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mojhsa [17]
3 years ago
11

What one is it I have have been struggling with this

Mathematics
1 answer:
Radda [10]3 years ago
7 0

Answer:

C is the correct answer.

Step-by-step explanation:

The reason it is C is because pi/the symbol on top is irrational.

Hope you have a good rest of your day :)

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Simplify (4x − 6) + (3x + 6). 7x 7x − 12 x 7x + 12
noname [10]

Answer:

<h2>7x</h2>

Step-by-step explanation:

\left(4x-6\right)+\left(3x+6\right)\\\mathrm{Remove\:parentheses}:\quad \left(a\right)=a\\=4x-6+3x+6\\Group\:like\:terms\\=4x+3x-6+6\\\mathrm{Add\:similar\:elements:}\:4x+3x=7x\\=7x-6+6\\-6+6=0\\\\=7x

5 0
3 years ago
A rectangle has a width of 2.5. When the length of this rectangle is decreased by 5 inches, the new area of the rectangle is 17.
kati45 [8]
<h3><u>Part A: An equation to represent this situation, where x represents the original length of the rectangle is:</u></h3>

17.5 = (x-5) \times 2.5

<h3><u>Part B: Original length of the rectangle is 12 inches</u></h3>

<em><u>Solution:</u></em>

Given that,

A rectangle has a width of 2.5

Therefore,

width = 2.5 inches

Let "x" be the original length of the rectangle

When the length of this rectangle is decreased by 5 inches

Therefore,

New length = x - 5

The new area of the rectangle is 17.5 square inches

Therefore,

<em><u>The area of rectangle is given as:</u></em>

Area = length \times width

17.5 = (x-5) \times 2.5\\\\17.5 = 2.5x - 12.5\\\\2.5x = 17.5 + 12.5\\\\2.5x = 30\\\\x = 12

Thus original length of the rectangle is 12 inches

7 0
3 years ago
How would you solve a problem like this
RUDIKE [14]

Answer:

ΔGJH ≅ ΔEKF

HL: GH and EF

SAS: FK and JH (or GH and EF)

ASA: ∠JGH and ∠FEK (or ∠EFK and ∠JHG)

ΔGFJ ≅ ΔEKH

SSS: KH and FJ

SAS: ∠KEH and ∠FGJ

Step-by-step explanation:

List whatever angles/sides need to be congruent for the two triangles to be congruent.

Prove ΔGJH ≅ ΔEKF using....

- HL (Hypotenuse + Leg)

           We already have two legs that are congruent (EK and GJ), so we just need the hypotenuses (GH and EF) to be equal.

- SAS (Side + Angle + Side)

         1 pair of sides (EK and JG) are equal, and m∠EKF = m∠GJH. So we need one more side. You can either use FK and JH or GH and EF.

- ASA (Angle + Side + Angle)

         1 pair of angles (∠EKF and ∠GJH) are already given as equal, and 1 pair of sides (EK and GJ) are equal. We just need one more pair of angles. So either ∠JGH and ∠FEK or ∠EFK and ∠JHG.

Prove ΔGFJ ≅ ΔEKH using...

- SSS (Side + Side + Side)

          Two pairs of sides (EK + GJ and EH + FG) are equal, so KH and FJ need to be equal.

- SAS (Side + Angle + Side)

          FG + EH and KE + GJ are equal. We need to use the angle in between them to use SAS, so ∠KEH and ∠FGJ need to be equal.

7 0
2 years ago
X+2y-5=0<br>solve coordinate
Marina86 [1]
x+2y-5=0\\&#10;2y=-x+5\\&#10;y=-\frac{1}{2}x+\frac{5}{2}
7 0
3 years ago
Please help I only have 2 days to complete this and I have lots of other work to do
EleoNora [17]

Answer:

i gotchu gimme a second pls

Step-by-step explanation:

4 0
2 years ago
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