What is the measure of

When y is 4, p is 0.5, and m is 2, x is 2. If x varies directly with the product of p and m and inversely with y, which equation

Radda [10]

<h3><u>Question:</u></h3>

When y is 4, p is 0.5, and m is 2, x is 2. If x varies directly with the product of p and m and inversely with y, which equation models the situation?

xpmy=8

xy/pm=8

xpm/y=0.5

x/pmy=0.5

<h3><u>Answer:</u></h3>

The equation models the situation is $\frac{x y}{p m}=8$

<h3><u>Solution:</u></h3>

Given that

x is 2, y is 4, p is 0.5, and m is 2

x varies directly with the product of p and m

x varies inversely with y

$\text {Product of } p \text { and } m=p \times m=p m$

x varies directly with the product of p and m

$=>x \propto p m$ ---- eqn 1

As x varies inversely with y,

$=>x \propto \frac{1}{y}$ ----- eqn 2

From (1) and 2, we can say that

$x \propto \frac{p m}{y}$

$\Rightarrow x=k \frac{p m}{y}$

where k is constant of proportionality

$\Rightarrow \frac{x y}{p m}=k$ ---- eqn 3

On substituting given values of x = 2, y = 4, p = 0.5 and m= 2 in eqn (3) we get

$\frac{x y}{p m}=\frac{2 \times 4}{0.5 \times 2}=k$

$\begin{array}{l}{\frac{x y}{p m}=\frac{8}{1}=k} \\\\ {=>\frac{x y}{p m}=8}\end{array}$

Hence correct option is second that is $\frac{x y}{p m}=8$

8 0

3 years ago

Read 2 more answers

Can someone please help me!

victus00 [196]

Answer:

33°

Step-by-step explanation:

11x = 1/2(x+63)

22x = x + 63

21x = 63

x = 3

IHJ = 11x

= 11(3) = 33°

8 0

4 years ago

Heidi has $4.55. She needs to buy bolts at the hardware store. Bolts cost $0.13 each.

MrRa [10]

Divide 4.55 by 0.13 (the cost of each bolt) to find out how many bolts Heidi can buy!

$4.55 ÷ $0.13 = 35

Heidi can buy 35 bolts with $4.55 !1

3 0

3 years ago

Read 2 more answers

The population of a community is known to increase at a rate proportional to the number of people present at time t. The initial

qaws [65]

Answer:

Step-by-step explanation:

Let P be the population of the community

So the population of a community increase at a rate proportional to the number of people present at a time

That is

$\frac{dp}{dt} \propto p\\\\\frac{dp}{dt} =kp\\\\ [k \texttt {is constant}]\\\\\frac{dp}{dt} -kp =0$

Solve this equation we get

$p(t)=p_0e^{kt}---(1)$

where p is the present population

p₀ is the initial population

If the initial population as doubled in 5 years

that is time t = 5 years

We get

$2p_o=p_oe^{5k}\\\\e^{5k}=2$

Apply In on both side to get

$Ine^{5k}=In2\\\\5k=In2\\\\k=\frac{In2}{5} \\\\\therefore k=\frac{In2}{5}$

Substitute $k=\frac{In2}{5}$ in $p(t)=p_oe^{kt}$ to get

$\large \boxed {p(t)=p_oe^{\frac{In2}{5}t }}$

Given that population of a community is 9000 at 3 years

substitute t = 3 in ${p(t)=p_oe^{\frac{In2}{5}t }}$

$p(3)=p_oe^{3 (\frac{In2}{5}) }\\\\9000=p_oe^{3 (\frac{In2}{5}) }\\\\p_o=\frac{9000}{e^{3(\frac{In2}{5} )}} \\\\=5937.8$

<h3>Therefore, the initial population is 5937.8</h3>

7 0

3 years ago

Guys I need help! I am given two chances to get this question right!

amid [387]

The four digit code is

1563

8 0

4 years ago