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kakasveta [241]
4 years ago
5

Please explain what I need to do 20 points

Mathematics
1 answer:
maxonik [38]4 years ago
6 0

Answer:

look at the graph and solve it

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the length of a rectagle is 5 in longer than its width. if the perimeter of the rectangle is 56 in, find its length and width
lisabon 2012 [21]

Answer:

Step-by-step explanation:

Let the width of the rectangle = x

As length is 5 inches longer than width, we have to add 5 to width

Length = x + 5

Perimeter of ractangle = 56 in

2* (length + width) = 56

2*( x + 5 + x) = 56

2* (2x + 5)    = 56

Use distributive property: a*(b +c) =(a*b) + (a * c)

2*2x + 2*5   = 56

4x  + 10 = 56

Subtract 10 from both sides

4x = 56- 10

4x = 46

Divide both sides by 4

x = 46/4

x = 11.5

Width = 11.5 in

length = 11.5 + 5

            = 16.5 in

3 0
2 years ago
I need Help With Question 3 Thank You
slava [35]

Central angle is twice an inscribed angle subtended by the same arc.

Therefore ∠CAB = 2∠CDB.

∠CAB = 45° → ∠CDB = 45° : 2 = 22.5°

Answer: C.

7 0
3 years ago
2. The two equations y = (x + 2)(x + 3) and y = x? + 5x + 6 are equivalent.
Nady [450]

Answer:

it b

Step-by-step explanation:

3 0
3 years ago
Simplify by combining like terms:<br><br> -5x-13+7y+26x
sveticcg [70]

Answer:

26x and -5x are like terms, so what you would do with these is 26x subtract -5x, which is 21x.

The final expression is:

  • -13 + 7y + 21x

The only pair of like terms in this expression is 26x and -5x.

Step-by-step explanation:

5 0
3 years ago
Evaluate the series
dalvyx [7]

Answer:

the value of the series;

\sum_{k=1}^{6}(25-k^2) = 59

C) 59

Step-by-step explanation:

Recall that;

\sum_{1}^{n}a_n = a_1+a_2+...+a_n\\

Therefore, we can evaluate the series;

\sum_{k=1}^{6}(25-k^2)

by summing the values of the series within that interval.

the values of the series are evaluated by substituting the corresponding values of k into the equation.

\sum_{k=1}^{6}(25-k^2) =(25-1^2)+(25-2^2)+(25-3^2)+(25-4^2)+(25-5^2)+(25-6^2)\\\sum_{k=1}^{6}(25-k^2) =(25-1)+(25-4)+(25-9)+(25-16)+(25-25)+(25-36)\\\sum_{k=1}^{6}(25-k^2) =24+21+16+9+0+(-11)\\\sum_{k=1}^{6}(25-k^2) = 59\\

So, the value of the series;

\sum_{k=1}^{6}(25-k^2) = 59

6 0
4 years ago
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