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love history [14]
4 years ago
11

The area of a rhombus is 40 in. If

Mathematics
2 answers:
polet [3.4K]4 years ago
7 0

Answer:

10 in.

Step-by-step explanation:

The area of a rhombus is the product of the lengths of the diagonals divided by 2.

Let the diagonals be x and y.

area = xy/2

Here you have

area = 40 in.^2

x = 8 in.

We are looking for y, the other diagonal.

xy/2 = area

(8 in.)y/2 = 40 in.^2

(8 in.)y = 80 in.^2

y = 10 in.

Answer: The other diagonal has length 10 in.

xz_007 [3.2K]4 years ago
5 0

Answer is provided in the image attached.

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Use the zero product property to find the solitions to the equation(x+2)(x+3)=12.
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Answer:

(x + 2) (x + 3) = x² + 3x + 2x + 6 = 12  (multiply the binomials)

(x + 2) (x + 3) = x² + 5x + 6 = 12

(x + 2) (x + 3) = x² + 5x + 6-12 = 0  (add the like term and bring the positive 12 to the other side which makes it negative)

x² + 5x + 6 = 0

Using the snowflake method -  (finding the factors of 6 that adds/subtracts to give you 5).  

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ZERO PROPERTY RULE:

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7 0
3 years ago
HELP ASAP PLEASE ONLY #4,6,8,10,12,14,16​
ExtremeBDS [4]

Answer:

Step-by-step explanation:

Oof, all 6?  That's not fun.  

I'm going to show you one for the first bunch, (4-12) and hopefully you can get the rest.  If not let me know and I can more carefully work you through some others on here.

4)  So first we need the angle.  How do we find that?  Well, we know it's somewhere between 270 and 360 degrees (or 3pi/2 and 2pi radians) since it's in the fourth quadrant.    The method to finding the angle is ifferent for each quadrant, but it's nice to know around where you'll get.  Anyway,, if we take 360 and subtract that little space  that is unlabeled  we will know  the rest.    Hopefully that makes sense, but let me know if not, it's important to understand.  Maybe think of what happens if you add the two together, you get around the whole circle then.  

Anyway, to find that little sliver, we are going to construct a right triangle wherethe x axis is one side.  If you draw a line connecting point p and the x axis you have this right triangle.  I am going to call this sliver we don't know s.  ANyway, now we use trig here.  

s = arcsin(3/5) (you can find 5 pretty easily but let me know if you don't see it.)

You may think to use -3 instead of 3, and  it wouldn't be a huge mistake, it will just give you the negative version of what we want.  

Anyway, now we know theta is 360-s (or 2pi-s).  You can solve s or leave it as arcsin(3/5) so you don't have to round.

Now we can solve the trig functions.  If you don't know the sum and difference trig identities  you will have to round.  They are pretty simple formula though.  For instance, sin(2pi - arcsin(3/5)) = sin(2pi)cos(arcsin(3/5))-cos(2pi)sin(arcsin(3/5)) = 0 - 3/5 = -3/5  You will also want to know that s = arcsin(3/5) = arccos(4/5) = arctan(3/4).  ROunding will get you close but not exact.

Again, let me know ifyou need any more help, they should just be a lot of doing the same thing over and over and over again though.  

14)  For 14 and 16 you just need to know how the graphs of sin, cos and tan look.  Is this something you have trouble with?  Like could you draw the graphs at the intervals of increasing/ decrasing and all.  if not I can give a quick explanation.

7 0
3 years ago
-1/3=j/4-10/3 solve for j
Thepotemich [5.8K]
J = 12

Workings in the picture below :)

3 0
3 years ago
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