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IRISSAK [1]
4 years ago
12

THe length of each side of a square was increased by 6 inches so the perimeter is now 52 inches .What was the original length of

each side of the square
Mathematics
2 answers:
damaskus [11]4 years ago
4 0
The original length was 7
Vsevolod [243]4 years ago
4 0
A square has 4 congruent sides.
Let x = the length of each side.
The perimeter of the original square is x + x + x + x = 4x
Each side is increased by 6 inches, so all sides combined
are increased by 4 * 6 in. = 24 in.
The perimeter is now 4x + 24.
We are told the new perimeter is 52 in.

4x + 24 = 52

4x = 28

x = 7

Answer: Each side measured 7 inches.
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Help Asap! 20 points!
VLD [36.1K]
Hello

A. will be your answer. 


Hoped This Helped! :D
5 0
3 years ago
In the diagram, the straight line ABC is parallel to EFG and DB is parallel to FC. Given that ABD=38 degrees and DFE=62 degrees.
Elanso [62]

Based on the calculations, the measure of angle BDF and CFG are 100° and 38° respectively.

<h3>The condition for two parallel lines.</h3>

In Geometry, two (2) straight lines are considered to be parallel if their slopes are the same (equal) and they have different y-intercepts. This ultimately implies that, two (2) straight lines are parallel under the following conditions:

m₁ = m₂

<u>Note:</u> m is the slope.

<h3>What is the alternate interior angles theorem?</h3>

The alternate interior angles theorem states that when two (2) parallel lines are cut through by a transversal, the alternate interior angles that are formed are congruent.

Based on the alternate interior angles theorem, we can infer and logically deduce the following properties from the diagram (see attachment):

  • <ABD = <BDH = 38°
  • <DFE = <HDF = 62°

For angle BDF, we have:

<BDF = <BDH + <HDF

<BDF = 38° + 62°

<BDF = 100°.

Since angles BDF and DFC are linear pair, they are supplementary angles. Thus, we have:

∠BDF + <DFC = 180°

<DFC = 180 - ∠BDF

<DFC = 180 - 100

<DFC = 80°.

For angle CFG, we have:

∠DFE + <DFC + <CFG= 180°

<CFG = 180° - ∠DFE - <DFC

<CFG = 180° - 62° - 80°

<CFG = 38°.

Read more on parallel lines here: brainly.com/question/3851016

#SPJ1

4 0
2 years ago
What is the slope of a line with endpoints at (3,4) and (2,5)?
Lesechka [4]

\bf (\stackrel{x_1}{3}~,~\stackrel{y_1}{4})\qquad (\stackrel{x_2}{2}~,~\stackrel{y_2}{5}) \\\\\\ slope = m\implies \cfrac{\stackrel{rise}{ y_2- y_1}}{\stackrel{run}{ x_2- x_1}}\implies \cfrac{5-4}{2-3}\implies \cfrac{1}{-1}\implies -1

6 0
3 years ago
Read 2 more answers
1. Graph the function f(x)=4cos(x). (Be sure to label your x-axis.)
Mekhanik [1.2K]

ANSWER

To graph the function


y=4cos(x) we need to plot some few points within one period. Since the interval is not given, we shall use [0,2 \pi]'


\left \begin{array}{cc}x&y=4cos(x)\\0 &4\\\frac{\pi}{2}&0 \\ \pi&-4\\\frac{3\pi}{2} &0\\2\pi&4\end{array}\right.


We plot the above points to obtain the graph as shown in the attachment.



6 0
4 years ago
If sin(A+B)=1/2 &amp; sin(A-B)=1/3 find sin(2A)
timofeeve [1]

Answer:

sin(2A) = (2√2 + √3) / 6

Step-by-step explanation:

2A = (A+B) + (A−B)

sin(2A) = sin((A+B) + (A−B))

Angle sum formula:

sin(2A) = sin(A+B) cos(A−B) + sin(A−B) cos(A+B)

sin(2A) = 1/2 cos(A−B) + 1/3 cos(A+B)

Pythagorean identity:

sin(2A) = 1/2 √[1 − sin²(A−B)] + 1/3 √[1 − sin²(A+B)]

sin(2A) = 1/2 √(1 − 1/9) + 1/3 √(1 − 1/4)

sin(2A) = 1/2 √(8/9) + 1/3 √(3/4)

sin(2A) = 1/3 √2 + 1/6 √3

sin(2A) = (2√2 + √3) / 6

6 0
3 years ago
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