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Alecsey [184]
2 years ago
9

Give an example of infinite solution

Mathematics
2 answers:
valkas [14]2 years ago
3 0

Answer:

?

Step-by-step explanation:

gladu [14]2 years ago
3 0
An example of an i finite solution is
3=3
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9+8{7•6-5[4+(3-2•1)]}=?​
34kurt

Answer:

145

Step-by-step explanation:

Ⓗⓘ ⓣⓗⓔⓡⓔ

9+8{7*6-5[4+(3-2*1)}]}

9+8{7*6-5[4+1]}

9+8{7*6-25}

9+8{42-25}

9+136

145

(っ◔◡◔)っ ♥ Hope this helped! Have a great day! :) ♥

Please, please give brainliest, it would be greatly appreciated, I only a few more before I advance, thanks!

5 0
3 years ago
Find the area of the parallelogram that has a base of 4m and a height of 5.5m.
Sveta_85 [38]

Answer:

22m²

Step-by-step explanation:

Area of parallelogram is simply the base length x height.

In this case, base length = 4m and height = 5.5m

hence,

Area = 4 x 5.5 = 22m²

3 0
3 years ago
Read 2 more answers
Solve for x and y 9/x-4/y=8
borishaifa [10]
You can solve <span>9/x-4/y=8 for x or for y but NOT at the same time.

Solving for x:  </span><span>9/x-4/y=8       Mult all 3 terms by xy to eliminate the fractions.

9(xy)/x - 4xy/y = 8xy         =>       9y - 4x = 8xy, or 9y = 4x + 8xy = 4x(1+2y)

                                                                        then 9y = x [ 4(1+2y) ]
                                                                                     9y
                                                            therefore x = ------------
                                                                                  4(1+2y)

Solve for y using a similar approach.

</span>
7 0
3 years ago
Whats -0.25 as a fraction
jonny [76]
-0.25 would be -25/100 which can be reduced to -1/4
3 0
3 years ago
a circle is inscribed in a square. the circumference of the circle is increading at a constant rate of 6 inches per second. As t
Burka [1]

Answer:

The rate at which Perimeter of the square is increasing is \frac{24}{\pi} \ in/secs.

Step-by-step explanation:

Given:

Circumference of the circle = 2\pi r

Rate of change of in circumference = 6 in/secs

We need to find the rate at which the perimeter of the square is increasing

Solution:

Now we know that;

\frac{d(2\pi r)}{dt} =6\\\\2\pi\frac{dr}{dt}=6\\\\\frac{dr}{dt}=\frac{6}{2\pi}\\\\\frac{dr}{dt}=\frac{3}{\pi}

Now we know that;

side of the square= diameter of the circle

side of the square = 2r

Now Perimeter of the square is given by 4 times length of the side.

P=4\times 2r =8r

Now we need to find the rate at which Perimeter is increasing so we will find the derivative of perimeter.

\frac{dP}{dt}= \frac{d(8r)}{dt}\\\\\frac{dP}{dt}= 8\times\frac{dr}{dt}

But \frac{dr}{dt} =\frac{3}{\pi}

So we get;

\frac{dP}{dt}= 8\times\frac{3}{\pi}\\\\\frac{dP}{dt}= \frac{24}{\pi}\  in/sec

Hence The rate at which Perimeter of the square is increasing is \frac{24}{\pi} \ in/secs.

5 0
3 years ago
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