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maw [93]
3 years ago
9

The diagonals of a rhombus are 21 m and 32m. what is the area of the rhombus

Mathematics
1 answer:
fredd [130]3 years ago
7 0
With area: multiply width and length

21•32 = 672
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Identify Slope and Y intercept for the equation X=4 M=B=
krek1111 [17]
Undefined and Undefined

Since the equation is x = 4, there is no y-intercept as x will never equal 0.

Additionally, since the equation is x = 4, there is no change in x value and as such the slope can not be determined.
6 0
3 years ago
(08.06 MC)
const2013 [10]

Answer:

  (3.5, 17)

Step-by-step explanation:

It would be nice to see the whole graph, so we can see where the functions cross.

Without that information, we can still eliminate unreasonable choices.

A) the quadratic at y=3.5 is well above the exponential

B) the most likely choice (3.5, 17)

C) at x=-8, the quadratic is above the exponential

D) neither graph goes anywhere near y = -8

8 0
3 years ago
Read 2 more answers
If angle 1 = 6n + 1 and angle 4 = 4n + 19, what is the angle 2?
kobusy [5.1K]
Angles 1 and 4 are alternate angles and thus have the same measurement. You will set their equations equal to each other and solve for the variable. 

6n + 1 = 4n + 19

Subtract 1 and 4n from both sides to get variables on one side and constants on the other. 

6n - 4n = 19 - 1

Combine like terms and solve for n.

2n = 18
n = 18 / 2
n = 9

Angles 1 and 2 are congruent as labeled in the graph. 

So angle 2 is also 6n + 1

Plug in for 9 for n

6(9) + 1
54 + 1
55

Option C is your answer. 


7 0
3 years ago
PLEASE HELP! I WILL GIVE BRAINLIEST! PLEASE SHOW YOUR WORK TOO :)!
Deffense [45]
2.75 as a fraction is 25/100 then you divided which you get 11/4 which is 2 3/4 so 2 3/5 is more hope that helped
4 0
3 years ago
Find all points having an x-coordinate of 2 whose distance from the point (-1,-2) is 5
mariarad [96]
The\ equation\ of\ the\ circle:(x-a)^2+(y-b)^2=r^2\\\\where\ (a;\ b)\ -the\ coordinates\ of\ the\ center;\ r-the\ radius\\-----------------------------\\(-1;-2)-the\ center\ of\ the\ circle\\r=5\\\\The\ equation:(x+1)^2+(y+2)^2=5^5\\\\Put\ x=2\ to\ the\ equation:\\\\(2+1)^2+(y+2)^2=25\\3^2+(y+2)^2=25\\9+(y+2)^2=25\\(y+2)^2=25-9\\(y+2)^2=16\iff y+1=\pm\sqrt{16}\\\\y+1=-4\ or\ y+1=4\ \ \ \ \ |subtract\ 1\ from\ both \sides\\y=-5\ or\ y=3\\\\Answer:(2;-5)\ and\ (2;\ 3).
5 0
3 years ago
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