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3 years ago

6&7 please...............

Mathematics

1 answer:

pychu [463]3 years ago

8 0

There the same so -37

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Find the point (x1,x2) that lies on the line x1 +5x2 =7 and on the line x1 - 2x2 = -2.

Savatey [412]

Given :

Two equation $x_1 +5x_2 =7$ and $x_1 - 2x_2 = -2$ .

To Find :

The point of intersection of these lines .

Solution :

We will use elimination method :

From equation 1 :

$x_1=7-5x_2$

Putting value of $x_1$ in equation 2 we get :

$(7-5x_2)-2x_2=-2\\\\7-7x_2=-2\\\\x_2=\dfrac{9}{7}$

Putting value of $x_2$ in equation 1 we get :

$x_1+5\times \dfrac{9}{7}=7\\\\x_1=\dfrac{4}{7}$

Therefore , point of interaction is $(\dfrac{4}{7},\dfrac{9}{7})$ .

Hence , this is the required solution .

7 0

3 years ago

Please help me, its urgent ❗❗❗❗❗❗❗❗❗❗❗❗

lorasvet [3.4K]

Answer:

the answer to this one is A.

Step-by-step explanation:

3 0

3 years ago

Write the decimal equivalent of the following: seven hundred seventy-five thousandths?

baherus [9]

Answer:

0.775

Step-by-step explanation:

think place values to the right

tenths, hundredths, thousandths

6 0

3 years ago

How do you do C and B

Hoochie [10]

"the argument" of a function, is simply "the expression being evaluated by that function".

$\bf a)\qquad (x+2)^3\impliedby \stackrel{function}{(\qquad )^3}\qquad \stackrel{argument}{x+2}
\\\\\\
b)\qquad f(2x-1)\impliedby \stackrel{function}{f(\qquad )}\qquad \stackrel{argument}{2x-1}
\\\\\\
c)\qquad \sqrt{x^2+4}\qquad \stackrel{function}{\sqrt{\qquad }}\qquad \stackrel{argument}{x^2+4}
\\\\\\
d)\qquad sin(5x)\impliedby \stackrel{function}{sin(\qquad )}\qquad \stackrel{argument}{5x}$

4 0

3 years ago

- If f ( x ) = x² + 2x - 1 and g ( x ) = 3x - 2 , then verify the following :

RideAnS [48]

Step-by-step explanation:

Hey there!

Given;

f ( x ) = x² + 2x - 1

g ( x ) = 3x - 2

To verify:

$( \frac{f}{g} )(2) = \frac{f(2)}{g(2)}$