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Brut [27]
3 years ago
11

50 points!!!! Easy two part multiple choice question, will mark brainliest

Mathematics
2 answers:
erastovalidia [21]3 years ago
6 0
Part A:
The answer is the second choice because that's the only interval where the function is increasing.

Part B:
The answer is the last choice.
x-intercepts are where the graph meets the x-axis.
y-intercepts are where the graph meets the y-axis.

Have an awesome day! :)
natulia [17]3 years ago
5 0

Part a: 2nd choice

part b:2nd choice

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Cuantos paquetes de 5/8 de kilo harán falta para envasar 125 kilos de harina
bonufazy [111]

Step-by-step explanation:

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4 0
3 years ago
Help answer this im struggling omg
ivanzaharov [21]
H^2 + 3.5 = 20^2

That is the formula
7 0
3 years ago
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Alabama papas must average 80 on five test . Scores on the first four test were 81 73 84 and 76 what is the lowest score she can
Salsk061 [2.6K]

She has to score at least 86 on her fifth test.

Further explanation:

We will use the average formula to find the fifth value.

Given

Average\ required=80\\x_1=81\\x_2=73\\x_3=84\\x_4=76\\x_5=?

We have to find x5

The formula for average is:

x=\frac{Sum\ of\ values}{Total\ number\ of\ values}\\Here,\\n=5\\So,\\x=\frac{x_1+x_2+x_3+x_4+x_5}{5}\\Putting\ values\\80=\frac{81+73+84+76+x_5}{5}

80*5=314+x_5\\400=314+x_5\\x_5=400-314\\x_5=86

She has to score at least 86 on her fifth test.

Keywords: Average, Word Problems

Learn more about average at:

  • brainly.com/question/9323337
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#LearnwithBrainly

7 0
3 years ago
Help please I really need answers fast
Dima020 [189]
1st: $6.44

2nd: 6.33 KM

8 0
2 years ago
Ad is tangent to circle o at d. find ab. round to the nearest tenth if necessary
Naddik [55]
Check the picture below.

\bf 0=AB^2+6AB-121\implies \stackrel{\textit{using the quadratic formula}}{AB=\cfrac{-6 \pm \sqrt{6^2-4(1)(-121)}}{2(1)}}
\\\\\\
AB=\cfrac{-6 \pm \sqrt{520}}{2}\implies AB=\cfrac{-6 \pm \sqrt{4\cdot 130}}{2}
\\\\\\
AB=\cfrac{-6 \pm \sqrt{2^2\cdot 130}}{2}\implies AB=\cfrac{-6 \pm 2\sqrt{130}}{2}
\\\\\\
AB=-3\pm\sqrt{130}\implies AB=
\begin{cases}
\boxed{-3+\sqrt{130}}\\\\
-3-\sqrt{130}
\end{cases}

since the distance AB cannot be a negative value, thus is not -3-√(130).

3 0
3 years ago
Read 2 more answers
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