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Debora [2.8K]
3 years ago
14

More than one unknown in a word problem can be represented in terms of the same variable assigned to first unknown. true of fals

e
Mathematics
2 answers:
Sergeeva-Olga [200]3 years ago
6 0
This is clearly false as it will just mess up the question.
Hitman42 [59]3 years ago
5 0

Answer:

True

Step-by-step explanation:

When you have a word problem with more than one unknown data, you can represent these unknowns in terms of the same variable assigned to the first unknown (it depends on the problem).

For example, if the word problem tells us that  Alex has 3 more years than Jennifer and that Ben has twice the age of Jennifer plus 4, you can represent all these unknowns (the three ages) in terms of the variable assigned to Jennifer's age. Let's see how this would be:

If you name x the age of Jennifer, then Alex's age would be x + 3 (3 more years than Jennifer), and Ben's age would be 2x + 4 (twice the age of Jennifer plus 4).

You can do this in some word problems to simplify the process of finding your unknowns because you'd be working with just one variable instead of many of them.

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Robbie has 2 times as many apples as Bob. After receiving 35 more, Robbie now has 7 times as many as Bob. How many apples does B
Hoochie [10]

Answer:  18.5 apples

Step-by-step explanation:

7 0
2 years ago
What are benchmarks when using rounding?
cricket20 [7]

Answer:

Benchmark can be defined as the standard or reference point against which something can be measured, compared, or assessed.

Step-by-step explanation:

7 0
3 years ago
By rounding to 1 significant figure, estimate the value of 48.7 x 61.2<br> 11.3
11111nata11111 [884]

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7 0
3 years ago
X-2y=18 2x+y=6 parallel perpendicular or neither
sergey [27]
<h3>Therefore they are perpendicular.</h3>

Step-by-step explanation:

A equation of line is

y =mx +c

Here the slope of the line is m.

Given equations are

x - 2y = 18

⇔-2y = -x +18

\Leftrightarrow y =\frac{1}{2} x -9............(1)

and 2x + y = 6

⇔y = -2x +6 ............(2)

Therefore the slope of equation (1) is(m_1)= \frac{1}{2}

Therefore the slope of equation (2) is(m_2)= -2

If two lines are perpendicular, when we multiply their slope we get -1.

therefore,

m_1. m_2 =\frac{1}{2}. (-2) = -1

Therefore they are perpendicular.

4 0
3 years ago
(10.02)
ollegr [7]
<h2>Hello!</h2>

The answer is:

C. Cosine is negative in Quadrant III

<h2>Why?</h2>

Let's discard each given option in order to find the correct:

A. Tangent is negative in Quadrant I: It's false, all functions are positive in Quadrant I (0° to 90°).

B. Sine is negative in Quadrant II: It's false, sine is negative in positive in Quadrant II. Sine function is always positive coming from 90° to 180°.

C. Cosine is negative in Quadrant III. It's true, cosine and sine functions are negative in Quadrant III (180° to 270°), meaning that only tangent and cotangent functions will be positive in Quadrant III.

D. Sine is positive in Quadrant IV: It's false, sine is negative in Quadrant IV. Only cosine and secant functions are positive in Quadrant IV (270° to 360°)

Have a nice day!

6 0
3 years ago
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