Make sure to show your work :))

PLS HELP ME ASAP FOR 1, 2, 3, and 4 (SHOW WORK!!) + LOTS OF POINTS!!

Lera25 [3.4K]

1/2in = 16 mi
2 3/4 in without whole numbers [(denominator ×whole) + numerator] / denominator
4×2 = 8 +3 = 11/4
Find common denominator for 1/2 and 11/4 (common D Is 4)
Multiply 1/2 by 2/2
2/4 in = 16 mi divide this by 2
2/4 ×1/2 = 1/4
16 /2 = 8
1/4 = 8
1/4 goes into 11/4 11 times
Multiply 8 times 11
88 miles (D)

End zone to end zone is = length so his scale is 6 in and in real life it is 360 ft
To find scale divide 360 by 6
1in = 60 ft (B)

160 ft wide
(divide 160 by 60 ft ) fraction
160/60 is a fraction now simplify
16/6
Common factors of both numbers is 2
Simplify by 2
8/3
Make into whole number fraction
3 goes into 8 two times (2×3=6)(2 becomes whole number) while there's is 2 left over( 8 -6=2) (becoming the numerator)
2 2/3 in (B)

120 ft long
80 ft wide

24 in is scale of "long"
Divide 120 by 24
120÷24=5
So 5in in scale equals 1ft in real life
To find width in scale divide 80 by 5
80/5= 16in (A)

If you have any trouble understanding feel free to let me know I will try to explain as best as I can :)

5 0

3 years ago

Emily is entering a bicycle race for charity. Her mother pledges $0.70 for every 0.5 mile she bikes. If Emily bikes 10 miles, ho

Jlenok [28]

Answer:

Emily's mother will donate $20

Step-by-step explanation:

0.5x20=10

7 0

3 years ago

Read 2 more answers

Write a rule to describe the function shown. x y −6 −4 −3 −2 0 0 3 2 y equals start fraction two over three end fraction x y equ

sertanlavr [38]

Answer:

$y=\frac{2}{3} x$ (y equals start fraction two over three end fraction x)

Step-by-step explanation:

Lest's organize the table values of our function first:

x y

-6 -4

-3 -2

0 0

3 2

The simplest kind of of function is a linear function of the form $y=mx+b$ where $m$ is the slope and $b$ is the y-intercept. Let's us check if our function is a linear one finding its equation and checking that all points are in the line.

The first step to find a linear equation is find the slope of the line; to do it, we are using the slope formula:

$m=\frac{y_2-y_1}{x_2-x_1}$

where

$m$ is the slope of the line

$(x_1,y_1)$ are the coordinates of the first point

$(x_2,y_2)$ are the coordinates of the second point

We know from our table that the first and second points are (-6, -4) and (-3, -2) respectively, so $x_1=-6$ , $y_1=-4$ , $x_2=-3$ , and $y_2=-2$ .

Replacing values:

$m=\frac{y_2-y_1}{x_2-x_1}$

$m=\frac{-2-(-4)}{-3-(-6)}$

$m=\frac{-2+4}{-3+6)}$

$m=\frac{2}{3)}$

Now that we have the slope of our line we can use the point slope formula to complete our linear function:

$y-y_1=m(x-x_1)$

where

$m$ is the slope

$(x_1,y_1)$ are the coordinates of the first point

Replacing values:

$y-y_1=m(x-x_1)$

$y-(-4)=\frac{2}{3)}(x-(-6))$

$y+4=\frac{2}{3)}(x+6)$

$y+4=\frac{2}{3} x+4$

$y=\frac{2}{3} x+4-4$

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

Now, to check if our function is valid, we can either check if each point satisfies the rule $y=\frac{2}{3} x$ , or we can graph it and check if each lies is in the graph.

Let's check both:

We already know that points (-6, -4) and (-3, -2) satisfy the rule (after all, those were the point we used to came out with the rule in the first place), so we just need to check the points (0, 0) and (3, 2)

- For (0, 0):

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

$0=\frac{2}{3}(0)$

$0=0$

The point satisfies the rule.

- For (3, 2):

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

$2=\frac{2}{3} (3)$

$2=2$

The point satisfies the rule.

Since all the points of our table satisfies the rule $y=\frac{2}{3} x$ , we can conclude that the rule describing the function shown is $y=\frac{2}{3} x$ .