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3 years ago

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:

6,174

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

PLEASE DO MARK ME AS BRAINLIEST UWU

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3 years ago

68+(-41)=? Solved and show how you got the answer.

mars1129 [50]

Answer:

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