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oksian1 [2.3K]
3 years ago
15

What times what equals 82 besides 8 and 4, 16 and 2?

Mathematics
2 answers:
artcher [175]3 years ago
7 0
I think 41 would count b/c 2 * 41 = 82 & you already have 2.
zvonat [6]3 years ago
4 0
The half of 82. Easy to explain. The answer is 41
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A student is asked to find the length of the hypotenuse of a right triangle. The length of one leg is 32 centimeters, and the le
andreyandreev [35.5K]

Answer:

Step-by-step explanation:

First off, since we're talking about finding the correct hypotenuse for the triangle, we know to use the Pythagorean Theorem:
a^2+b^2=c^2

Since we know from the question, 32 and 22 are not the hypotenuse we know that those two values would be a and b. Now, all we do is stick these both into the equation and solve!
32^2+22^2=c^2

1024+484=c^2

1508=c^2

\sqrt{1508}=c

c=38.83cm

5 0
2 years ago
HELP PLEASE SHOW STEPS I HAVE NO IDEA WHAT TO DO. I HAVE BEEN TRYING TO FIGURE IT OUT BUT I CAN’T.
weqwewe [10]

Answer: x^{2} +23x+49

Step-by-step explanation:

To find the area of the shaded region, you must find the area of the larger rectangle [(x+10)×(2x+5)] and subtract the area of the smaller rectangle [(x+1)×(x+1)] from it.

First, you simplify the larger rectangle's area:

(x+10)×(2x+5) = 2x^{2} +25x+50

Then, simplify the smaller rectangle's area:

(x+1)×(x+1) = x^{2} +2x+1

Finally, subtract the smaller rectangle's area from the larger rectangle's area:

(2x^{2} +25x+50) - (x^{2} +2x+1) = x^{2} +23x+49

Therefore, the final answer is x^{2} +23x+49.

I hope this helps!

8 0
2 years ago
Read 2 more answers
6x squared -54=0<br><br> Solve for square root
Katen [24]
Sowe are going to work backward first were going to get rid off the negative 54 by adding it to the 0 making it 6x^2 =54 now were going to move the 6 by dividing both sides by 6 making it x^2=9 3 squared =9 so the variable is 3
7 0
3 years ago
Which set of points lies on the given graph?
beks73 [17]
Set A.

The other points do not lie on the given graph.

7 0
3 years ago
What is the solution set of the equation using the quadratic formula? x2+6x+10=0 {−3+2i,−3−2i} {−6+2i,−6−2i} {−3+i,−3−i} {−2i,−4
olga2289 [7]

Answer:

The solution set of the quadratic function x^{2}+6\cdot x +10 is \{-3+i,-3-i\} .

Step-by-step explanation:

Let be a second-order polynomial (quadratic function) is standard form and equalized to zero:

a\cdot x^{2}+b\cdot x + c = 0

Its roots can be determined by the Quadratic Formula in terms of its polynomial coefficients, which states that:

x_{1,2} = \frac{-b\pm\sqrt{b^{2}-4\cdot a\cdot c}}{2\cdot a}

Given that a = 1, b = 6 and c = 10, the roots of the polynomial are, respectively:

x_{1,2} = \frac{-6\pm \sqrt{6^{2}-4\cdot (1)\cdot (10)}}{2\cdot (1)}

x_{1,2} = -3\pm i

x_{1} = -3+i

x_{2} = -3 -i

The solution set of the quadratic function x^{2}+6\cdot x +10 is \{-3+i,-3-i\} .

8 0
3 years ago
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