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3 years ago

Please help with number 6 ( at the top)

Mathematics

1 answer:

castortr0y [4]3 years ago

8 0

30 + 8d = 102
8d = 102 - 30
8d = 72
d = 72/8
d = 9....you would have to walk 9 dogs

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Plz help with this I don’t really get it

Greeley [361]

Answer:

Step-by-step explanation:

Associative Property of Multiplication is a(bc) = (ab)c

d is the only one that follows the rule

6 0

2 years ago

Is the ordered pair (-2, 1) a solution to the equation - 3x + y = 5? <br>No <br>Yes

Otrada [13]

Answer:

Step-by-step explanation:

Plug (-2,1) into -3x+y=5. This will make it -3(-2)+(1)=5. This is equal to 6+1=5. 7=5. 7 is not equal to 5. Therefore, (-2,1) is not a solution to the equation -3x+y=5.

If this helps please mark as brainliest

6 0

2 years ago

Help please I’ll make it brainliest!!!!!

mr Goodwill [35]

Step-by-step explanation:

it is very simple, once you remember that "kilo" means 1000, "mili" means 1/1000, and "centi" means 1/100.

and therefore 1 cg = 10 mg, or 1 cm = 10 mm.

"deci" means 1/10.

and therefore 1 dm = 100 mm, 1dg = 100 mg.

9.32 kg = 9.32×1000 = 9320 g

1.429 g = 1.429/1000 = 0.001429 kg

287 g = 287/1000 = 0.287 kg

4.6 L = 4.6×1000 = 4600 mL

0.119 L = 0.119×1000 = 119 mL

9936 mL = 9936/1000 = 9.936 L

26793 mL = 26793/1000 = 26.793 L

0.06 L = 0.06×1000 = 60 mL

170 cg = 170×10 = 1700 mg

10.

2674 cm = 2674/100 = 26.74 m

11.

9.05 mm = 9.05/100 = 0.0905 dm

12.

2 L = 2×1000 = 2000 mL

13.

62.4 L = 62.4×1000 = 62400 mL

14.

99.9 mm = 99.9/1000 = 0.0999 m

15.

4.34 g = 4.34×100 = 434 cg

16.

10 km = 10×1000 m = 10×1000×1000 = 10000000 mm

17.

65 cL = 65/100 = 0.65 L

18.

105 mL = 105/1000 = 0.105 L

19.

0.27 g = 0.27×100 = 27 cg

20.

7777 m = 7777/1000 = 7.777 km

4 0

1 year ago

A species of beetles grows 32% every year. Suppose 100 beetles are released into a field. How many beetles will there be in 10 y

siniylev [52]

Given that a species of beetles grows 32% every year.

So growth rate is given by

r=32%= 0.32

Given that 100 beetles are released into a field.

So that means initial number of beetles P=100

Now we have to find about how many beetles will there be in 10 years.

To find that we need to setup growth formula which is given by

$A=P(1+r)^n$ where A is number of beetles at any year n.

Plug the given values into above formula we get:

$A=100(1+0.32)^n$

$A=100(1.32)^n$

now plug n=10 years

$A=100(1.32)^{10}=100(16.0597696605)=1605.97696605$

Hence answer is approx 1606 beetles will be there after 20 years.

Now we have to find about how many beetles will there be in 20 years.

To find that we plug n=20 years

$A=100(1.32)^{20}=100(257.916201549)=25791.6201549$

Hence answer is approx 25791 beetles will be there after 20 years.

Now we have to find time for 100000 beetles so plug A=100000

$A=100(1.32)^n$

$100000=100(1.32)^n$

$1000=(1.32)^n$

$log(1000)=n*log(1.32)$

$\frac{\log\left(10000\right)}{\log\left(1.32\right)}=n$

33.174666862=n

Hence answer is approx 33 years.

4 0

3 years ago

Read 2 more answers

Need help now!! 20 points!!

Gelneren [198K]

When you have something like this, all you need to do is substitute the values, the last is for what value of x

For the first one;

((x^2+1)+(x-2))(2)

(x^2+x-1)(2)

(2)^2+(2)-1

4+2-1

For the second one;

((x^2+1)-(x-2))(3)

(x^2-x+3)(3)

(3)^2-(3)+3

9-3+3

For the last one;

3(x^2+1)(7)+2(x-2)(3)

3((7)^2+7)+2((3)-2)

3(49+7)+2(3-2)

3(56)+2(1)

168+2

170

3 0

2 years ago