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bagirrra123 [75]
3 years ago
11

What is the area of figure abcd? a trapezoid abcd is drawn with length of parallel sides ab and cd equal to 10 inches and 12 inc

hes, respectively. the length of one of the non-parallel sides bc is 6 inches. side bc is perpendicular to dc. a perpendicular line drawn from a to side cd meets cd at e. 54 square inches 60 square inches 66 square inches 72 square inches
Mathematics
1 answer:
Mama L [17]3 years ago
4 0

Answer: 60 square inches

Step-by-step explanation:

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Can someone plz help hereI got confused
Rainbow [258]

Answer:

Option 4

Step-by-step explanation:

-½ means it's -0.5, and then -0.4 then -3/10 = -0.3 and then -0.23, therefore it goes -½, -0.4, -3/10, then -0.23

8 0
3 years ago
How is the sum expressed in sigma notation?<br><br> 11 + 17 + 23 + 29 + 35 + 41
Advocard [28]

Answer:

The given sums 11+17+23+29+35+41  can be written as in sigma notation is  \sum\limits_{i=1}^{6}5+6i

Therefore 11+17+23+29+35+41=\sum\limits_{i=1}^{6}5+6i

Step-by-step explanation:

Given series is 11+17+23+29+35+41

To find the given sum expressed in sigma notation :

11+17+23+29+35+41  can be written as the given sums in sigma notation is  \sum\limits_{i=1}^{6}5+6i

That is

11+17+23+29+35+41=\sum\limits_{i=1}^{6}5+6i

Now expand the sums and to verify that whether the summation is true or not :

\sum\limits_{i=1}^{6}5+6i=(5+6(1))+(5+6(2))+(5+6(3))+(5+6(4))+(5+6(5))+(5+6(6))

=(5+6)+(5+12)+(5+18)+(5+24)+(5+30)+(5+36)

=11+17+23+29+35+41

Therefore 11+17+23+29+35+41=\sum\limits_{i=1}^{6}5+6i

8 0
3 years ago
Show that there do not exist scalars c1, c2, and c3 such that c1(1, 0, 1, 0) + c2(1, 0, -2, 1) + c3(2, 0, 1, 2) = (1, -2, 2, 3)
Aloiza [94]

Write the system in augmented-matrix form:

c_1(1,0,1,0)+c_2(1,0,-2,1)+c_3(2,0,1,2)=(1,-2,2,3)

\iff\left[\begin{array}{ccc|c}1&1&2&1\\0&0&0&-2\\1&-2&1&2\\0&1&2&3\end{array}\right]

Row reduce this matrix:

  • Add -1(row 1) to row 3:

\left[\begin{array}{ccc|c}1&1&2&1\\0&0&0&-2\\0&-3&-1&1\\0&1&2&3\end{array}\right]

  • Add 3(row 4) to row 3:

\left[\begin{array}{ccc|c}1&1&2&1\\0&0&0&-2\\0&0&5&10\\0&1&2&3\end{array}\right]

  • Multiply row 3 by 1/5:

\left[\begin{array}{ccc|c}1&1&2&1\\0&0&0&-2\\0&0&1&2\\0&1&2&3\end{array}\right]

  • Add -2(row 3) to row 4:

\left[\begin{array}{ccc|c}1&1&2&1\\0&0&0&-2\\0&0&1&2\\0&1&0&-1\end{array}\right]

  • Add -2(row 3) and -1(row 4) to row 1:

\left[\begin{array}{ccc|c}1&0&0&-2\\0&0&0&-2\\0&0&1&2\\0&1&0&-1\end{array}\right]

This matrix tells us that c_1=-2, c_2=-1, and c_3=2, but clearly 0c_1+0c_2+0c_3=0\neq-2, so there is no solution.

3 0
3 years ago
Find the midpoint of the segment with the following endpoints.<br> (9,8) and (3,2)
BaLLatris [955]

Answer:

(6, 5)

Step-by-step explanation:

add x's divide by 2 add y's divide by two

8 0
3 years ago
Classify this triangle by its sides.
ivann1987 [24]

Answer:

isosceles but not equilateral

<h2>stay safe healthy and happy.</h2>

4 0
3 years ago
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