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RSB [31]
3 years ago
5

A square has an area of 49 cm2. what is the length of each side?​

Mathematics
2 answers:
KiRa [710]3 years ago
8 0

Answer:

7

Step-by-step explanation:

because 7 cm as the side length is always the square root of the area in a square.

Maru [420]3 years ago
3 0

Answer:

7x7=49, so 7

Step-by-step explanation:

You might be interested in
Is 22 ounces equal to 6 pound
serious [3.7K]
NO

16 ounce = 1 pound

22 ounce = 1.375 pound ≠ 6 pound


7 0
3 years ago
Select the correct answer.
anygoal [31]

Answer:

B: c(3c^2-4)

Step-by-step explanation:

1. Substitute 1 for c: (24^6-32^4)/(8^3)

2. Solve substitution: -1

3. Compare -1 to calculated answers when c=1

4. c(3c^2-4) when c=1 is -1

6 0
2 years ago
X^2=14x-49. i just dont know how to do the steps to solve this equation
KonstantinChe [14]
x^2=14x-49 \\ \\ x^2 - 14x + 49 = 0 \ / \ move \ terms \\ \\ x^2 - 2(x)(7) + 7^2 = 0 \ / \ rewrite \\ \\ (x - 7)^2 = 0 \ / \ square \ of \ difference \ method \\ \\ x - 7 = 0 \ / \ square \ root \ each \ side \\ \\ x = 7 \ / \ add \ 7 \\ \\

The answer is: x = 7.
5 0
3 years ago
Solve on the interval [0/2pi]<br><br> 1-cos(theta) = (1/2)
Airida [17]

Answer:

Final answer is x=\frac{\pi}{3} and x=\frac{5\pi}{3}.

Step-by-step explanation:

Given equation is 1-\cos\left(\theta\right)=\frac{1}{2}

Now we need to find the solution of  1-\cos\left(\theta\right)=\frac{1}{2} in given interval [0, 2\pi ].

1-\cos\left(\theta\right)=\frac{1}{2}

-\cos\left(\theta\right)=\frac{1}{2}-1

-\cos\left(\theta\right)=-\frac{1}{2}

\cos\left(\theta\right)=\frac{1}{2}

which gives x=\frac{\pi}{3} and x=\frac{5\pi}{3} in the given interval.

Hence final answer is x=\frac{\pi}{3} and x=\frac{5\pi}{3}.

6 0
3 years ago
Where were the two cars in relation to each other when they began traveling?
Elza [17]

Answer:

The answer is A.

Step-by-step explanation:

As can be seen in the graph, car B starts at the "20" mark on the vertical axis, signifying that it starts 20 miles east of car A.

Hope this helped!

3 0
3 years ago
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