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alexira [117]
2 years ago
12

Find the functions y-intercept algebraically

Mathematics
1 answer:
grigory [225]2 years ago
3 0
To find the y intercept using the equation of the line, plug in 0 for the x variable and solve for y. If the equation is written in the slope-intercept form, plug in the slope and the x and y coordinates for a point on the line to solve y. Hope this helped! :)
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An oblong box has a volume equal to lwh, where l is the length, w is the width, and h is the height. If the volume is 24 cubic f
rjkz [21]
<span>An oblong box has a volume equal to lwh, where l is the length, w is the width, and h is the height. If the volume is 24 cubic feet, solve for the height in terms of the other sides.

Given:
volume of 24 cubic feet

Required:
height

Solution:
V = 24 cubic feet
assume that the length, weight and height of the box are all equal
so l = w = h

24 = l^3
l = 2.88 feet</span>
3 0
2 years ago
Read 2 more answers
a direct variation function contains the points ( -8, -6 ) and (12,9 ) . Which equation represents the function ?
9966 [12]

Answer:

y = \frac{3}{4} x

Step-by-step explanation:

Given that the quantities vary directly then the equation relating them is

y = kx ← k is the constant of variation

To find k use either of the 2 given points

Using (12, 9), that is x = 12 when y = 9, then

k = \frac{y}{x} = \frac{9}{12} = \frac{3}{4}

y = \frac{3}{4} x ← equation of variation

8 0
1 year ago
What is the sum of 9.8 and 573.54
Setler79 [48]
583.34 Youre Welcome!
4 0
1 year ago
In the problems below, f(x) = log₂x and
kramer

Answer:

(1,0)

Step-by-step explanation:

The given functions are:

f(x) = log₂x

and

g(x) = log₁₀x

We know that logarithm of 1 is always zero.

This means that irrespective of the base, the y-values of both functions will be equal to 0 at x=1

Therefore the point the graphs of f and g have in common is (1,0)

3 0
1 year ago
Juana tiene en su tienda un costal con 28 libras de azucar.Hizo seis paquetes de 2,5 kg cada uno,de los cuales vendio cinco.¿Cua
Likurg_2 [28]

Answer:

Al comienzo Juana tiene 28 libras de azúcar.

Ella hace 6 paquetes de 2.5kg cada paquete.

Entonces, entre los 6 paquetes, hay:

6*2.5kg = 15kg de azúcar.

Ahora el problema es que tenemos dos unidades distintas, kilogramos y libras.

Aca podemos usar la relación:

1k = 2.2lb.

Entonces 15kg de azúcar son 15 veces 2.2 libras:

15kg = 15*2.2lb = 33lb.

Esto nos da mas de lo que hay originalmente en el costal.

Ahora si lo pensamos como:

Solo vendió 5 paquetes de 2.5kg, entonces el sobrante queda en el costal inicial:

5*2.5kg = 12.5kg

De vuelta, 1kg = 2.2lb

12.5kg = 12.5*2.2lb = 27.5lb

En este caso hay un sobrante de:

28lb - 27.5lb = 0.5lb

6 0
2 years ago
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