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3 years ago

At what value of x do the graphs of the equations below intersect?

Mathematics

2 answers:

Verizon [17]3 years ago

7 0

Answer with Step-by-step explanation:

The point of intersection of the graphs of system of equation is the solution to the system of equations.

So, we need to find the solution of the system of equations:

2x – y = 6

5x + 10y = –10

Multiplying first equation by 10 and add it to second equation.

5x+10y+10(2x-y)= -10+60

5x+10y+20x-10y=50

25x=50

⇒ x=2

Hence, value of x where the graphs of the equations

2x – y = 6

5x + 10y = –10 intersect is:

x=2

Bogdan [553]3 years ago

3 0

Answer:

Step-by-step explanation:

The intersection is the point where two equations meet. It is calculated by substituting terms into the equations involved. For the given systems of equation, calculations are as follows:

2x - y = 6

y = 2x - 6

We substitute the equation above to the second equation.

5x + 10y = –10

5x + 10( 2x - 6 )= –10

Simplifying,

5x + 20x - 60 = -10

25x = 50

x = 2

Therefore, the intersection has the value of x equal to 2.

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3 years ago

Consider the following system of equations:

IrinaK [193]

Answer:

x = 6 ; y = -9

Step-by-step explanation:

8 x + 2 y = 30 ..........equ No 1

7 x + 2 y = 24......... equ No 2

8 x = 30 - 2y

∴ x = $\frac{ 30 - 2 y}{8}$

substituting the value of x in equ No 2

7 $\frac{30 - 2 y}{8}$ + 2 y = 24

7 ( 30 -2 y) + 2 y × 8 = 24 × 8

7 × 30 - 14 y + 16 y = 192

210 + 2 y = 192

2 y = 192 - 210

2 y = - 18

∴ y = - 9

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3 years ago

Find sin2x, cos2x, and tan2x if sinx=-15/17 and x terminates in quadrant III

vodka [1.7K]

Given:

$\sin x=-\dfrac{15}{17}$

x lies in the III quadrant.

To find:

The values of $\sin 2x, \cos 2x, \tan 2x$ .

Solution:

It is given that x lies in the III quadrant. It means only tan and cot are positive and others are negative.

We know that,

$\sin^2 x+\cos^2 x=1$

$(-\dfrac{15}{17})^2+\cos^2 x=1$

$\cos^2 x=1-\dfrac{225}{289}$

$\cos x=\pm\sqrt{\dfrac{289-225}{289}}$

x lies in the III quadrant. So,

$\cos x=-\sqrt{\dfrac{64}{289}}$

$\cos x=-\dfrac{8}{17}$

Now,

$\sin 2x=2\sin x\cos x$

$\sin 2x=2\times (-\dfrac{15}{17})\times (-\dfrac{8}{17})$

$\sin 2x=-\dfrac{240}{289}$

And,

$\cos 2x=1-2\sin^2x$

$\cos 2x=1-2(-\dfrac{15}{17})^2$

$\cos 2x=1-2(\dfrac{225}{289})$

$\cos 2x=\dfrac{289-450}{289}$

$\cos 2x=-\dfrac{161}{289}$

We know that,

$\tan 2x=\dfrac{\sin 2x}{\cos 2x}$

$\tan 2x=\dfrac{-\dfrac{240}{289}}{-\dfrac{161}{289}}$

$\tan 2x=\dfrac{240}{161}$

Therefore, the required values are $\sin 2x=-\dfrac{240}{289},\cos 2x=-\dfrac{161}{289},\tan 2x=\dfrac{240}{161}$ .

7 0

3 years ago

If 2^(x+3) - 2x = k(2^x) what is the value of k

svlad2 [7]

Answer:

$\large\boxed{C)\ 7}$

Step-by-step explanation:

Use:

$a^n\cdot a^m=a^{n+m}$

$2^{x+3}-2^x=2^x\cdot2^3-2^x=8\cdot2^x-2^x=(8-1)\cdot2^x=7\cdot2^x\\\\2^{x+3}-2^x=k(2^x)\\\\7\cdot2^x=k(2^x)\Rightarrow k=7$

$8\cdot2^x-2^x=8\cdot2^x-1\cdot2^x=(8-1)(2^x)=7\cdot2^x\\\\\text{distributive property:}\ a(b-c)=ab-ac$

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