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uysha [10]
3 years ago
11

There is one real solution if the radicand is

Mathematics
2 answers:
koban [17]3 years ago
7 0

If the radicand is even is the answer. Hope this helped could I possibly get brainliest?

suter [353]3 years ago
6 0

Answer:

1st one is positive (There are two real solutions if the radicand is___) 2nd one is zero(There is one real solution if the radicand is____) and 3rd one is negative (There are no real solutions if the radicand is____)

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Solve the system <br> thank you
Lunna [17]

Answer:

infinite solutions

Step-by-step explanation:

replace x with 3y + 9 in the second equation

9 - (3y + 9) = -3y ➡ 9 -3y -9 = -3y ➡9-9 = 3y-3y ➡ 0 = 0

this means whatever numver you write instead of y there will be a solution.

6 0
3 years ago
En una pasteleria realizan 15 posteles cada 30 min
artcher [175]

Answer:

60

Step-by-step explanation:

había multiplicado 4 y 15 espero que esto ayude

4 0
3 years ago
Simplify the following polynomial
Amiraneli [1.4K]

Answer:

-3x^4 - 13x^3 + 14x - 7

Step-by-step explanation:

(5x^4 – 9x^3 + 7x – 1) + (-8x^4 + 4x^2 – 3x + 2) - (-4x^3 + 5x - 1)(2x – 7)

simplify multiplied terms

(-4x^3 + 5x - 1)(2x – 7)

(-4x^3+10x-8)

group like terms together

(5x^4-8x^4) + (-9x^3-4x^3) + (7x-3x+10x) + (-1+2-8)

simplify grouped terms

-3x^4 - 13x^3 + 14x - 7

3 0
3 years ago
Guys help me pls I need it immediately
julsineya [31]

Answer:

Hey there!

First, we want to find the side length of the large triangle.

1/2bh=y cm^2

bh=2y cm^2

Since b and h are the same here, we have b^2= 2y, b=\sqrt{2y}.

And we have FC=\frac{\sqrt{2y} }{2}.

Area of shaded triangle: \frac{1}{2} (\frac{\sqrt{2y} }{2})(\frac{\sqrt{2y} }{2}).

This is equal to: \frac{1}{2}(\frac{2y}{4}), or y/4.

Let me know if this helps :)

6 0
3 years ago
Find the limit of the sequence of partial sums whose general term is <img src="https://tex.z-dn.net/?f=a_n%3D%5Cfrac%7B100%5En%7
nasty-shy [4]

Answer:

0

Step-by-step explanation:

If ∑aₙ converges, then lim(n→∞)aₙ = 0.

Using ratio test, we can determine if the series converges:

If lim(n→∞) |aₙ₊₁ / aₙ| < 1, then ∑aₙ converges.

If lim(n→∞) |aₙ₊₁ / aₙ| > 1, then ∑aₙ diverges.

lim(n→∞) |(100ⁿ⁺¹ / (n+1)!) / (100ⁿ / n!)|

lim(n→∞) |(100ⁿ⁺¹ / (n+1)!) × (n! / 100ⁿ)|

lim(n→∞) |(100 / (n+1)|

0 < 1

The series converges.  Therefore, lim(n→∞)aₙ = 0.

7 0
3 years ago
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