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RUDIKE [14]
3 years ago
8

Replace * with a monomial so that the trinomial may be represented by a square of a binomial:

Mathematics
1 answer:
vodka [1.7K]3 years ago
5 0

Please find attached photograph.

The answers are:

1) 2.5ab

2) 1/100 b^2

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Find the output, hhh, when the input, ttt, is 353535.
kow [346]

9514 1404 393

Answer:

  43

Step-by-step explanation:

Put the value where t is and do the arithmetic.

  h = 50 -t/5

  h = 50 -35/5 = 50 -7 = 43

The output, h, is 43 when the input is 35.

4 0
3 years ago
How do you write 12.375 in expanded form
viva [34]
12.375 = (1 x 10) + (2 x 1) + (3/10) + (7/100) + (5/1000). Hope this helps..
7 0
4 years ago
Read 2 more answers
Line a and b intersect, forming two adjacent angles, ∠1 and ∠2. If m ∠1= 7x-8 and
german

Answer:

x = 14; y = 8

Step-by-step explanation:

A perpendicular intersection means an intersection between 2 lines, forming angles of 90°. Then:

7x - 8 = 90°

11y + 2 = 90°

7x - 8 = 90

7x = 90 + 8

7x = 98

x = 98/7

x = 14

11y + 2 = 90

11y = 90 - 2

11y = 88

y = 88/11

y = 8

I hope I've helped you.

5 0
4 years ago
Lincoln Middle School had a canned food drive. The 6th graders brought in 5.75 boxes of canned food, the 7th graders brought in
makkiz [27]

Answer:

The total number of boxes of canned food brought in by the three grades is 17.63 boxes

Step-by-step explanation:

To find the total number of canned foods brought in by the three grades, we will have to add the number of canned foods brought by each of the grades to each other. Careful attention has to be paid to the decimal points.

Canned foods brought by 6th graders: 5.75 boxes

Canned foods brought by 7th graders: 6.5 boxes

Canned foods brought by 6th graders: 7/1.3 boxes

We may have a slight challenge adding in 7/1.3 boxes to the other figures. What we can do now is to evaluate it to a decimal value by dividing it up.

7/ 1.3 will be the same thing as 7 divided by 1.3 = 5.38 boxes.

hence, we have 5.75 +6.5+ 5.38= 17.63 boxes

The total number of boxes of canned food brought in by the three grades is 17.63 boxes

3 0
3 years ago
Which equations are correct?
stepan [7]

Answer:

3y^5(2y^3+3)=−6y^8−9y^5

−2c^6(5c^3+6)=−10c^9−12c^6

are the correct equations.

Step-by-step explanation:

If you simply all these equations you will get following equations :

when bases are same and are in multiplication we add the exponents.

−4b^4(5b^2+6)=−20b^6−24b^4

−5x^3(3x^3+2)=−15x^6−10x^3

−3y^5(2y^3+3)=−6y^8−9y^5

−2c^6(5c^3+6)=−10c^9−12c^6

so equation 3 and 4 are correct.

5 0
3 years ago
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