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Korolek [52]
3 years ago
8

How do I solve the system of equations x^2 +y=0 and x^2+y=0? I have to use substitution

Mathematics
1 answer:
Wittaler [7]3 years ago
5 0
Equations
• x² + y = 0 ... (1)
• x² - 4x - y = 0 ... (2)

Answer -> (x, y) = (2, -4)

Step-by-step solution:

From equation 2,
-> y = x² - 4x ... (3)

• Substitute y into equation 1

-> x² + (x² - 4x) = 0
-> 2x² - 4x = 0
-> 2x² = 4x
-> x²/x = 4/2
-> x = 2

• Put x = 2 in equation 3

-> y = (2)² - 4(2)
-> y = 4 - 8
-> y = -4
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mariarad [96]

Answer:

D) 1

Step-by-step explanation:

When you start raising i to certain powers, you begin to notice a pattern.

i^1=i \\\\i^2=-1 \\\\i^3=-i \\\\i^4=1 \\\\i^5=i \\\\ i^6=-1 \\\\ i^7=-i \\\\i^8= 1

This cycle repeats forever. Since 84 is a multiple of 4, i^84 must be 1. Hope this helps!

7 0
4 years ago
NEED HELP YOU GUYS!!!!
AURORKA [14]

Answer:

x1=

\frac{ - 2 +  \sqrt{100} }{ - 48}

x2=

\frac{ - 2 +  \sqrt{100} }{ - 48}

5 0
3 years ago
You owe $1,945.61 on a credit card that has an 11.2% APR. The minimum payment due is $156.00. You decide to pay $300.00. How muc
djyliett [7]
We are given
P = <span>$1,945.61
r = 11.2%
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After one month, the interest saved by paying more than the minimum is
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8 0
3 years ago
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A rectangular solid with a square base has a volume of 4096 cubic inches - determine the dimensions that yield the minimum surfa
Bumek [7]

Answer:

side of base, a = 10.1 inches, height, h = 40.1 inches

Step-by-step explanation:

Volume of rectangular solid, V = 4096 cubic inches

Let the side of base is a and the height is h.

V = a^2h\\\\4096 = a^2 h ..... (1)

surface area of the solid  

S = 2a^2 + 4 ah \\\\S = 2a^2 + 4 \times a \times \frac{4096}{a^2}    from (1)\\\S = 2a^2 + \frac{4096}{a}\\\\\frac{dS}{da}= 4 a - \frac{4096}{a^2}\\\\4 a - \frac{4096}{a^2} = 0 \\\\a^3 = 1024\\\\a =10.1 inches

So, h = 40.2 inches

5 0
3 years ago
The proof ABC ≅ DCB that is shown.
Schach [20]

The <em><u>correct answer</u></em> is:

AAS

Explanation:

AAS stands for "angle-angle-side."  This states that if two angles and a non-included side of one triangle are congruent to the corresponding two angles and non-included side of another triangle, then the triangles are congruent.

In these triangles, we have ∠CAB ≅ ∠CDB given to us to begin with.  Throughout the proof, we find that ∠ABC ≅ ∠DCB.  We also have that CB is congruent to itself.  This is two angles and a side not included, so this is AAS.

4 0
3 years ago
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