Find the area of the figure below, formed from a triangle and a parallelogram.

How can this quadrilateral be classified? Select from the drop-down menus to complete the statements. This quadrilateral has Cho

Ilia_Sergeevich [38]

Yes, it can since it is parallel.

7 0

3 years ago

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

2 years ago

Find the surface area Will give brainliest to first answer!!!!

user100 [1]

120 I think I’m not sure

7 0

2 years ago

A bean plant grows at a constant rate for a month. After 10 days, the plant is 25 centimeters tall. After 20 days, the plant is

wariber [46]

Y = 2x + 5

If you take the height at 10 days, 25 centimeters, after 10 more days the height is 45 centimeters, meaning for those 10 extra days the been plant grown 20 centimeters. Since we're told the plant is growing at a constant rate, this shows the bean plant is growing 2 centimeters per day. We can represent this with y = 2x. (After 10 days, the bean plant will be 20 centimeters, after 20 days, the bean plant will be 40 centimeters, etc.)

However, this is not completely true yet. As you can see, after the first 10 days the plant is not 20 centimeter, it's 25 centimeters. We already know the rate in which the plant is changing, but now we need to find the height that the plant was originally, before it started growing.

After the first 10 days, the plant is 25 centimeters tall. Since we know that the plant is growing 2 centimeters per day, we can subtract 20 from 25 to find the original height of the bean plant.

25 - 20 = 5
The bean plant was originally 5 centimeters.

This makes our final equation y = 2x + 5.
2x is the slope, and 5 is the y intercept.

Hope this helps1!

5 0

2 years ago

A map has a scale of 1 in. : 38 ft use the given map distance to find the actual distance

Dvinal [7]

Answer:

for q 19 its x=209ft and for 21 its 66.5 ft i forgot q 20

Step-by-step explanation:

8 0

2 years ago