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3 years ago

Yo can someone help me out

Mathematics

1 answer:

grin007 [14]3 years ago

6 0

Yes. I completely agree with that person ^^

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Someone please help me will give BRAILIEST!!!!!!

Misha Larkins [42]

The answer would be 1/5

8 0

3 years ago

Solve 41-9+(8×2.3)-15+(2.1×4)

dsp73

(8x2.3)= (18.4)
(2.1x4)= (8.4)
Now you should have:
41-9+18.4-15+8.4
9+18= 27
15+8.4= 23.4
Should have:
41-27-23.4= -9.4

6 0

3 years ago

The difference of two numbers is 16 the greater number is five less than four times a small number find the two numbers

Elis [28]

Answer:

The numbers are 23 and 7

Step-by-step explanation:

1. Set up the equations

Let x = the greater number

and y = the smaller number. Then

(1) x - y = 16

4y = 4 times the smaller number and

4y - 5 = 5 less than 4 times the smaller number. Then

(2) x = 4y - 5

You have a system of two equations:

$\begin{cases}(1) & x - y = 16\\(2) & x = 4y - 5\end{cases}$

2. Solve the equations

$\begin{array}{lrcll}(3) & 4y - 5 - y & = & 16 &\text{Substituted (2) into (1)}\\& 3y - 5 & = & 16 &\text{Simplified}\\ & 3y & = & 21 &\text{Added 5 to each side}\\(4) &y & = & \mathbf{7} &\text{Divided each side by 3} \\& x - 7 & = & 16 & \text{Substituted (4) into (1)}\\& x& = & \mathbf{23} & \text{Added 7 to each side}\\\end{array}$

The larger number is 23; the smaller number is 7.

3. Check

$\begin{array}{rclcrcl}23 - 7& = & 16 & \qquad &23&=&4(7) - 5\\16 & = & 16 & \qquad &23& = &28 - 5\\&&& \qquad &23& = &23\\\end{array}$

OK.

3 0

3 years ago

You take a bite of a cookie and begin to crush it between your teeth making smaller bits in your mouth.

timofeeve [1]

Physical, you aren't changing the substance, you are just changing the shape of the cookie

5 0

3 years ago

the half life of polonium is 139 days. you have a brand new sample of polonium that will not be useful to you after 80% of the m

melamori03 [73]

The half-life of Po is 139 days, so its decay rate is

0.5 = e^(139k) ==> k = ln(0.5)/139

After some time t, 80% of the material will have decayed and you'll be left with a sample 20% of its original mass:

0.2 = e^(kt) ==> t = ln(0.2)/k = ln(0.2)/ln(0.5) * 139 = 322.748

so your sample is usable for about 323 days.

7 0

3 years ago