator.

Mistakenly, she had these steps

(3t^2-4t+1)-(t^2+2t+2)

3t^2-4t+1-t^2+2t+2 .... her mistake made here

(3t^2-t^2)+(-4t+2t)+(1+2)

2t^2-2t+3

All of this applies to the numerator. The denominator stays at t+3 the entire time. So effectively we can ignore it on a temporary basis.

Here's what Morgan should have for her steps when simplifying the numerator.

(3t^2-4t+1)-(t^2+2t+2)

3t^2-4t+1-t^2-2t-2 ..... distribute the negative

(3t^2-t^2)+(-4t-2t)+(1-2)

2t^2-6t-1

Note in the second step, the negative outside flips the sign of each term in the second parenthesis.

Therefore,

\frac{3t^2-4t+1}{t+3}-\frac{t^2+2t+2}{t+3}\\\\\frac{(3t^2-4t+1)-(t^2+2t+2)}{t+3}\\\\\frac{3t^2-4t+1-t^2-2t-2}{t+3}\\\\\frac{2t^2-6t-1}{t+3}\\\\

which means \frac{3t^2-4t+1}{t+3}-\frac{t^2+2t+2}{t+3}=\frac{2t^2-6t-1}{t+3}, \ \ \text{ where } t \ne -3\\\\

Side notes:

  • The fractions can only be subtracted since the denominators are the same.
  • We have t \ne -3 to avoid a division by zero error.
  • Rational expressions are a fraction, or ratio, of two polynomials.
5 0
2 years ago
Please help me with this problem<br><br> please show steps so i know for next time
vladimir2022 [97]
<h3>Answer:  2/3, -4/3</h3>

=========================================================

Explanation:

You could use the AC method to factor this, but the quadratic formula is the most efficient route in my opinion. This will avoid any guess-and-check.

Compare the original equation to the form a\text{x}^2 + b\text{x} + c  = 0

We have a = 9, b = 6, and c = -8

Those values lead to...

\text{x} = \frac{-b\pm\sqrt{b^2-4ac}}{2a}\\\\\text{x} = \frac{-6\pm\sqrt{(6)^2-4(9)(-8)}}{2(9)}\\\\\text{x} = \frac{-6\pm\sqrt{324}}{18}\\\\\text{x} = \frac{-6\pm18}{18}\\\\\text{x} = \frac{-6+18}{18} \ \text{ or } \ \text{x} = \frac{-6-18}{18}\\\\\text{x} = \frac{12}{18} \ \text{ or } \ \text{x} = \frac{-24}{18}\\\\\boldsymbol{\text{x} = \frac{2}{3} \ \text{ or } \ \text{x} = -\frac{4}{3}}\\\\

Side notes:

  • 2/3 = 0.667 approximately
  • -4/3 = 1.333 approximately
  • Since your teacher did not give rounding instructions, I'll assume s/he wants the fraction form of each x value (rather than the decimal form). Be sure to follow all instructions given, and ask for clarification if need.
  • To confirm the solutions, replace every copy of x with either 2/3 or -4/3 (pick one value only). Simplifying the left hand side should lead to 0. I'll let you check each answer.
5 0
2 years ago
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