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77julia77 [94]
3 years ago
5

Graph x+y=6 4x+y=20I NEED HELP ​

Mathematics
1 answer:
Tcecarenko [31]3 years ago
7 0

Answer:Hi is this a simultaneous equation,

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Find the value of x:<br> please help fast
DedPeter [7]

Answer:

BC×AD=K

CKD= 180-32-79= 2x +180- 5x

-> x = (-32-79) : (-3) =37

4 0
3 years ago
6x+9/x-5 x is not equal to 5 find inverse
algol [13]
You have f(x)=(<span>6x+9)/(x-5) and want to find the inverse function?  Please note that the parentheses are obligatory here.

1.  Replace "f(x)" with y:   y = </span>(6x+9)/(x-5)
<span>
2.  Interchange the variables x and y:     x</span> = (6y+9)/(y-5)

3.  Solve this result for y:     x(y-5) = (6y+9), or xy-5x = 6y+9.  Then:
                                                             
                                            xy-6y = 9 + 5x, or y(x-6) = 9+5x
                                                             9+5x
                     Then                      y = --------------                       
                                                             x-6

Here x may not = 6, because division by zero would occur otherwise.
3 0
3 years ago
PLEASE HELP I have been trying to figure this out for a really long time and I want to get the best grade on my test !
Shalnov [3]

Answer:

see below

Step-by-step explanation:

-4^{2} =-16\\\\-\sqrt{4} =-2\\\\\sqrt{4^{2} } =4\\\\-2\sqrt{4} =-4\\\\(-4)^{2} =16\\\\\sqrt{1/4} =1/2

so the order

-16

-4

-2

1/2

4

16

in root forms and exponent

-4^{2} \\\\-2\sqrt{4} \\\\-\sqrt{4} \\\\\sqrt{\frac{1}{4} } \\\\\sqrt{4^{2} } \\\\(-4 )^{2}

8 0
3 years ago
The length of a rectangle is 4 less than 2 times the width. The perimeter is 40 units.
bezimeni [28]

Answer:

Let l be length and w be width.

First equation:

l = 2w-4

Second equation:

perimeter = 40 units

Perimeter of a rectangle: 2(l+w)

2(2w-4+w)=40

2(3w-4)=40

6w-8=40

6w=40+8

6w=48

w=48/6

w=8

Therefore the width=8;

length= 2w-4 = 2(8)-4  = 16-4  = 12

perimeter = 40 units

8 0
3 years ago
The pizza has an area of 78.5 in². What is the minimum width of a placemat that can be placed underneath it so that the pizza do
andrey2020 [161]

Answer:

<h2>   10 in</h2>

Step-by-step explanation:

This problem bothers on the mensuration of flat shapes, circles and square.

we know that the pizza has a  circular profile

and the place-mat has a square profile.

Step one:

let us solve for the radius of the pizza

area-of- pizzza= \pi r^2\\\\  78.5= 3.142*r^2\\\\r^2= 78.5/3.142\\\\r^2= 24.98\\\\r= \sqrt{24.98} \\\\r=4.998\\\\r= 5 in

Step two:

from inspection we know that the width of the place-mat is same as the diameter of the pizza

hence diameter of pizza

= 2r

=2*5

=10 in

3 0
3 years ago
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