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sdas [7]
3 years ago
7

Create an example of a linear system that would best be solved using the substitution method. Explain your reasoning.

Mathematics
1 answer:
Readme [11.4K]3 years ago
5 0
In your change cup you have 28 coins.  All of which are nickels and quarters.  Together they add up to $4.  How many coins do you have of each. 
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Solve the following exponential equations.<br> 10^(3x−5) = 7^x
zalisa [80]

Answer:

x = 2.3202

Step-by-step explanation:

Given equation:

10^{(3x-5)} = 7^x

on taking log both sides, we get

\log(10^{(3x-5))} =\log(7^x)

now,

using the property of log function

log(aᵇ) = b × log(a)

therefore,

we get

(3x-5)log(10) = xlog(7)

now,

log(10) = 1

and

log(7) = 0.84509

thus,

( 3x - 5 ) × 1 = 0.84509x

or

3x - 0.84509x - 5 = 0

or

2.15491x = 5

or

x = 2.3202

4 0
3 years ago
What is the value of x in the equation -6x = 5x+ 22?<br> -22<br> 0 -2<br> 02<br> O 22
Ivahew [28]

Answer:

x = -2

Step-by-step explanation:

Solve  -6x = 5x + 22

add 6x to both sides

0 = 6x + 5x + 22

0 = 11x + 22

11x = -22

x = -22/11

x = -2

3 0
3 years ago
Instruction
natulia [17]

Answer:

Step-by-step explanation:

Your question has typographical errors. The equation is 2a + b = 15.7, not 2e+b-15.7

2a + b = 15.7

b = 15.7  - 2a

since a > b, a = 6.3 cm

b = 15.7 - 2·6.3 = 3.1 cm

3 0
3 years ago
Round 10,031.464 to the nearest hundred.
insens350 [35]

Answer:

10,031.46

Rounded to the nearest 0.01 or

the Hundredths Place.

Step-by-step explanation:

7 0
3 years ago
A vehicle factory manufactures cars. The unit cost (the cost in dollars to make each car) depends on the number of cars made. If
Nataly [62]

Answer:

The function is missing in the question. The function is $ C(x) = 0.8x^2 -544x +97410 $

The answer is 4930

Step-by-step explanation:

Unit cost of a car is given as

$ C(x) = 0.8x^2 -544x +97410 $

Cost will be minimum when

x = -(-544)/ 2 x 0.8

  = 340

Therefore, minimum cost for unit car is

$ C(x) = 0.8x^2 -544x +97410 $

$ = 0.8(340)^2 -544(340) +97410 $

=  4930

4 0
3 years ago
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