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ollegr [7]
2 years ago
14

Match this system of linear equations to the correct description of its solution set.

Mathematics
1 answer:
Misha Larkins [42]2 years ago
8 0

Answer:

Infinite number of solutions.

Step-by-step explanation:

Are the equations written correctly?  They are identical.  Since they overlap, every point is a solution for as long as the lines reach.  Since no one is stopping them, the answers are infinite.

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Find the least common denominator for the group of fractions.<br> 5/12, 9/14, 2/5, 1/10
makkiz [27]

9514 1404 393

Answer:

  420

Step-by-step explanation:

The prime factors of each denominator are ...

  12 = 2²×3

  14 = 2×7

  5 = 5

  10 = 2×5

The unique factors are 2²×3×5×7 = 420.

The least common denominator of these fractions is 420.

_____

<em>Additional comment</em>

  • 5/12 = 175/420
  • 9/14 = 270/420
  • 2/5 = 168/420
  • 1/10 = 42/420
5 0
2 years ago
Stop saying free trial, when kids need answers for help, don’t just say oh to get the answer sign up for a trial, that’s not how
tino4ka555 [31]
Dudeeeeee literally tho
3 0
3 years ago
Read 2 more answers
Eva has $10 for lunch. Her meal cost $8.60. She wants to leave a 20% tip. Does she have enough money? Explain your reasoning.
snow_tiger [21]
No she doesn't have enough money. All you do is multiply 8.60 by .2 which come out to 1.72 and 8.60 plus 1.73 which equals 10.33 so she needs 33 more cents tone able to tip 20%.
6 0
3 years ago
The hypotenuse of a 90-45-45 triangle is 20√2. What is the short leg and the long leg&gt;
stepan [7]

Answer:

Both legs would be 20

Step-by-step explanation:
In the image below, it shows the relationship between the legs and the hypotenuse in every 45 45 90 triangle. If we know that the hypotenuse is 20, then the 2 legs are both 20.

I am slightly confused with the problem since they asked for the "short" leg and the "long" leg. Both legs should be the same length because a 45 45 90 triangle is isosceles, with the two legs being the same length.

6 0
2 years ago
Exercise 3.5. For each of the following functions determine the inverse image of T = {x ∈ R : 0 ≤ [x^2 − 25}.
masya89 [10]

a. The inverse image of f(x) is f⁻¹(x) = ∛(x/3)

b. The inverse image of g(x) is g^{-1}(x) = e^{x}

c. The inverse image of <u>h</u>(x) is h⁻¹(x) = x + 9

<h3 /><h3>The domain of T</h3>

Since T = {x ∈ R : 0 ≤ [x^2 − 25} ⇒ x² - 25 ≥ 0

⇒ x² ≥ 25

⇒ x ≥ ±5

⇒ -5 ≤ x ≤ 5.

<h3>Inverse image of f(x)</h3>

The inverse image of f(x) is f⁻¹(x) = ∛(x/3)

f : R → R defined by f(x) = 3x³

Let f(x) = y.

So, y = 3x³

Dividing through by 3, we have

y/3 = x³

Taking cube root of both sides, we have

x = ∛(y/3)

Replacing y with x we have

y = ∛(x/3)

Replacing y with f⁻¹(x), we have

So,  the inverse image of f(x) is f⁻¹(x) = ∛(x/3)

<h3>Inverse image of g(x)</h3>

The inverse image of g(x) is g^{-1}(x) = e^{x}

g : R+ → R defined by g(x) = ln(x).

Let g(x) = y

y =  ln(x)

Taking exponents of both sides, we have

e^{y} = e^{lnx} \\e^{y} = x

Replacing x with y, we have

y = e^{x}

Replacing y with g⁻¹(x), we have

So, the inverse image of g(x) is g^{-1}(x) = e^{x}

<h3>Inverse image of h(x)</h3>

The inverse image of <u>h</u>(x) is h⁻¹(x) = x + 9

h : R → R defined by h(x) = x − 9

Let y = h(x)

y = x - 9

Adding 9 to both sides, we have

y + 9 = x

Replacing x with y, we have

x + 9 = y

Replacing y with h⁻¹(x), we have

So, the inverse image of <u>h</u>(x) is h⁻¹(x) = x + 9

Learn more about inverse image of a function here:

brainly.com/question/9028678

5 0
2 years ago
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