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netineya [11]
2 years ago
15

HELPPP MEE PLEASEEE I DONT WANNA FAIL SUMMER SCHOOL NO FILES OR LINKS WOULD BE DEEPLY APPRECIATED

Mathematics
2 answers:
Sav [38]2 years ago
7 0
1) 1:4. 2) 5:14 3) 2:3 4) $2 for every 1 pound
scoray [572]2 years ago
3 0

Answer:

1) 1/4

2) 1 1/2

3) 9/5

4) 4 pounds

Step-by-step explanation:

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a person standing at a ridge in the grand canyon throws a penny upeard and toward the pit of the canyon. The height of the penny
Lelechka [254]
I believe its 80 if you derive it out, but there is no units of measurement.
4 0
3 years ago
C=r(5-d)/y We need to solve for d.
GrogVix [38]

Answer:

5-Cy/r  = d

Step-by-step explanation:

C=r(5-d)/y

Multiply each side by y

Cy=r(5-d)/y *y

Cy=r(5-d)

Divide each side by r

Cy/r=r(5-d)/r

Cy/r=(5-d)

Subtract 5 from each side

Cy/r - 5 = 5-d-5

Cy/r - 5 = -d

Multiply by -1

-Cy/r + 5 = d

5-Cy/r  = d

5 0
2 years ago
Read 2 more answers
What is the smallest multiple of 18 of the form 2A945B, where A and B are digits?
musickatia [10]

Answer:

219456

Step-by-step explanation:

It must end in 0,2,4,6,8 (divisible by 2 rule) and the digits must sum to a multiple of 9(divisible by 9 rule).

The digit sum is 2+A+9+4+5+B=20+A+B.

We want to make A as low as possible because it's a higher digit than B.

If A is as low as possible, we have that the sum is 20+0+7=27. Uh-oh. B has to be even! So A=0 doesn't work.

If A=1, we have that the sum is 20+1+6=27. Yay!

So, the answer is \boxed{219456}

6 0
2 years ago
Need some help with this... <br> Any unsuitable response will be reported. Thank you!
Anika [276]
Rectangle side length is 20

20 times 5=100
Sq 10x10=100
4 0
3 years ago
A rectangular block of copper metal mass is 1896 g. The dimensions of the points
vagabundo [1.1K]

Answer:

8.921513269...g/cm

Step-by-step explanation:

the density has this formula: D= m/V

so to solve the V, which refers to volume, you do: 8.4×5.5×4.6= 212.52, and the m, which refers to mass, you already have it. So the last thing to do is:

D= 1896g/212.52= 8.921513269...g/cm

I hope I've been helpful... good luck...!!!

8 0
3 years ago
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