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3 years ago

What is the answer please help

Mathematics

1 answer:

matrenka [14]3 years ago

6 0

Answer:

Step-by-step explanation:

7 shows up twice

all other numbers show up only once.

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Are the ratios 2:3 and 8:12 equivalent? Justify you answer.

Lunna [17]

Answer:Yes

Step-by-step explanation:You are multiplying the ratios by 4

3 0

2 years ago

Rob mixed 3.7 grams of salt into a pot of soup he was cooking. Before he served the soup, Rob added 0.5 grams of salt. How much

saul85 [17]

Answer:

4.2 grams in total

Step-by-step explanation:

<h2>Remember to line up the decimal point when adding numbers that contain decimals.</h2>

3.7

+0.5

_______

4.2

5 0

3 years ago

Solve the system of linear equations by graphing. y = –5/2 x – 7 x + 2y = 4 What is the solution to the system of linear equatio

Anestetic [448]

Keywords

system; linear equations; graphing; solution

we have

$y=-\frac{5}{2}x-7$ -------> equation A

$x+2y=4$

$x=4-2y$ ----> equation B

The <u>system</u> of <u>linear equations</u> is composed of equation A and equation B

Substitute the equation B in equation A

$y=-\frac{5}{2}[4-2y]-7$

$y=-10+5y-7$

$4y=17$

$y=17/4=4.25$

Find the value of x

$x=4-2[17/4]=-4.5$

The<u> solution</u> is the point $(-4.5,4.25)$

Using a <u>graphing</u> tool

see the attached figure

therefore

the answer is the option A

$(-4.5,4.25)$

4 0

2 years ago

Read 2 more answers

True Value Rentals charges $18 to rent a moving truck for 3 hours, with an additional

Readme [11.4K]

It will cost $42 because it’s $12 for a hour

5 0

2 years ago

A large tank is filled to capacity with 300 gallons of pure water. brine containing 5 pounds of salt per gallon is pumped into t

Licemer1 [7]

If $A(t)$ is the amount of salt in the tank at time $t$ , then the rate at which this amount changes over time is given by the ODE

$A'(t)=\dfrac{3\text{ gal}}{1\text{ min}}\cdot\dfrac{5\text{ lb}}{1\text{ gal}}-\dfrac{3\text{ gal}}{1\text{ min}}\cdot\dfrac{A(t)\text{ lb}}{300+(3-3)t\text{ gal}}$

$A'(t)+\dfrac1{100}A(t)=15$

We're told that the tank initially starts with no salt in the water, so $A(0)=0$ .

Multiply both sides by an integrating factor, $e^{t/100}$ :

$e^{t/100}A'(t)+\dfrac1{100}e^{t/100}A(t)=15e^{t/100}$
$\left(e^{t/100}A(t)\right)'=15e^{t/100}$
$e^{t/100}A(t)=1500e^{t/100}+C$
$A(t)=1500+Ce^{-t/100}$

Since $A(0)=0$ , we have

$0=1500+C\implies C=-1500$

so that the amount of salt in the tank over time is given by

$A(t)=1500(1-e^{-t/100})$

After 10 minutes, the amount of salt in the tank is

$A(10)=1500(1-e^{-1/10})\approx142.74\text{ lb}$

8 0

3 years ago