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antiseptic1488 [7]
2 years ago
10

Solve the equation x² =6

Mathematics
1 answer:
Klio2033 [76]2 years ago
3 0

I tried doing some equation hope it helps you a little

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Wendy has $27 to buy seed for her birdfeeders. Each bag of seed costs $5.
Alekssandra [29.7K]

Answer:

5

Step-by-step explanation:

If Wendy has $27 to buy seeds, and each seed bag costs $5, then:

5 · 5 = 25

5 · 6 = 30

27<30

27>25

So, Wendy is able to buy 5 bags and remain with 2 dollars.

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2 years ago
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ILL MARK BRAINIEST IF U DO THIS RIGHT!!!
barxatty [35]

Answer:

D because even though the flat fee is 150 paying 5$ a hour it will cost less

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3 years ago
Can you guyz help me with this question step by step please ☺️
zlopas [31]
2 1/2 x 3 = 7 1/2 = 7 4/8

10 11/8
- 7 4/8
3 7/8
A is the correct answer

7 0
2 years ago
The model represents x2 – 9x + 14. An algebra tile configuration showing only the Product spot. 24 tiles are in the Product spot
klio [65]

I didn't get all the part with the tiles, but here's the general answer:

given a polynomial

p(x)=ax^2+bx+c

we have that x-k is a factor of p(x) if and only if k is a root of p(x), i.e. if

p(k)=ak^2+bk+c=0

So, given the polynomial

p(x)=x^2-9x+14

We can check if x-9 is a factor by evaluating p(9):

p(9)=81-81+14=14\neq 0

So, x-9 is not a factor.

Similarly, we can evaluate p(2),\ p(-5),\ p(-7) to check if x-2,\ x+5,\ x+7 are factors:

p(2)=4-18+14=0,\quad p(-5)=25+45+14=84\neq 0,\quad p(-7)=49+63+14=126 \neq 0

So, only x-2 is a factor of x^2-9x+14

4 0
3 years ago
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Divide.<br> a.1 − 2i/2i<br> b. 5 − 2i/5 + 2i<br> c.√3 − 2i/−2 − √3i
Umnica [9.8K]

Answer:

(a) \frac{-1i}{2}-1[/tex]

(b) \frac{20i+21}{29}

(c) i

Step-by-step explanation:

We have to perform division

(a) \frac{1-2i}{2i}

So after division

\frac{1-2i}{2i}=\frac{1}{2i}-\frac{2i}{2i}=\frac{-1i}{2}-1

(b) We have given expression \frac{5-2i}{5+2i}

After rationalizing \frac{5-2i}{5+2i}\times \frac{5-2i}{5-2i}=\frac{(5-2i)^2}{25+4}=\frac{25+4i^2+20i}{29}=\frac{20i+21}{29}

(c) We have given expression \frac{\sqrt{3}-2i}{-2-\sqrt{3i}}

After rationalizing

\frac{\sqrt{3}-2i}{-2-\sqrt{3i}}\times \frac{-2+\sqrt{3i}}{-2+\sqrt{3i}}=\frac{-2\sqrt{3}+7i+2\sqrt{3}}{7}=\frac{7i}{7}=i

5 0
3 years ago
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