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3 years ago

What is the solution to -4x+5y=14 and 7x+3y=-1 I need the answer now.

2 answers:

8 0

X= -1 and y=2
Proof:
-4x+5y=14 (equation 1)
7x+3y= -1 (equation 2)

3y= -7x -1
y= -7/3x -1/3 (equation 3)
(3) into (1)
-4x+5(-7/3x-1/3)=14
-4x-35/3x-5/3=14
-47/3x-5/3=14
-47/3x=47/3
x= (-47/3)/ (47/3)
x= -1
sub x= -1 into (3)
y= -7/3(-1)-1/3
y= 2

Serjik [45]3 years ago

7 0

One way to solve is to try to cancel out some terms
we till try to cancel out the y terms

-4x+5y=14
7x+3y=-1
find lcm of 3 and 5 the lcm=15
-4x+5y=14
multily by 3
-12x+15y=42

7x+3y=-1
multiply both sides by -5
-35x-15y=5
add 2 equations together
-12x-35x+15x-15x=42+5
add like terms
-47x=47
divide both sides by 47
-x=1
multpiply by -1
x=-1

subsitute -1 for x in one quation

7x+3y=-1
x=-1
7(-1)+3y=-1
-7+3y=-1
add 7 to both sides
3y=6
divide both sides by 3
y=2

x=-1
y=2

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7. The hexagon GIKMPR is regular. The dashed line segments form 30 degree angles. what is the image of oh after a rotation of 18

velikii [3]

Answer:

The image of OH is ON.

Step-by-step explanation:

Te figure GIKMPR is a regular hexagon. The number of vertices of a regular hexagon is 6. The central angle between any two consecutive vertices is 60 degree.

The dashed line segments form 30 degree angles.

If we rotate the hexagon 180 degree about O, then the each point shifts to 6th place from its original place.

Since we rotate the hexagon 180 degree about O, so the image and preimage lies on a straight line. Because a straight line make angle of 180 degree.

The line OH and ON lies on a straight line therefore the image of OH is ON.

3 0

3 years ago

Consider the following function. f(x) = 16 − x2/3 Find f(−64) and f(64). f(−64) = f(64) = Find all values c in (−64, 64) such th

VARVARA [1.3K]

Answer:

This does not contradict Rolle's Theorem, since f '(0) = 0, and 0 is in the interval (−64, 64).

Step-by-step explanation:

The given function is

$f(x)=16-\frac{x^2}{3}$

To find $f(-64)$ , we substitute $x=-64$ into the function.

$f(-64)=16-\frac{(-64)^2}{3}$

$f(-64)=16-\frac{4096}{3}$

$f(-64)=-\frac{4048}{3}$

To find $f(64)$ , we substitute $x=64$ into the function.

$f(64)=16-\frac{(64)^2}{3}$

$f(64)=16-\frac{4096}{3}$

$f(64)=-\frac{4048}{3}$

To find $f'(c)$ , we must first find $f'(x)$ .

$f'(x)=-\frac{2x}{3}$

This implies that;

$f'(c)=-\frac{2c}{3}$

$f'(c)=0$

$\Rightarrow -\frac{2c}{3}=0$

$\Rightarrow -\frac{2c}{3}\times -\frac{3}{2}=0\times -\frac{3}{2}$

$c=0$

For this function to satisfy the Rolle's Theorem;

It must be continuous on $[-64,64]$ .

It must be differentiable on $(-64,64)$ .

and

$f(-64)=f(64)$ .

All the hypotheses are met, hence this does not contradict Rolle's Theorem, since f '(0) = 0, and 0 is in the interval (−64, 64) is the correct choice.

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3 years ago

Read 2 more answers

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Butoxors [25]

<span>3x(x^2+4)
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hope it helps</span>

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3 years ago

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