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Jet001 [13]
3 years ago
14

What improper fraction goes with 1 and three fourths

Mathematics
1 answer:
Katyanochek1 [597]3 years ago
7 0
7/4 because you times 4 and 1.then add 3 and you get 7 so 7/4
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The diagonals of a rectangle are always
guapka [62]

Answer:

The diagonals of a rectangle are always congruent

7 0
2 years ago
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Write as<br> logarithm with base 4:<br> 4<br> -5<br> 2<br> -4
NeX [460]

Answer:

log_4(256)=4

log_4(1/1024)=-5

log_4(16)=2

log_4(1/256)=-4

Step-by-step explanation:

We want to write a number, x, such that

Log_4(y)=x.

In exponential form that is 4^x=y.

So first number is x=4.

4^4=256 which means log_4(256) is 4 as a logarithm with base 4.

The second number is x=-5.

4^-5=1/4^5=1/1024 which means log_4(1/1024) is -5 as a logarithm with base 4.

The third number is x=2.

4^2=16 so log_4(16) is 2 as a logarithm with base 4.

The fourth number is x=-4.

Since 4^4=256 then 4^-4=1/256 which means -4 as a logarithm with base 4 is log_4(1/256).

8 0
2 years ago
Hey hun can any of y'all help me?
11111nata11111 [884]
\bf \cfrac{5}{\frac{1}{6}+\frac{1}{x+1}}\qquad \cfrac{}{\impliedby LCD\textit{ will be 6(x+1)}}\implies \cfrac{5}{\frac{1(x+1)+1(6)}{6(x+1)}}&#10;\\\\\\&#10;\cfrac{5}{\frac{x+1+6}{6(x+1)}}\implies \cfrac{\frac{5}{1}}{\frac{x+7}{6(x+1)}}\implies \cfrac{5}{1}\cdot \cfrac{6(x+1)}{x+7}\implies \cfrac{30(x+1)}{x+7}

and you can expand the numerator if you wish, it won't be simplified further though.
6 0
3 years ago
Problem<br> 3x + 5 + 2x = 12 + 4x
Ratling [72]
X=7 All you need to do is combine like terms and then move the appropriate terms to the appropriate side.
8 0
3 years ago
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A farmer has 1800 m of fencing. He needs to create
NeTakaya

Answer:

67,500 m²

Step-by-step explanation:

ASSUMING the fields look like this __________________

                                                            |                    |                   |

                                                            |                    |                   | W

                                                            |_________|_________|

                                                                                L                                                            

Let L be the length of the combined field and W be the width

2L + 3W = 1800

2L = 1800 - 3W

L = 900 - 1.5W

A = LW

A = (900 - 1.5W)W

A = 900W - 1.5W²

Area will be maximized when the derivative equals zero.

dA/dW = 900 - 3W

        0 = 900 - 3W

     3W = 900

       W = 300 m

L = 900 - 1.5(300)

L = 450 m

A = LW = 450(300) = 135,000 m²

so each sub field is 135000/2 = 67,500 m²

7 0
2 years ago
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