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3 years ago

In Harold’s class, there are 7 boys for every 6 girls. Identify three equivalent ratios to compare boys to girls.

Mathematics

2 answers:

jekas [21]3 years ago

8 0

Answer:

7:6

14:12

21:18

28:24

trapecia [35]3 years ago

5 0

7:6 14:12 21:18 the are the answers to your question

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How can you use models to explain why 3.1 = 3.10

Andrej [43]

3.1 is like a example when it is hiding a zero behind it and you find this when you do a math problem and you don't put zero in front of a number like 100-89 you don't write "011" so that is why it is like that

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2 years ago

Does anyone know this??

Evgesh-ka [11]

The answer is letter A

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3 years ago

Jennifer jogged at a rate of 3/5 mile every 6 minutes. what was her jogging rate in miles per hour

alexgriva [62]

Answer:

she jogged for 35 hours

Step-by-step explanation:

6×5 6+5=11 11+6=17 17+6=23 23+6=29 29+6=35

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3 years ago

(3^2+4−2)(2^2−3−4) can you guys please solve this for me please.

Degger [83]

Answer:

-33

Step-by-step explanation:

$(3^2+4-2)(2^2-3-4)$

$(9+4-2)(4-3-4)$

$(13-2)(1-4)$

$(11)$ · $(-3)$

(-33)

6 0

3 years ago

Integrate this two questions

tankabanditka [31]

Simplify the integrands by polynomial division.

$\dfrac{t^2}{1 - 3t} = -\dfrac19 \left(3t + 1 - \dfrac1{1 - 3t}\right)$

$\dfrac t{1 + 4t} = \dfrac14 \left(1 - \dfrac1{1 + 4t}\right)$

Now computing the integrals is trivial.

$\displaystyle \int \frac{t^2}{1 - 3t} \, dt = -\frac19 \int \left(3t + 1 - \frac1{1-3t}\right) \, dt \\\\ = -\frac19 \left(\frac32 t^2 + t + \frac13 \ln|1 - 3t|\right) + C \\\\ = \boxed{-\frac{t^2}6 - \frac t9 - \frac{\ln|1-3t|}{27} + C}$

where we use the power rule,

$\displaystyle \int x^n \, dx = \frac{x^{n+1}}{n+1} + C ~~~~ (n\neq-1)$

and a substitution to integrate the last term,

$\displaystyle \int \frac{dt}{1-3t} = -\frac13 \int \frac{du}u \\\\ = -\frac13 \ln|u| + C \\\\ = -\frac13 \ln|1-3t| + C ~~~ (u=1-3t \text{ and } du = -3\,dt)$

$\displaystyle \int \frac t{1+4t} \, dt = \frac14 \int \left(1 - \frac1{1+4t}\right) \, dt \\\\ = \frac14 \left(t - \frac14 \ln|1 + 4t|\right) + C \\\\ = \boxed{\frac t4 - \frac{\ln|1+4t|}{16} + C}$

using the same approach as above.

5 0

1 year ago