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pav-90 [236]
3 years ago
14

What are the solutions to x2 + x - 5≤0

Mathematics
1 answer:
Juliette [100K]3 years ago
8 0

I have no idea what that x is doing, so imma take it out.

2+x-5≤0

-3+x≤0

+3     +3

x≤3

---

hope it helps

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How can you use substitution to solve a system of linear equations?
makvit [3.9K]
Step 1:
Solve one of the equations for either x = or y = .
Step 2:
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Step 3:
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Step 4:
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Example 1: Solve the following system by substitution
Substitution Method Example
Solution:
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solution step 1
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solution step 2
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The solution is: (x, y) = (10, -5)
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3 years ago
Which letter has at least one line of symmetry? <br> a.D<br> b.J<br> c.G<br> d.L
iren2701 [21]
The correct answer is A. D

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8 0
3 years ago
HELPPPP i’ll give brainliest if it’s right!!!
STALIN [3.7K]

Answer:

the answer is A and D

Step-by-step explanation:

5 0
2 years ago
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The number of fiction books in a libary is %150 of the number of non-fiction books. the libary has 4,000 books in total. how man
Nitella [24]

Answer:

The number of

Fiction books = x = 2400 fiction books

Non fiction books = y = non fiction books

Step-by-step explanation:

Let the number of

Fiction books = x

Non fiction books = y

The libary has 4,000 books in total.

Hence: x + y = 4000........ Equation 1

The number of fiction books in a libary is %150 of the number of non-fiction books.

x = 150% of y

x = 150/100 × y

x = 1.5y

We substitute 1.5y for x in Equation 1

x + y = 4000

1.5y + y = 4000

2.5y = 4000

y = 4000/2.5

y = 1600 non fiction books

Solving for x

x = 1.5y

x = 1.5 × 1600

x = 2400 fiction books.

Therefore,

The number of

Fiction books = x = 2400 fiction books

Non fiction books = y = non fiction books

5 0
2 years ago
Math help questions <br>​
Evgen [1.6K]

After 1 year, the initial investment increases by 7%, i.e. multiplied by 1.07. So after 1 year the investment has a value of $800 × 1.07 = $856.

After another year, that amount increases again by 7% to $856 × 1.07 = $915.92.

And so on. After t years, the investment would have a value of \$800 \times 1.07^t.

We want the find the number of years n such that

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Solve for n :

856 \times 1.07 ^n = 1400

1.07^n = \dfrac74

\log_{1.07}\left(1.07^n\right) = \log_{1.07}\left(\dfrac74\right)

n \log_{1.07}(1.07) = \log_{1.07} \left(\dfrac74\right)

n = \log_{1.07} \left(\dfrac74\right) = \dfrac{\ln\left(\frac74\right)}{\ln(1.04)} \approx \boxed{8.3}

4 0
1 year ago
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