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horsena [70]
2 years ago
14

PLEASE HELP IM FAILING ALGEBRA!!(answer all 3 problems) (SHOW WORK)

Mathematics
1 answer:
Lena [83]2 years ago
5 0

Hope this helps You ;)

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Help!!!!!!!!!!! please!!!!!!!!!
ohaa [14]

Answer: The first one is not a function the second is a function the third one is not one and the fourth one is

Step-by-step explanation:

3 0
3 years ago
What is the product?
ZanzabumX [31]

Answer:

The product is:

\left[\begin{array}{cc}15 & 14\\-1 & 9\end{array}\right]

Step-by-step explanation:

For this problem you need to multiply the first row only for the two first column of the others matrix and get the desired result:

\left[\begin{array}{ccc}1&3&1\\-2&1&0\end{array}\right] \times \left[\begin{array}{cc}2&-2\\3&5\\4&1\end{array}\right]

1 \times 2 + 3 \times 3 + 1 \times -2 = 15

So the value of the element in the position a_{11} is 15

1 \times -2 + 3 \times 5 + 1 \times 1 = 14

So the value of the element in the position a_{12} is 14

Then with these two values you can determinate the result matrix.

\left[\begin{array}{cc}15 & 14\\-1 & 9\end{array}\right]

3 0
2 years ago
F(3) =-2x^2+9x+5 calculate the following
ElenaW [278]

Answer:

50

Step-by-step explanation:

2(3)^2+9(3)+5

2(9)+27+5

18+27+5

=50

8 0
3 years ago
A triangle with sides 11m , 13m and 18m is a right triangle.<br> ​<br> A<br> True<br> B<br> False
Julli [10]

\bold{\huge{\pink{\underline{ Solution}}}}

\bold{\underline{ Given :-}}

  • <u>A </u><u>triangle </u><u>with </u><u>sides </u><u>11m</u><u>, </u><u> </u><u>13m </u><u>and </u><u>18m</u>

\bold{\underline{ To \: Find :-}}

  • <u>We</u><u> </u><u>have </u><u>to </u><u>check </u><u>it </u><u>whether </u><u>it </u><u>is </u><u>right </u><u>angled </u><u>triangle </u><u>or </u><u>not</u><u>? </u>

\bold{\underline{ Let's \: Begin :-}}

\sf{\red{ In \:right \:angled\ : triangle, }}

According to the Pythagoras theorem, The sum of the squares of perpendicular height and the square of the base of the triangle is equal to the square of hypotenuse that is sum of the squares of two small sides equal to the square of longest side of the triangle.

<u>We </u><u>imply</u><u> </u><u>it </u><u>in </u><u>the </u><u>given </u><u>triangle </u><u>,</u>

\sf{\red{ ( Perpendicular)² + (Base)² = (Hypotenuse)²}}

\sf{(AB)² + (BC)² = (AC)²}

\sf{ (11)² + (13)² = (18)² }

\sf{ 121 + 169 ≠  324 }

\sf{ 290 ≠ 324  }

<u>From </u><u>Above </u><u>we </u><u>can </u><u>conclude </u><u>that</u><u>, </u>

The sum of the squares of two small sides that is perpendicular height and base is not equal to the square of longest side that is Hypotenuse

\sf{\blue{ Hence,\: Your\: answer \:is \:false }}

5 0
2 years ago
Thanks! praise u if u answer
lakkis [162]
B I think hopes this helps
7 0
3 years ago
Read 2 more answers
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