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Dmitry [639]
3 years ago
8

sam decides to build a square garden. if the area of the garden is 4x2 28x 49 square feet, what is the length of one side of the

garden? (2x 7) feet (7x 2) feet (2x − 7) feet (7x − 2) feet
Mathematics
2 answers:
AleksandrR [38]3 years ago
5 0

Answer:

(2x + 7) feet

Step-by-step explanation:

just took the test this is the correct answer

jekas [21]3 years ago
4 0
To find the side lengths, you have to factor 4x^2 + 28x + 49. This equation factored is (2x+7)(2x+7), so each side length is (2x+7) feet.
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Jeremy had 5 Roosevelt dimes, 3 Kennedy half dollars, and 8 silver dollars in his coin collection worth a total of $231.
andre [41]
The linear equation would be, 5d + 3h + 8s = 231
4 0
2 years ago
What is the perimeter???
vekshin1

Answer:

90 cm

Step-by-step explanation:

The top and the bottom are the same length, 45 cm

The horizontal piece at the bottom is 45-25 = 20 cm

The left and the right are the same length 30cm

The vertical piece at the bottom is 30 -22 =8 cm

Starting at the bottom left corner and going clockwise, we add up all the sides to find the perimeter

22 +45+30+20+8+25

150

The perimeter is 150 cm

7 0
3 years ago
the cost of a ticket t to a concert with a 3% sales tax can be represented by the expression t+0.03t. simplify the expression. t
Brilliant_brown [7]
72.00 x 3% = 2.16
72.00 + 2.16 = 74.16 total
7 0
3 years ago
This is an integration question.
Alla [95]

Answer:(1,1),\ \dfrac{1}{3}

Step-by-step explanation:

Given

Equation of the curves are y=x^2,\ y^2=x

The intersection of the curve is

\Rightarrow y^4-y=1\\\\\Rightarrow y(y^3-1)=0\\\\\Rightarrow y=0,1\\

So, x coordinates are x=0,1

points of intersection are(0,0),(1,1)

So, the area bounded between the curves

\Rightarrow I=\int_{0}^{1}\left (  \sqrt{x}-x^2\right )dx\\\\\Rightarrow I=\int_{0}^{1}\sqrt{x}dx-\int_{0}^{1}x^2dx\\\\\Rightarrow I=\left ( \frac{2}{3}x^{\frac{3}{2}} \right )_0^1-\left ( \frac{1}{3}x^3 \right )_0^1\\\\\Rightarrow I=\frac{2}{3}\left ( 1-0 \right )-\frac{1}{3}\left ( 1^3-0 \right )\\\\\Rightarrow I=\frac{2}{3}-\frac{1}{3}\\\\\Rightarrow I=\frac{1}{3}

The area bounded by them is \frac{1}{3}

8 0
2 years ago
7) In 2011, there were 1,000 students at Jefferson High. In 2017, there were 1,324 students. What is
ale4655 [162]

Answer:

54

Step-by-step explanation:

5 0
2 years ago
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