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3 years ago

Given m∥n .

Mathematics

1 answer:

Zielflug [23.3K]3 years ago

3 0

Answer:

Kitna Lamba ha sorry plz

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In na hour , a jogger burns about 4 calories pero pound of body weight. How Manu calories will a jogger who weighs 151 pounds in

Alex777 [14]

This is a rate*unit problem.

So, 4 calories per pound x 151 pounds

$4\times151=604$

He will burn 604 calories in one hour

6 0

3 years ago

Solve the system of equations by elimination.<br> 2x - y + 3z=-1<br> x + 2y - 4z=-1<br> y – 2z=0

MrRa [10]

Answer:

x = -1, y = 2 and z = 1

Step-by-step explanation:

The given system of equations are :

2x - y + 3z= -1 ....(1)

x + 2y - 4z = -1 ......(2)

y – 2z = 0 .....(3)

Equation (3) can be written as :

y = 2z

Use y = 2z in equation (2)

x + 2(2z) - 4z = -1

x + 4z - 4z = -1

x = -1

Put the value of x in equation (1) :

-2 -y +3z = -1

-y+3z = 1 ....(4)

Adding equation (3) and (4)

y-2z+(-y+3z)=1

z = 1

Now put z = 1 in equation (4)

-y+3=1

-y = -2

y = 2

Hence, the values of x,y and z are -1, 2 and 1 respectively.

6 0

3 years ago

For with equation would you add 12 as the first step in the solution?<br>

slavikrds [6]

Answer:

No, you would multiply 12

Step-by-step explanation:

7 0

3 years ago

Is -2 a solution to 4x^(3)-x^(2)+16x-4

sergiy2304 [10]

Answer:

Not a solution

Step-by-step explanation:

In order for $x=-2$ to be a solution, the result after substituting must be 0:

$4(-2)^3-(-2)^2+16(-2)-4\stackrel{?}{=}0$

$4(-8)-4-32-4\stackrel{?}{=}0$

$-32-40\stackrel{?}{=}0$

$-72\neq0$

Therefore, $x=-2$ is not a solution to $4x^3-x^2+16x-4$

3 0

3 years ago

Find the product of all real values of r for which 1/2x=r-x/7

Dahasolnce [82]

Answer:

$r = \±\sqrt{14$

$Product = -14$

Step-by-step explanation:

Given

$\frac{1}{2x} = \frac{r - x}{7}$

Required

Find all product of real values that satisfy the equation

$\frac{1}{2x} = \frac{r - x}{7}$

Cross multiply:

$2x(r - x) = 7 * 1$

$2xr - 2x^2 = 7$

Subtract 7 from both sides

$2xr - 2x^2 -7= 7 -7$

$2xr - 2x^2 -7= 0$

Reorder

$- 2x^2+ 2xr -7= 0$

Multiply through by -1

$2x^2 - 2xr +7= 0$

The above represents a quadratic equation and as such could take either of the following conditions.

(1) No real roots:

This possibility does not apply in this case as such, would not be considered.

(2) One real root

This is true if

$b^2 - 4ac = 0$

For a quadratic equation

$ax^2 + bx + c = 0$

By comparison with $2x^2 - 2xr +7= 0$

$a = 2$

$b = -2r$

$c =7$

Substitute these values in $b^2 - 4ac = 0$

$(-2r)^2 - 4 * 2 * 7 = 0$

$4r^2 - 56 = 0$

Add 56 to both sides

$4r^2 - 56 + 56= 0 + 56$

$4r^2 = 56$

Divide through by 4

$r^2 = 14$

Take square roots

$\sqrt{r^2} = \±\sqrt{14$

$r = \±\sqrt{14$

Hence, the possible values of r are:

$\sqrt{14$ or $-\sqrt{14$

and the product is:

$Product = \sqrt{14} * -\sqrt{14}$

$Product = -14$

8 0

3 years ago