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nika2105 [10]
3 years ago
12

WHATS THE AREA OF THE TRAPIZOID?!

Mathematics
1 answer:
Helen [10]3 years ago
3 0

Answer: 45.. i think.

Step-by-step explanation: I couldn't really get the measurements from this photo so here is what I got. I used the formula A=a+b over 2 times h. I used 7=a, b=11,h=5. I added 7 and 11 to get 18, then divide that by 2, which equals 9. Then 9 times 5 = 45. I hope this somewhat helps.

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Carly left for school at 8:15.she left school at 2:45.if she arrived at 8:45,how many hours was she in the school
Ne4ueva [31]
She was in school for 6 hours.
3 0
2 years ago
Which answer is equal to the quotient in the expression below? (2x3 - x2) (x2)
serg [7]

(2x3 - x2) / (x2)

= 2x^3/x^2 - x^2/x^2

=2x - 1

answer

2x - 1

4 0
3 years ago
Which statement is true about the factorization of 30x2+40xy+51y2
icang [17]
The complete question is
Which statement is true about the factorization of 30x² + 40xy + 51y²<span>?

A. The factorization of the polynomial is 10(3x2 + 4xy + 5y2).
B. The polynomial can be rewritten as the product of a trinomial and xy.
C. The greatest common factor of the polynomial is 51x2y2.
D. The polynomial is prime, and the greatest common factor of the terms is 1.

we know that

case A) </span>is not right because 10 is not a common factor of the three terms.
case B) is not right because the original polynomial is already a trinomial
case C) is not right because the terms do not contain 51x^2y^2
<span>case D) is right 
because
</span><span>Factors of 30 are-----> 1,2,3,5,6,10,15,30
</span>Factors of 40 are-----> 1,2,4,5,8,10,20,40
Factors of 51 are-----> 1,51
<span>so
</span><span>The "Greatest Common Factor" is the largest of the common factors
</span><span>the GFC is 1

therefore

the answer is the option
</span>D. The polynomial is prime, and the greatest common factor of the terms is 1<span>



</span>
8 0
3 years ago
The volume of a cylinder is given by the formula y = ²h, where r is the radius of the cylinder and h is the height.
Anna11 [10]
Option 3 is the best answer
4 0
2 years ago
Applying Rate of change
olga nikolaevna [1]
\bf \begin{array}{ccllll}&#10;minutes(x)&Mbs\ left(y)\\&#10;-----&-----\\&#10;2&38.4\\&#10;8&9.6&#10;\end{array}\\\\&#10;-----------------------------\\\\&#10;slope = {{ m}}= \cfrac{rise}{run} \implies &#10;\cfrac{{{ y_2}}-{{ y_1}}}{{{ x_2}}-{{ x_1}}}\impliedby \textit{rate of change}&#10;\\\\\\&#10;m=\cfrac{9.6-38.4}{8-2}\implies m=-\cfrac{24}{5}\iff m= -4.8

now, notice, the rate of change for downloading is negative, because, "y" is decreasing,  namely the Mbs to be downloaded, are less and less and less as the minutes go by, because the file is almost fully downloaded

so is -4.8

now... at 8minutes, there are 9.6Mbs to download
bear in mind that 9.6 is just 4.8 * 2
that simply means, another minute, another 4.8 Mbs, and another minute and another 4.8 Mbs and the file is done

so, the file downloaded really in 10minutes

now, we know the rate is -4.8 or -24/5,   let us nevermind the sign for now

since we know the file is downloading at 24Mbs in 5minutes, a rate of 24/5

how much is it for 10 minutes?   well 10 is really just 5 * 2

so if it downloads at a rate of 24Mbs per 5mins, in 10 minutes it downloaded 24*2 or 48Mbs
7 0
3 years ago
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