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xz_007 [3.2K]
3 years ago
11

EXTRA POINTS!NNeed help with Algebra 1 question soon please!

Mathematics
1 answer:
VikaD [51]3 years ago
4 0
If in the first quarter they scored 1/6 of their points, this would mean in the first quarter they scored 6 points (36/6=6). If in the second quarter they scored 1/4 of their points, they scored 9 points (36/4=9). If in the third quarter they scored 1/3 of their points, they scored 12 points (36/3=12).

12+9+6=27.
36 - 27 = 9

the answer is c, 9
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What is (-33x-78)+(5x8x9x10)divided by9.75
kherson [118]

Answer:

<em>6,174</em>

Step-by-step explanation:

(-33 * - 78) + (5 * 8 * 9 * 10)

Remove the brackets:

-33 * - 78 + 5 * 8 * 9 * 10

Solve like so:

-33 * - 78 = 2,574

5 * 8 * 9 * 10 = 3,600

(2,574) + (3,600)

2,574 + 3,600 = 6,174

<em>PLEASE</em><em> </em><em>DO</em><em> </em><em>MARK</em><em> </em><em>ME</em><em> </em><em>AS</em><em> </em><em>BRAINLIEST UWU</em>

7 0
3 years ago
68+(-41)=?<br> Solved and show how you got the answer.
mars1129 [50]

Answer:

27

Step-by-step explanation:

68 + (-41) = ?

Well, adding a negative is the same as subtracting a positive, so...

68 + (-41)   =   68 - 41

68 - 41 = 27

7 0
3 years ago
Read 2 more answers
If 5 potatoes together have a mass of 1 kilogram and 8 pears together have a mass of 1200 grams, which has the greater mass, a p
lord [1]
The Potato has more mass
7 0
3 years ago
Point A has coordinates (-5, 3). If point (1,6) is
gtnhenbr [62]

Answer:

(3, 7)

Step-by-step explanation:

 (4/3)P -(4/3)A + A = B . . . . . . add A

 (4P -A)/3 = B . . . . . . . . . . . . . simplify

Then the coordinates of point B are ...

 B = (4(1, 6) -(-5, 3))/3 = (9, 21)/3

 B = (3, 7)

8 0
3 years ago
A particular isotope has a​ half-life of 74 days. If you start with 1 kilogram of this​ isotope, how much will remain after 150
shtirl [24]

\bf \stackrel{150~days}{\textit{Amount for Exponential Decay using Half-Life}} \\\\ A=P\left( \frac{1}{2} \right)^{\frac{t}{h}}\qquad \begin{cases} A=\textit{accumulated amount}\\ P=\textit{initial amount}\dotfill &1\\ t=\textit{elapsed time}\dotfill &150\\ h=\textit{half-life}\dotfill &74 \end{cases} \\\\\\ A=1\left( \frac{1}{2} \right)^{\frac{150}{74}}\implies A=1\left( \frac{1}{2} \right)^{\frac{75}{37}}\implies \boxed{A\approx 0.24536}


\bf \stackrel{300~days}{\textit{Amount for Exponential Decay using Half-Life}} \\\\ A=P\left( \frac{1}{2} \right)^{\frac{t}{h}}\qquad \begin{cases} A=\textit{accumulated amount}\\ P=\textit{initial amount}\dotfill &1\\ t=\textit{elapsed time}\dotfill &300\\ h=\textit{half-life}\dotfill &74 \end{cases} \\\\\\ A=1\left( \frac{1}{2} \right)^{\frac{300}{74}}\implies A=1\left( \frac{1}{2} \right)^{\frac{150}{37}}\implies \boxed{A\approx 0.060202}

6 0
3 years ago
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