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3 years ago

Math question, please help!

Mathematics

1 answer:

valkas [14]3 years ago

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Suppose point z is halfway between point w and x a standard numberline, and point x is halfway between points z and y. Where is

Delicious77 [7]

Answer:

The point w is located at -4/3

Step-by-step explanation:

Point z is halfway between point w and x

Point x is halfway between point z and y

Point z is located at 1/3 and point y is located at 11/3

Please refer to the number line attached

Then the x is located at the mid-point of z and y

$x = \frac{z+y}{2} \\\\x = \frac{\frac{1}{3}+ \frac{11}{3} }{2}$

$x = 2$

Since z is the mid-point of w and x

$z = \frac{w+ x }{2}\\\\\frac{1}{3} = \frac{w+ 2 }{2}\\\\\frac{2}{3} = w+ 2 \\\\w = \frac{2}{3} - 2\\\\w = -\frac{4}{3}$

Therefore, the point w is located at -4/3

6 0

3 years ago

Is the ratio 50 yards to 18 seconds a rate? Explain.

Anna11 [10]

Answer:50:18

Step-by-step explanation:

I think that this is the answer because 50 yards can be ran in 18 seconds so for every 18 seconds you get 50 yards

7 0

4 years ago

Given below are lease terms at the local dealership. What is the total

tiny-mole [99]

Answer:

Depreciation fee = $5,850

Step-by-step explanation:

Kindly check attached picture for detailed calculation

3 0

3 years ago

Graph each absolute value function. State the domain, range, and y-intercept.

Inessa05 [86]

Answer:

i. D:All real numbers

ii.R: $y\le-2$

iii. Y-int: $b=-2.5$

Step-by-step explanation:

The given function is

$y=-\frac{1}{2} |x-1|-2$

The domain of this function are the values of x that makes the function defined.

The absolute value function is defined for all real values.

The domain is all real numbers.

ii. The range refers to the values of y for which x is defined.

$y=-\frac{1}{2} |x-1|-2$

The vertex of this function is

$(1,-2)$

The function is reflected in the x-axis.

The vertex is therefore the maximum point on the graph of the function.

The range is therefore;

$y\le-2$

Or

$(-\infty,2]$

iii. To find the y-intercept, we substitute $x=0$ into the funtion.

$y=-\frac{1}{2} |0-1|-2$

$y=-\frac{1}{2} (1)-2$

$y=-2.5$

The y-intercept is $b=-2.5$

The graph is shown in the attachment.

3 0

3 years ago

1. Una escalera de 4 m de longitud se apoya sobre una pared vertical. Si la distancia entre la base de la escalera a la pared es

enot [183]

Answer:

puedes resolver de dos maneras si diste teorema de Pitágoras lo aplicas

hipotenusa al cuadrado = cateto 2 +cateto 2 ( el 2 significa al cuadrado)

sustituyes

hipotenusa= 4m

cateto= 2,5

hay que hallar el otro cateto que nos daría la altura a la que está la escalera

despejamos y tenemos

cateto 2= hip2 -cat2

cat2=(4)2-(2,5)2=

= 16-6,25=9,75

luego hallamos la raíz cuadrada de 9,75= 3,12

la altura a la que se encuentra es 3,12m

-si aun no diste Pitágoras podes representarlo en una hoja utilizando cm en lugar de metros( a escala). trazas el triángulo rectángulo la base te la da la distancia a la cual se encuentra la escalera de la pared es decir 2,5cm trazas la hipotenusa de 4cm de manera que coincida con el cateto opuesto , y mides el valor de este ,será de 3,12cm no olvides que la respuesta la debes dar en metros ya que es la unidad de medida que te da.

4 0

3 years ago