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noname [10]
3 years ago
6

The figure below shows a trapezoid, ABCD, having side AB parallel to side DC. The diagonals AC and BD intersect at point O.

Mathematics
2 answers:
Ymorist [56]3 years ago
4 0

the answer is A. half the length of AB I just took the quiz and I got it right hope this helps!

Elan Coil [88]3 years ago
3 0
The answer should be A. half of the length of AB.
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A rhombus and a square have one and the same side of 6 cm. The area of the rhombus is 4/5 of the area of the square. Find the he
Arada [10]
You will find the area the square and then find out what 4/5 of that area is.

A = bh
     6 cm x 6 cm 
A = 36 square cm

4/5 of 36 square cm
4/5 x 36
28 4/5 square cm

The area of the rhombus is 28 4/5 square cm.  Use this to solve for the height of the rhombus.

A = bh
<u>28 4/5</u> = <u>6 x h</u>
6            6
h = 4 4/5 cm

The height of the rhombus is 4 4/5 cm.
7 0
3 years ago
How to write 53.02 in word form
Troyanec [42]
Fifty-three and two hundredths
*I could be wrong*
3 0
3 years ago
Read 2 more answers
Complete the square to solve the equation below.
sasho [114]
Hello there!

x² + x = 7/4
x² + - 7/4 = 0

Now we gonna use the quadratic formula to find x
a= 1
b=1
c = -1.75

x = -b+/-√b² -4ac all of them divide by 2a
x = -(1)+/-√(1)² - (4)(1)(-1.75) all of them divide by 2(1)
x= -1+/-√8 all of them divide by 2
x = -1/2 + √2 or x = -1/2 - √2

The correct option is option C

I hope that helps!
4 0
3 years ago
9x^2+ 8x + 3 = 0 solve using quadratic formula
pav-90 [236]
18x + 8x = -3
26x = -3
x = -3/26
3 0
3 years ago
I’m need this worked out step by step by tonight
Pie

9514 1404 393

Answer:

  -3 ≤ x ≤ 19/3

Step-by-step explanation:

This inequality can be resolved to a compound inequality:

  -7 ≤ (3x -5)/2 ≤ 7

Multiply all parts by 2.

  -14 ≤ 3x -5 ≤ 14

Add 5 to all parts.

  -9 ≤ 3x ≤ 19

Divide all parts by 3.

  -3 ≤ x ≤ 19/3

_____

<em>Additional comment</em>

If you subtract 7 from both sides of the given inequality, it becomes ...

  |(3x -5)/2| -7 ≤ 0

Then you're looking for the values of x that bound the region where the graph is below the x-axis. Those are shown in the attachment. For graphing purposes, I find this comparison to zero works well.

__

For an algebraic solution, I like the compound inequality method shown above. That only works well when the inequality is of the form ...

  |f(x)| < (some number) . . . . or ≤

If the inequality symbol points away from the absolute value expression, or if the (some number) expression involves the variable, then it is probably better to write the inequality in two parts with appropriate domain specifications:

  |f(x)| > g(x)   ⇒   f(x) > g(x) for f(x) > 0; or -f(x) > g(x) for f(x) < 0

Any solutions to these inequalities must respect their domains.

8 0
3 years ago
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