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2 years ago

Help pleaseeeeeee!!!!!!

Mathematics

1 answer:

Mandarinka [93]2 years ago

8 0

Answer:

1. 3x7

2. 4A

3. 2M

4. 4X

5. 7D (left) 7x4 (right)

6. 2Y (left) 3Y (right)

Step-by-step explanation:

very simple for me, just take the variable or number on the left or right and multiply by the number on the top or bottom. And for problems with 2 variables do 2X or 2Y whatever the variable may be. Hope this helped

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What is the true solution to the equation below?

Naddik [55]

Yeeee

assuming your equaiton is
$2ln(e^{ln(2x)})-ln(e^{ln(10x)})=ln(30)$

remember some nice log rules
$log_a(b)=c$ translates to $a^c=b$
and
$a^{log_a(b)}=b$
and
$xlog_c(b)=log_c(b^x)$
and
$ln(x)=log_e(x)$
and
$log(a)-log(b)=log(\frac{a}{b})$
and
if $log(a)=log(b)$ then a=b

so

we can simplify a bit of stuff here

the $e^{ln(2x)} \space\ and \space\ the \space\ e^{ln(10x)}$ can be simplified to $2x \space\ and \space\ 10x$

so we gots now

$2ln(2x)-ln(10x)=ln(30)$
$ln((2x)^2)-ln(10x)=ln(30)$
$ln(4x^2)-ln(10x)=ln(30)$
$ln(\frac{4x^2}{10x})=ln(30)$
same base so
$\frac{4x^2}{10x}=30$
$\frac{2x}{5}=30$
times both sides by 5
$2x=150$
divide both sides by 2
$x=75$
answer is x=75

5 0

3 years ago

Tom is 8 years younger than John. Let j represent John's age in years. Write an algebraic expression to represent Tom's age in y

AlexFokin [52]

Answer:

(j-8)years is the answer to the question

5 0

2 years ago

Suppose your friend's parents invest $15,000 in an account paying 5% compounded annually. What will the balance be after 10 ye

Bad White [126]

Answer:

ofofxotdurz8sues8re8r8zirrizririzrir9

8 0

3 years ago

A randomized study compared the effectiveness of two mosquito repellents. The treatment group using Spray A reported a mean of 3

BlackZzzverrR [31]

Answer:

The difference of the means is not significant because the re-randomizations show that it is within the range of what could happen by chance.

Step-by-step explanation:

Given that:

The treatment group using Spray A reported a mean of 3.5 mosquito bites.

The treatment group using Spray B reported a mean of 5.6 mosquito bites.

After the data are re-randomized; the differences of the means are shown in the dot plot. The dot plot is attached in the file below;

The best conclusion that can be make based on the data from the dot plot is :

The difference of the means is not significant because the re-randomizations show that it is within the range of what could happen by chance.

5 0

3 years ago

Suppose that a large mixing tank initially holds 500 gallons of water in which 50 pounds of salt have been dissolved. Another br

Lina20 [59]

Answer:

The differential equation for the amount of salt A(t) in the tank at a time t > 0 is $\frac{dA}{dt}=12 - \frac{2A(t)}{500+t}$ .

Step-by-step explanation:

We are given that a large mixing tank initially holds 500 gallons of water in which 50 pounds of salt have been dissolved. Another brine solution is pumped into the tank at a rate of 3 gal/min, and when the solution is well stirred, it is then pumped out at a slower rate of 2 gal/min.

The concentration of the solution entering is 4 lb/gal.

Firstly, as we know that the rate of change in the amount of salt with respect to time is given by;

$\frac{dA}{dt}= \text{R}_i_n - \text{R}_o_u_t$

where, $\text{R}_i_n$ = concentration of salt in the inflow $\times$ input rate of brine solution

and $\text{R}_o_u_t$ = concentration of salt in the outflow $\times$ outflow rate of brine solution

So, $\text{R}_i_n$ = 4 lb/gal $\times$ 3 gal/min = 12 lb/gal

Now, the rate of accumulation = Rate of input of solution - Rate of output of solution

= 3 gal/min - 2 gal/min

= 1 gal/min.

It is stated that a large mixing tank initially holds 500 gallons of water, so after t minutes it will hold (500 + t) gallons in the tank.

So, $\text{R}_o_u_t$ = concentration of salt in the outflow $\times$ outflow rate of brine solution

= $\frac{A(t)}{500+t} \text{ lb/gal } \times 2 \text{ gal/min}$ = $\frac{2A(t)}{500+t} \text{ lb/min }$

Now, the differential equation for the amount of salt A(t) in the tank at a time t > 0 is given by;

= $\frac{dA}{dt}=12\text{ lb/min } - \frac{2A(t)}{500+t} \text{ lb/min }$

or $\frac{dA}{dt}=12 - \frac{2A(t)}{500+t}$ .

4 0

3 years ago