Ask question

Ask question

All categories

2 years ago

6

Does anyone know this answer??

1 answer:

xxMikexx [17]2 years ago

4 0

I believe by the first equation they are referring two a representation of the total amount of money made up by the two types of coins. I found this to be 0.05x + 0.10y = 3.50

In my work I solved for the amount of each coin- I assume this is what the final step of your problem entails. If you need any help understanding/ I misinterpreted something please feel free to comment and ask :)

#5c coins or x= 36

#10c coins or y= 17

You might be interested in

Drag the phrases into the box to match each expression. 3/4x+1/2 3/4−1/2x 3/4−(x+1/2) 20 pointss!!! i know i must be cra

fomenos

Answer:

$\frac{3}{4}x + \frac{1}{2}$ = One-half more than three-fourths of a number

$\frac{3}{4}-\frac{1}{2}x$ = Three-fourths minus one-half of a number

$\frac{3}{4}-(x+\frac{1}{2})$ = three-fourths minus the sum of a number and one-half

Step-by-step explanation:

Suppose x be an unknown number.

∵ 3/4th of a number = $\frac{3}{4}x$

$\frac{3}{4}x + \frac{1}{2}$ = One-half more than three-fourths of a number

Now, half of a number = $\frac{1}{2}x$

$\frac{3}{4}-\frac{1}{2}x$ = Three-fourths minus one-half of a number

Sum of a number and one-half = $x+\frac{1}{2}$

$\frac{3}{4}-(x+\frac{1}{2})$ = three-fourths minus the sum of a number and one-half

6 0

3 years ago

-k=7<br>what is k equal to

kifflom [539]

-k=7

divide each side by -1

-k/-1 = 7/-1

k = -7

3 0

3 years ago

Read 2 more answers

Please help asap. Thanks.

GarryVolchara [31]

400 each unit is forty

6 0

2 years ago

60 POINTS FOR THE THISS!!!

Goryan [66]

Yes because they are constant throughout the table.

4 0

2 years ago

Read 2 more answers

Find the area each sector. Do Not round. Part 1. NO LINKS!!<br><br>

sladkih [1.3K]

Answer:

$\textsf{Area of a sector (angle in degrees)}=\dfrac{\theta}{360 \textdegree}\pi r^2$

$\textsf{Area of a sector (angle in radians)}=\dfrac12r^2\theta$

17) Given:

$\theta$ = 240°
r = 16 ft

$\textsf{Area of a sector}=\dfrac{240}{360}\pi \cdot 16^2=\dfrac{512}{3}\pi \textsf{ ft}^2$

19) Given:

$\theta=\dfrac{3 \pi}{2}$
r = 14 cm

$\textsf{Area of a sector}=\dfrac12\cdot14^2 \cdot \dfrac{3\pi}{2}=147 \pi \textsf{ cm}^2$

21) Given:

$\theta=\dfrac{ \pi}{2}$
r = 10 mi

$\textsf{Area of a sector}=\dfrac12\cdot10^2 \cdot \dfrac{\pi}{2}=25 \pi \textsf{ mi}^2$

23) Given:

$\theta$ = 60°
r = 7 km

$\textsf{Area of a sector}=\dfrac{60}{360}\pi \cdot 7^2=\dfrac{49}{6}\pi \textsf{ km}^2$

3 0

2 years ago