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Svetach [21]
3 years ago
11

I need help please???

Mathematics
1 answer:
Olegator [25]3 years ago
4 0

Answer:

i think it is E

Step-by-step explanation:

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What is the missing constant term in the perfect square that starts with x2+10x
Art [367]

Answer:

+25

Step-by-step explanation:

(x+5)²=x²+10x+25

3 0
3 years ago
Read 2 more answers
(1 point)
guapka [62]

The statements A, B, D, and E are defined.

When we multiply matrices, the number of columns of the 1st matrix must equal the number of rows of the 2nd matrix.

<h3>What is the subtraction of the matrix?</h3>

When we add or subtract matrices, the number of rows and columns must be the same.

A) If C is 7x3 and A is 3x2, that means C has 3 columns and A has 3 rows. So C * A is defined.

B) When you transpose a matrix, the number of rows and columns are switched. So C^T has 3 rows and 7 columns. So C^T is defined.

C) B has 2 columns and C has 7 rows, so B * C is not defined.

D) Because A and B have the same number of rows and columns, we can add them. So A + B is defined.

E) B^T has 2 rows and 3 columns, and C^T has 3 rows and 7 columns. So B^T has the same number of columns as the number of rows as C^T. So B^T * C^T is defined.

F) C and A have different numbers of rows and columns, so C-A is not defined.

To learn more about the matrices visit:

brainly.com/question/94574

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7 0
2 years ago
Which expression is equivalent to x-6?
Rufina [12.5K]

Answer:

X^-2 x X^3  or C

Step-by-step explanation:

why because a negative and a positive when multiplied make a negative

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3 years ago
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Factorise the following expression :​
jok3333 [9.3K]

Answer:

4p²+5pq-6pq²

taking common

p(4p+5q-6q²)

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3 years ago
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The diagram shows the rectangle ABCD, where A is (3,2) and B is (1,6).
In-s [12.5K]
(i)
A(3,2),B(1,6)&#10;\\&#10;\\m_{AB}= \frac{6-2}{1-3}=-2&#10;\\&#10;\\m_{BC}m_{AB}=-1&#10;\\ &#10;\\m_{BC}= \frac{-1}{m_{AB}} = \frac{-1}{-2}= \frac{1}{2}  &#10;\\&#10;\\B(1,6)\Rightarrow x_1=1,y_1=6&#10;\\&#10;\\BC:y-y_1=m_{BC}(x-x_1)&#10;\\&#10;\\y-6= \frac{1}{2}(x-1)&#10;\\&#10;\\2y-12=x-1&#10;\\x-2y+11=0

(ii)
AC:y=x-1&#10;\\BC:x-2y+11=0&#10;\\&#10;\\C=AC\cap BC&#10;\\&#10;\\x-2(x-1)+11=0&#10;\\x-2x+2+11=0&#10;\\x=13&#10;\\y=x-1=13-1=12&#10;\\C(13,12)

(iii)
A(3,2)B(1,6),C(13,12)&#10;\\P=2(d(A,B)+d(B,C))&#10;\\&#10;\\d(A,B)= \sqrt{(1-3)^2+(6-2)^2}= \sqrt{20}  =2 \sqrt{5} &#10;\\&#10;\\d(B,C)= \sqrt{(13-1)^2+(12-6)^2}= \sqrt{180}= 6\sqrt{5}  &#10;\\ &#10;\\P=2(2 \sqrt{5}+6 \sqrt{5})=2\times8 \sqrt{5} =16 \sqrt{5}
6 0
3 years ago
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