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Tpy6a [65]
2 years ago
11

Solve by elimination -2x + 6y = 16 - 4x - 3y = 2 please help

Mathematics
1 answer:
prohojiy [21]2 years ago
7 0

Answer:  

(−2,2)

Hope it helps!

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I need help (40pts)I will give brainliest & I put the screenshot of question
gizmo_the_mogwai [7]

Answer:

168 ft

Step-by-step explanation:

To calculate the area of a parallelogram, the formula is Area=base*height

In this case, the base is 14ft and the height is 12ft

so,

14ft*12ft=168ft^2

He needs 168ft or fabric to cover the area

5 0
2 years ago
-1/3 + 5e = -3/4 <br> Pls show your work
Citrus2011 [14]

Answer:

it has. no solution

Since 5e- 1/3 = 3/4  is false, there is no solution.

4 0
2 years ago
In what ratio of line x-y-2=0 divides the line segment joining (3,-1)and (8,9)​
NISA [10]

Answer:

The ratio in which the line x-y-2=0 divides the line joining A(3,-1) and B(8,9) is in the ratio 2:3

Step-by-step explanation:

Let (x, y) be the coordinates of point of intersection.

Hence x=(a*8+1*3)(a+1) = (8a+3)/(a+1)

and

y = {a*9+1*(-1)}/(a+1)=(9a-1)/(a+1)

Since this point lies on the line x-y-2=0

Hence (8a+3)/(a+1)-(9a-1)/(a+1)-2=0

i.e. 8a+3–9a+1–2(a+1)=0

Or 8a+3–9a+1–2a-2=0

i.e.-3a+2=0

Hence a=2/3

hence the ratio in which the line x-y-2=0 divides the line joining A(3,-1) and B(8,9) in the ratio 2/3:1

i.e. 2:3

6 0
3 years ago
Mark cast a shadow that is 4 feet long. At the same time, a nearby tree casts a shadow that is 20 feet long. If mark is 6 feet t
Trava [24]

Answer:

30 ft

Step-by-step explanation:

We can use the given information to set up a ratio...

\frac{Mark's\ height}{Mark's\ shadow}=\frac{Tree's\ height}{Tree's\ shadow}

\frac{6\ ft}{4\ ft}=\frac{x}{20\ ft}

We can solve this ratio using cross multiplication (multiply the numerator of the first fraction by the denominator of the second and the numerator of the second fraction by the denominator of the first):

120 = 4x

Divide both sides by 4

x=30 ft

5 0
2 years ago
(cos x - (sqrt 2)/2)(sec x -1)=0
Darya [45]

(\cos x-\frac{\sqrt{2}}{2})(\sec x-1)=0 [/tex]

=(\cos x-\frac{1}{\sqrt{2}})(\sec x-1)=0 [/tex]

\frac{(\sqrt{2}\cos x-1)}{\sqrt{2}}(\frac{1}{\cos x\ }-1)=0

(Reciprocal Identity)

(\frac{^{\sqrt{2}\\cos  x-1}}{^{\sqrt{2}}})(\frac{1-\cos x}{\cos x})=0

\frac{^{(\sqrt{2}\cos x-1})}{\sqrt{2}}\frac{(1-\cos x)}{\cos x}=0

(\sqrt{2}\cos x-1}){(1-\cos x)}=0 (ZeroProduct Property)

\sqrt{2}\cos x-1=0

\sqrt{2}\cos x=1

\cos x=\frac{1}{\sqrt{2}}

x=\frac{\Pi }{4}

and

1-\cos x=0

\cos x=1

x=0

x=0 and x=\frac{\Pi }{4} are the solutions.

6 0
3 years ago
Read 2 more answers
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