All categories

2 years ago

What is the value of the expression (3a-b)/(6(b+a) when a = 4 and b = -2?

Mathematics

1 answer:

Radda [10]2 years ago

6 0

Answer:

5/8 or 0.625

I hope this helped!

You might be interested in

Write each sum as a product of two factors for 6•7+3•7

lapo4ka [179]

65 is the product of this

6 0

2 years ago

John has $20.00 to spend for his mother's birthday. A string of pearls and a pair of pearl earrings costs $25.00. They are on sa

Phoenix [80]

Answer:

2.50

Step-by-step explanation:

30% of 25 is 7.50 take that away from 25 gives you 17.50 take that away from 20 gives you 2.50

4 0

2 years ago

Given that h(x) = 3x −19, find the value of x that makes h(x) = 71.

rosijanka [135]

Keywords:

<em>Equation, variable, value, clear </em>

For this case we have an equation with a variable of the form $y = h (x)$ . Where $h (x) = 3x-19$ . Given the value of $h (x) = 71$ , we want to find the value of the variable "x". So, we have:

$h (x) = 3x-19\\71 = 3x-19$

We must clear "x", for this, we add "19" to both sides of the equation:

$71 + 19 = 3x-19 + 19\\90 = 3x$

We divide between "3" on both sides of the equation:

$\frac {90} {3} = \frac {3x} {3}\\30 = x$

Thus, the value of the variable "x" is 30.

Answer:

$x = 30$

Option A

6 0

2 years ago

Read 2 more answers

13. Greg is going into business selling homemade blankets.

Orlov [11]

Answer:

all positive inegers

Step-by-step explanation:

8 0

2 years ago

a circle has a center (3,5) and a diameter AB. The coordinates of A are (-4,6). what are the coordinates of B?

Kruka [31]

Find the length of the radius.

$r= \sqrt{(3-(-4))^2+(5-6)^2}
\\r = \sqrt{7^2+(-1)^2}
\\r= \sqrt{50} 
\\ r=5 \sqrt{2}$

Find the length of the diameter.

d = 2r = 2 × 5√2 = 10√2

Point B must lie on a line AC, where C is a center of a circle.
Find equation of line AC.
A(–4, 6), C(3, 5)

$y-y_1= \frac{y_2-y_1}{x_2-x_1}(x-x_1)
\\y-6= \frac{5-6}{3-(-4)} (x-(-4))
\\y-6=- \frac{1}{7} (x+4)
\\7y-42=-x-4
\\x+7y-38=0$

The distance from B(x, y) to C(3, 5) is 5√2.
$\sqrt{(x-3)^2+(y-5)^2}=5 \sqrt{2} 
 \\(x-3)^2+(y-5)^2=(5 \sqrt{2})^2 
\\(x-3)^2+(y-5)^2=50$

Solve system of equations.
$x+7y-38=0
\\(x-3)^2+(y-5)^2=50
\\
\\x=38-7y
\\(38-7y-3)^2+(y-5)^2=50
\\(35-7y)^2+(y-5)^2=50
\\1225-490y+49y^2+y^2-10y+25-50=0
\\50y^2-500y+1200=0
\\y^2-10y+24=0
\\y^2-6y-4y+24=0
\\y(y-6)-4(y-6)=0 \\(y-6)(y-4)=0 \\y_1=6,y_2=4
$

$x_1=38-7y_1=38-7 \times 6 = 38-42=-4
\\x_2=38-7y_2=38-7 \times 4 = 38-28=10
$

Point B could have coordinates
$\\
\\(x_1,y_1)=(-4,6),(x_2,y_2)=(10,4)
$

But (–4, 6) are the coordinates of point A.
Therefore, point B has coordinates (10,4).

8 0

3 years ago