Ask question

Ask question

All categories

2 years ago

10

Write two division equations for each multiplication equation. 15⋅25=6 15 ⋅ 2 5 = 6 6⋅43=8 6 ⋅ 4 3 = 8 16⋅78=14

1 answer:

Gemiola [76]2 years ago

6 0

Answer:

$(a)$ $\frac{30}{5} = 6$ and $\frac{60}{10} = 6$

$(b)$ $\frac{24}{3} = 8$ and $\frac{96}{12} = 8$

$(c)$ $\frac{112}{8} = 14$ and $\frac{28}{2} = 14$

Step-by-step explanation:

Given

$(a)\ 15 * \frac{2}{5} = 6$

$(b)\ 6 * \frac{4}{3} = 8$

$(c)\ 16 * \frac{7}{8} = 14$

Required

Write 2 division equations

$(a)\ 15 * \frac{2}{5} = 6$

Express 15 as 15/1

$\frac{15}{1} * \frac{2}{5} = 6$

Multiply numerator and denominator

$\frac{30}{5} = 6$ --- This is (1)

Multiply the fraction by a common term (e.g. 2)

So, we have:

$\frac{30*2}{5*2} = 6$

$\frac{60}{10} = 6$ ---- (2)

$(b)\ 6 * \frac{4}{3} = 8$

Express 6 as 6/1

$\frac{6}{1} * \frac{4}{3} = 8$

Multiply numerator and denominator

$\frac{24}{3} = 8$ --- This is (1)

Multiply the fraction by a common term (e.g. 4)

So, we have:

$\frac{24*4}{3*4} = 8$

$\frac{96}{12} = 8$ ---- (2)

$(c)\ 16 * \frac{7}{8} = 14$

Express 16 as 16/1

$\frac{16}{1} * \frac{7}{8} = 14$

Multiply

$\frac{112}{8} = 14$ ---- This is (1)

Divide the fraction by a common term (e.g. 4)

$\frac{112/4}{8/4} = 14$

$\frac{28}{2} = 14$

You might be interested in

To convert 8.5 yards to feet which ratio could you multiply by

uranmaximum [27]

8.5 yd/1 * 3ft/1yd ................

7 0

2 years ago

Read 2 more answers

How many unique values of x solve x2 + 14x + 49 = 0?

Natalka [10]

I think x= -7 but I'm not to sure

3 0

3 years ago

Use the figure to find n please.

ira [324]

Answer:

n = 5

Step-by-step explanation:

Since this is a right triangle, we can use trig

tan theta = opp /adj

tan 30 = n / 5 sqrt(3)

5 sqrt(3) tan 30 = n

5 sqrt(3) * 1/ sqrt(3) = n

5 = n

4 0

2 years ago

Read 2 more answers

A coin is loaded so that the probability of heads is 0.55 and the probability of tails is 0.45. suppose the coin is tossed twice

Vladimir79 [104]

The probability is 0.235

8 0

3 years ago

A given field mouse population satisfies the differential equation dp dt = 0.5p − 410 where p is the number of mice and t is the

ohaa [14]

Answer:

a) $t = 2 *ln(\frac{82}{5}) =5.595$

b) $t = 2 *ln(-\frac{820}{p_0 -820})$

c) $p_0 = 820-\frac{820}{e^6}$

Step-by-step explanation:

For this case we have the following differential equation:

$\frac{dp}{dt}=\frac{1}{2} (p-820)$

And if we rewrite the expression we got:

$\frac{dp}{p-820}= \frac{1}{2} dt$

If we integrate both sides we have:

$ln|P-820|= \frac{1}{2}t +c$

Using exponential on both sides we got:

$P= 820 + P_o e^{1/2t}$

Part a

For this case we know that p(0) = 770 so we have this:

$770 = 820 + P_o e^0$

$P_o = -50$

So then our model would be given by:

$P(t) = -50e^{1/2t} +820$

And if we want to find at which time the population would be extinct we have:

$0=-50 e^{1/2 t} +820$

$\frac{820}{50} = e^{1/2 t}$

Using natural log on both sides we got:

$ln(\frac{82}{5}) = \frac{1}{2}t$

And solving for t we got:

$t = 2 *ln(\frac{82}{5}) =5.595$

Part b

For this case we know that p(0) = p0 so we have this:

$p_0 = 820 + P_o e^0$

$P_o = p_0 -820$

So then our model would be given by:

$P(t) = (p_o -820)e^{1/2t} +820$

And if we want to find at which time the population would be extinct we have:

$0=(p_o -820)e^{1/2 t} +820$

$-\frac{820}{p_0 -820} = e^{1/2 t}$

Using natural log on both sides we got:

$ln(-\frac{820}{p_0 -820}) = \frac{1}{2}t$

And solving for t we got:

$t = 2 *ln(-\frac{820}{p_0 -820})$

Part c

For this case we want to find the initial population if we know that the population become extinct in 1 year = 12 months. Using the equation founded on part b we got:

$12 = 2 *ln(\frac{820}{820-p_0})$

$6 = ln (\frac{820}{820-p_0})$

Using exponentials we got:

$e^6 = \frac{820}{820-p_0}$

$(820-p_0) e^6 = 820$

$820-p_0 = \frac{820}{e^6}$

$p_0 = 820-\frac{820}{e^6}$

8 0

3 years ago