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3 years ago

A ball is dropped out a wimdow that is 36 feet high. how long does it take for the ball to hit the ground

1 answer:

4 0

36 feet to metres = 10.97
first find the final velocity
vf^2 = vi^2 + 2ad
vf^2 = 0 + 2(9.81)(10.97)
vf^2 = 215.2314
vf = 14.6708

vf = vi + at
at = vf - vi
t = (vf - vi)/a
t = (14.6708)/9.81
t = 1.4955seconds

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The equation y = \large 1\frac{1}{2}x represents the number of cups of dried fruit, y, needed to make x pounds of granola. Deter

Natasha2012 [34]

Answer:

$(1\frac{1}{2},1)$ - False

$(4,6)$ - True

$(18,12)$ -- False

$(0,0)$ -- True

$(2\frac{1}{2},3\frac{3}{4})$ -- True

Step-by-step explanation:

The points are

$(1\frac{1}{2},1)$ , $(4,6)$ , $(18,12)$ , $(0,0)$ and $(2\frac{1}{2},3\frac{3}{4})$ ---- missing from the question

Given

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

Required

Determine if each of the points would be on $y = 1\frac{1}{2}x$

To do this, we simply substitute the value of x and of each point in $y = 1\frac{1}{2}x$ .

(a) $(1\frac{1}{2},1)$

In this case;

$x = 1\frac{1}{2}$ and $y = 1$

$y = 1\frac{1}{2}x$ becomes

$y = 1\frac{1}{2} * 1\frac{1}{2}$

$y = \frac{3}{2} * \frac{3}{2}$

$y = \frac{9}{4}$

$y = 2\frac{1}{4}$

The point $(1\frac{1}{2},1)$ won't be on the graph because the corresponding value of y for $x = 1\frac{1}{2}$ is $y = 2\frac{1}{4}$

(b) $(4,6)$

In this case;

$x = 4$

$y = 6$

$y = 1\frac{1}{2}x$ becomes

$y = 1\frac{1}{2} * 4$

$y = \frac{3}{2} * 4$

$y = \frac{3* 4}{2}$

$y = \frac{12}{2}$

$y = 6$

The point $(4,6)$ would be on the graph because the corresponding value of y for $x = 4$ is $y = 6$

(c) $(18,12)$

In this case:

$x = 18;y = 12$

$y = 1\frac{1}{2}x$ becomes

$y = 1\frac{1}{2} * 18$

$y = \frac{3}{2} * 18$

$y = \frac{3* 18}{2}$

$y = \frac{54}{2}$

$y = 27$

The point $(18,12)$ wouldn't be on the graph because the corresponding value of y for $x = 18$ is $y = 12$

(d) $(0,0)$

In this case;

$x =0; y = 0$

$y = 1\frac{1}{2}x$ becomes

$y = 1\frac{1}{2} * 0$

$y = 0$

The point $(0,0)$ would be on the graph because the corresponding value of y for $x = 0$ is $y = 0$

(e) $(2\frac{1}{2},3\frac{3}{4})$

In this case:

$x = 2\frac{1}{2}$ ; $y = 3\frac{3}{4}$

$y = 1\frac{1}{2}x$ becomes

$y = 1\frac{1}{2} * 2\frac{1}{2}$

$y = \frac{3}{2} * \frac{5}{2}$

$y = \frac{15}{4}$

$y = 3\frac{3}{4}$

The point $(2\frac{1}{2},3\frac{3}{4})$ would be on the graph because the corresponding value of y for $x = 2\frac{1}{2}$ is $y = 3\frac{3}{4}$

3 0

3 years ago

What is 0.05 repeating as a fraction ?

grin007 [14]

This is the best I got Require to make 2 equations with the same repeating part and subtract them to eliminate the repeating part.

begin by letting x = 0.5555555................. (1)

To obtain the same repeating part after the decimal point need to multiply by 10

hence 10x = 5.555555........................(2)

It is important to obtain 2 equations in x, where the recurring part after the decimal points are exactly the same.

now subtract (1) from (2) to obtain fraction

(2) - (1) : 9x=5⇒x=59

8 0

3 years ago

Which data set contains an outlier?

Bezzdna [24]

The data set which contains an outlier from the given answer choices is; Choice D; {16, 42, 45, 45, 46, 48}.

<h3>What is an outlier?</h3>

An outlier in a set of data values is a data value which differs significantly from other data points and hence, tends to affect the mean of such set of data values significantly.

On this note, choice D is the set of data values which contains an outlier.