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bulgar [2K]
3 years ago
12

Hello please help i’ll give brainliest

Mathematics
1 answer:
Liula [17]3 years ago
7 0

Answer:

option 4

Explaination;

in order to sustain themselves

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Evaluate the discriminant of each equation. Tell how many solutions each equation has and whether the solutions are real or imag
Elena L [17]
All these equations are in the form of ax^2 + bx + c = 0, where a, b, and c are some numbers. the discriminants of equations like this are equal to b^2 - 4ac. if the discriminant is negative, there are two imaginary solutions. if the discriminant is positive, there are two real solutions. if the discriminant is 0, there is one real solution.

<span>x^2 + 4x + 5 = 0
</span>b^2 - 4ac
4^2 - 4(1)(5)
16-20
-4, two imaginary solutions.

<span>x^2 - 4x - 5 = 0
</span>b^2 - 4ac
(-4)^2 - 4(1)(-5)
16 + 20
36, two real solutions.

<span>4x^2 + 20x + 25 = 0
</span>b^2 - 4ac
20^2 - 4(4)(25)
400 - 400
0, one real solution.
6 0
3 years ago
Help needed for assignment
bearhunter [10]
The answer is C........
3 0
3 years ago
The expression -7y is a<br><br> a. term<br> b. variable<br> c. constant
Jet001 [13]

Answer:

the answer is a) term

Step-by-step explanation:

hope this helps! :)

brainliest?

8 0
3 years ago
Read 2 more answers
in 2009 a total of R36 000 was invested in two accounts. One account earned 7% annual interest and the other earned 9% .The tota
Akimi4 [234]

a = amount invested at 7%

b = amount invested at 9%

we know the amount invested was ₹36000, thus we know that whatever "a" and "b" are, a + b = 36000.  We can also say that

\begin{array}{|c|ll} \cline{1-1} \textit{a\% of b}\\ \cline{1-1} \\ \left( \cfrac{a}{100} \right)\cdot b \\\\ \cline{1-1} \end{array}~\hspace{5em}\stackrel{\textit{7\% of a}}{\left( \cfrac{7}{100} \right)a}\implies 0.07a~\hfill \stackrel{\textit{9\% of b}}{\left( \cfrac{7}{100} \right)b}\implies 0.09b

since we know the interest earned from the invested was ₹2920, then we say that 0.07a + 0.09b = 2920.

\begin{cases} a + b = 36000\\\\ 0.07a+0.09b=2920 \end{cases} \\\\[-0.35em] ~\dotfill\\\\ \stackrel{\textit{using the 1st equation}}{a + b = 36000\implies \underline{b = 36000-a}}~\hfill \stackrel{\textit{substituting on the 2nd equation}}{0.07a~~ + ~~0.09(\underline{36000-a})~~ = ~~2920} \\\\\\ 0.07a+3240-0.09a=2920\implies 3240-0.02a=2920\implies -0.02a=-320 \\\\\\ a=\cfrac{-320}{-0.02}\implies \boxed{a=16000}~\hfill \boxed{\stackrel{36000~~ - ~~16000}{20000=b}}

5 0
3 years ago
Help me, please!! I wasn't there for the lesson.
liberstina [14]
The first option would be the right answer. Martha has 33, where jackson has 22
6 0
4 years ago
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