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rjkz [21]
4 years ago
13

What is the first step in solving a quadratic equation with the form given below (ax+b)^2=c

Mathematics
2 answers:
kondaur [170]4 years ago
8 0

Answer:

rearrange all of the terms

Step-by-step explanation:


Colt1911 [192]4 years ago
8 0

The first step is to arrange the TERMS correcty


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Does anyone know the missing leg of the length?
n200080 [17]

Answer:

Hello! answer: 16

Step-by-step explanation:

These are a bit different instead your solving for B so you have to work backwords so... 20 × 20 = 400 12 × 12 = 144 400 - 144 = 256 now we just find out what multiplied by itself = 256 16 × 16 = 256 therefore b = 16 Hope that helps!

5 0
3 years ago
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Which expressions have products that are positive?
Mandarinka [93]
(1.2)(-3.5)(2.7)(-0.8
3 0
3 years ago
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There are 19 boxes of cherries. This is 5 less than twice the number of boxes of kiwis. How many boxes of kiwis are there?
sasho [114]
All you have to do is 19+5 because it is 5 less. 19 is 5 less than the kiwis. so 19+5= 24.  


There is 24 boxes of kiwis
8 0
3 years ago
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Find the sum of the positive integers less than 200 which are not multiples of 4 and 7​
taurus [48]

Answer:

12942 is the sum of positive integers between 1 (inclusive) and 199 (inclusive) that are not multiples of 4 and not multiples 7.

Step-by-step explanation:

For an arithmetic series with:

  • a_1 as the first term,
  • a_n as the last term, and
  • d as the common difference,

there would be \displaystyle \left(\frac{a_n - a_1}{d} + 1\right) terms, where as the sum would be \displaystyle \frac{1}{2}\, \displaystyle \underbrace{\left(\frac{a_n - a_1}{d} + 1\right)}_\text{number of terms}\, (a_1 + a_n).

Positive integers between 1 (inclusive) and 199 (inclusive) include:

1,\, 2,\, \dots,\, 199.

The common difference of this arithmetic series is 1. There would be (199 - 1) + 1 = 199 terms. The sum of these integers would thus be:

\begin{aligned}\frac{1}{2}\times ((199 - 1) + 1) \times (1 + 199) = 19900 \end{aligned}.

Similarly, positive integers between 1 (inclusive) and 199 (inclusive) that are multiples of 4 include:

4,\, 8,\, \dots,\, 196.

The common difference of this arithmetic series is 4. There would be (196 - 4) / 4 + 1 = 49 terms. The sum of these integers would thus be:

\begin{aligned}\frac{1}{2}\times 49 \times (4 + 196) = 4900 \end{aligned}

Positive integers between 1 (inclusive) and 199 (inclusive) that are multiples of 7 include:

7,\, 14,\, \dots,\, 196.

The common difference of this arithmetic series is 7. There would be (196 - 7) / 7 + 1 = 28 terms. The sum of these integers would thus be:

\begin{aligned}\frac{1}{2}\times 28 \times (7 + 196) = 2842 \end{aligned}

Positive integers between 1 (inclusive) and 199 (inclusive) that are multiples of 28 (integers that are both multiples of 4 and multiples of 7) include:

28,\, 56,\, \dots,\, 196.

The common difference of this arithmetic series is 28. There would be (196 - 28) / 28 + 1 = 7 terms. The sum of these integers would thus be:

\begin{aligned}\frac{1}{2}\times 7 \times (28 + 196) = 784 \end{aligned}.

The requested sum will be equal to:

  • the sum of all integers from 1 to 199,
  • minus the sum of all integer multiples of 4 between 1\! and 199\!, and the sum integer multiples of 7 between 1 and 199,
  • plus the sum of all integer multiples of 28 between 1 and 199- these numbers were subtracted twice in the previous step and should be added back to the sum once.

That is:

19900 - 4900 - 2842 + 784 = 12942.

8 0
3 years ago
Guests had to choose their meals from beef, chicken or vegetarian.
faltersainse [42]
1/3+5/12=4/12+5/12=9/12=3/4 of guests choose meat
1-3/4=1/4 of guests are vegetarians
so
23 guests is 1/4 of all guests
23*4=92 all guests 
6 0
3 years ago
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