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nadezda [96]
4 years ago
15

Find the area of the figure.

Mathematics
1 answer:
OLga [1]4 years ago
6 0

Answer:

63in

Step-by-step explanation:

You add 5in plus 4in since they are both on the same side which would equal 9, then you go to the bottom and add 5in plus 2in which equals 7in, then you use the formula for area which is length times width and you get 63in.

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What is the slope of a line parallel to y = 4/3x-2
den301095 [7]

I'm not that sure but I want to help

Step-by-step explanation:

y=4/3×-2

y=2/3×

Divide 2/3×

4 0
3 years ago
Simplify (8z - 10) ÷ (-2) + 5(z - 1).       plz show work 
Bess [88]
(8z - 10) ÷ (-2) + 5(z - 1) = 8z - 10 ÷ -2 + 5z - 5 = 8z + 5 + 5z - 5 = 13z
6 0
4 years ago
Need help solving for X
Stels [109]
I hope this helps you



x=108/6


x=18

4 0
3 years ago
The first term of a geometric sequence is 32, and the 5th term of the sequence is 818 .
sammy [17]

Answer:

32,24,18,\frac{27}{2} ,\frac{81}{8}

Step-by-step explanation:

Let x, y , and z be the numbers.

Then the geometric sequence is 32,x,y,z,\frac{81}{8}

Recall that  term of a geometric sequence  are generally in the form:

a,ar,ar^2,ar^3,ar^4

This implies that:

a=32 and ar^4=\frac{81}{8}

Substitute a=32 and solve for r.

32r^3=\frac{81}{8}

r^4=\frac{81}{256}

Take the fourth root to get:

r=\sqrt[4]{\frac{81}{256} }

r=\frac{3}{4}

Therefore x=32*\frac{3}{4} =24

y=24*\frac{3}{4} =18

z=18*\frac{3}{4} =\frac{27}{2}

8 0
3 years ago
Polynomial of degree 4 has 1 positive real root that is bouncer and 1 negative real root that is a bouncer. How many imaginary r
Rainbow [258]

Answer:

<h3>The given polynomial of degree 4 has atleast one imaginary root</h3>

Step-by-step explanation:

Given that " Polynomial of degree 4 has 1 positive real root that is bouncer and 1 negative real root that is a bouncer:

<h3>To find how many imaginary roots does the polynomial have :</h3>
  • Since the degree of given polynomial is 4
  • Therefore it must have four roots.
  • Already given that the given polynomial has 1 positive real root and 1 negative real root .
  • Every polynomial with degree greater than 1  has atleast one imaginary root.
<h3>Hence the given polynomial of degree 4 has atleast one imaginary root</h3><h3> </h3>

8 0
3 years ago
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