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mina [271]
2 years ago
8

How many intergers from 10 to 3a - 4

Mathematics
2 answers:
VMariaS [17]2 years ago
6 0

Answer:

Step-by-step explanation:

10 , 9,8,7,6,5,4,3,2,1,0,-1,-2-3,-4

Hope this helps

Mama L [17]2 years ago
3 0

pretty sure my answer was incorrect but I cant delete it

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Can someone please help? the other box options are the same as the first:)
Nina [5.8K]

Answer:

0.5 in the first box and 1 or -1 in the second

4 0
3 years ago
The sum of Pete's and Sam's ages is 30. Five years ago, Pete was 3 times as old as Sam. How old is Sam?
RUDIKE [14]
P-5=3s-15 because Pete is 20 and Sam is 10. 20-5= 15 and 3(10)-15 also equals 15. The answer is C.
4 0
4 years ago
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Help please!! I will give crown! :D
GREYUIT [131]

Answer:

C. Please mark Brainliest!!!!!!

Step-by-step explanation:

You move it 5 over, so it is 3.844x10^5

6 0
3 years ago
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What is the product?
fredd [130]

Answer:

\frac{4k + 2}{k^{2}-4 }  ×  \frac{k-2}{2k+1}   =  \frac{2}{k + 2}

Step-by-step explanation:

\frac{4k + 2}{k^{2}-4 }  ×  \frac{k-2}{2k+1}

To solve the above, we need to follow the steps below;

4k+2 can be factorize, so that;

4k +2 = 2 (2k + 1)

k² - 4  can also be be expanded, so that;

k² - 4 = (k-2)(k+2)

Lets replace  4k +2  by  2 (2k + 1)

and

k² - 4 by  (k-2)(k+2)   in the expression  given

\frac{4k + 2}{k^{2}-4 }  ×  \frac{k-2}{2k+1}

\frac{2(2k+ 1)}{(k-2)(k+2)}   ×  \frac{k-2}{2k+1}

(2k+1) at the numerator will cancel-out (2k+1) at the denominator, also (k-2) at the numerator will cancel-out (k-2) at the denominator,

So our expression becomes;  

\frac{2}{k + 2}

Therefore, \frac{4k + 2}{k^{2}-4 }  ×  \frac{k-2}{2k+1}   =  \frac{2}{k + 2}

3 0
3 years ago
Read 2 more answers
A Norman window is a window with a semi-circle on top of regular rectangular window. (See the picture.) What should be the dimen
Vikki [24]

Answer:

bottom side (a) = 3.36 ft

lateral side (b) = 4.68 ft

Step-by-step explanation:

We have to maximize the area of the window, subject to a constraint in the perimeter of the window.

If we defined a as the bottom side, and b as the lateral side, we have the area defined as:

A=A_r+A_c/2=a\cdot b+\dfrac{\pi r^2}{2}=ab+\dfrac{\pi}{2}\left (\dfrac{a}{2}\right)^2=ab+\dfrac{\pi a^2}{8}

The restriction is that the perimeter have to be 12 ft at most:

P=(a+2b)+\dfrac{\pi a}{2}=2b+a+(\dfrac{\pi}{2}) a=2b+(1+\dfrac{\pi}{2})a=12

We can express b in function of a as:

2b+(1+\dfrac{\pi}{2})a=12\\\\\\2b=12-(1+\dfrac{\pi}{2})a\\\\\\b=6-\left(\dfrac{1}{2}+\dfrac{\pi}{4}\right)a

Then, the area become:

A=ab+\dfrac{\pi a^2}{8}=a(6-\left(\dfrac{1}{2}+\dfrac{\pi}{4}\right)a)+\dfrac{\pi a^2}{8}\\\\\\A=6a-\left(\dfrac{1}{2}+\dfrac{\pi}{4}\right)a^2+\dfrac{\pi a^2}{8}\\\\\\A=6a-\left(\dfrac{1}{2}+\dfrac{\pi}{4}-\dfrac{\pi}{8}\right)a^2\\\\\\A=6a-\left(\dfrac{1}{2}+\dfrac{\pi}{8}\right)a^2

To maximize the area, we derive and equal to zero:

\dfrac{dA}{da}=6-2\left(\dfrac{1}{2}+\dfrac{\pi}{8}\right )a=0\\\\\\6-(1-\pi/4)a=0\\\\a=\dfrac{6}{(1+\pi/4)}\approx6/1.78\approx 3.36

Then, b is:

b=6-\left(\dfrac{1}{2}+\dfrac{\pi}{4}\right)a\\\\\\b=6-0.393*3.36=6-1.32\\\\b=4.68

3 0
3 years ago
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