Find cordninates of the midpoint of the segment with each pair of endpoints

Fill in the blank and which goes where, drag and drop.. 80% 95% 33 1/3% 15%

scoundrel [369]

1. = 80 2 = 95 3 = 33 4 = 15

4 0

3 years ago

Solve 27 and 29, will give BRAINLIST!!

otez555 [7]

Answer:

see below

Step-by-step explanation:

27.

We can use ratios to solve

152 dollars 532 dollars

---------------- = ----------------

4 days x days

Using cross products

152x = 4*532

152x =2128

Divide each side by 152

152x/152 = 2128/152

x =14

14 days

29.

1 inch 2/3 inches

------------ = ----------------

6.5 ft x ft

Using cross products

1x = 2/3 * 6.5

x = 4.333333

c = 4 1/3 ft

5 0

3 years ago

Read 2 more answers

Find the perimeter of a rectangle whose length is 3 cm greater than its height, and whose area is 40 cm².

jonny [76]

Answer: $26\ cm$

Step-by-step explanation:

The perimeter of a rectangle can be calcualated with this formula:

$P=2l+2h$

Where "l" is the lenght and "h" is the height.

The area of a rectangle can be found with this formula:

$A=lh$

Where "l" is the lenght and "h" is the height.

In this case we know that:

$A=40\ cm^2$

Therfore, we can susbsitute them into $A=lh$ and solve for "h" in order to find its value:

$40=(h+3)h\\\\40=h^2+3h\\\\h^2+3h-40=0$

Find two number whose sum is 3 and whose product is -40. These are -5 and 8. Then, factorizing, we get:

$(h-5)(h+8)=0\\\\h_1=5\\\\h_2=-8$

The positive value is the height. Then:

$h=5\ cm$

Since the length is 3 centimeters greater than its height, we get that this is: Then:

$l=5\ cm+3\ cm\\\\l=8\ cm$

Substituting values into $P=2l+2h$ , we get that the perimeter is:

$P=2(8\ cm)+2(5\ cm)\\\\P=26\ cm$

8 0

4 years ago

Read 2 more answers

Solve for P I am so confused

Jlenok [28]

Answer:

3q/2

Step-by-step explanation:

3p+3q=p

2p=-3q

p=-3q/2

4 0

3 years ago

Read 2 more answers

Given that 8 bolts and 6 nuts weigh 138grams and 3 bolts and 5 nuts weigh 71 grams. Find the weight of;(a)one bolt and one nut (

Viktor [21]

Answer:

(a) Weight of a bolt = 12 grms and weight of a nut is 7 gms. Weight of one bolt and one nut = 12 + 7 = 19 gms

(b) 69 gms

(c) 209 gms

Step-by-step explanation:

This problem relates to solution of 2 unknowns using simultaneous equations.

Let B be the weight of a single bolt and N be the weight of a single nut

Since 8 bolts and 6 nuts weigh 138 grams we get one of the equations as

8B + 6N = 138 (1)

The second equation in the unknowns is

3B + 5N = 71 (2)

To solve, we eliminate one of the unknowns by making its coefficients the same and subtracting one from the other

Multiplying (1) by 5 ===> 40B + 30N = 690 (3)

Multiplying (2) by 6 ===> 18B + 30N = 426 (4)

(3) - (4) eliminates the N variables and yields

22B = 264 ==> B = 264/22 ==> B = 12

So the weight of a single bolt is 12 grams

We can find the weight of a single nut by substituting this value of B into any of the equations (1), (2), (3) or (4) and solving for N. Let's use equation (2)

3(12) + 5N = 71 ==> 36 + 5N = 71 ==> 5N = 71-36 = 35 ==> N= 7

So the weight of a single nut is 7 grams

Weight of one bolt and one nut is the sum of the above individual weights = 12 + 7 = 19 gms

To solve (b) and (c) we could set up two other equations and plug in values for B and N

(b) 4B + 3N = 4.12 + 3.7 = 69
However, an alternate way is to perceive that 4B + 3N is exactly half of 8B + 6N so the value of that must be 138/2 = 69

(c) If we add both equations (1) and (2), we get

11B + 11N = 138 + 71 = 209 which is the equation for the total weight of 11 bolts and 11 nuts

5 0

2 years ago