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zalisa [80]
4 years ago
9

Which of the following constructions is this?

Mathematics
2 answers:
ohaa [14]4 years ago
8 0

Answer:

acute hope it helps

Step-by-step explanation:

mark as brainiest

juin [17]4 years ago
5 0

Construction of 30° acute angle

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What is the perimeter of an area
Inessa [10]

Answer:

The perimeter is the length of the outline of a shape. To find the perimeter of a rectangle or square you have to add the lengths of all the four sides. x is in this case the length of the rectangle while y is the width of the rectangle. The area is measurement of the surface of a shape.

Step-by-step explanation:

4 0
4 years ago
what is the length, in units, of the hypotenuse of a right triangle if each of the two legs is 2 units?
kykrilka [37]
To figure this out, you'd use the a^2+b^2=c^2 formula. Since you have your two sides, you want to find the value for c. So, you'd plug in two for a and b.

2^2+2^2=c^2

4+4=c^2

Add up the two fours to get 8 and square root it, and you get 2.828427124...
4 0
3 years ago
Read 2 more answers
Tyrell Otis deposited his IRS refund check for $760 at 5% and made no other deposits or withdrawals. How much simple interest di
Helga [31]

Answer:

He earned $114 in interest.

The final amount in the account is $874

Step-by-step explanation:

Initial balance (b): $760

Interest rate (i): 0.05

Period (n): 3 months

Simple interest applications yield the same amount in interest every period. Therefore, the total simple interest earned can be defined by:

I = n*B*i\\I= 3*760*0.05\\I=114

The final amount (A) in the account is the initial balance plus the interest earned in 3 months.

A = 760 +114\\A=874

He earned $114 in interest and the final amount in the account is $874

6 0
3 years ago
Use the quadratic formula to solve each equation.
Elden [556K]

Answer:

1) x_1 = \frac{2 - 2\sqrt{13}}{2}= 1-\sqrt{13}

x_2 = \frac{2 + 2\sqrt{13}}{2}= 1+\sqrt{13}

2) x_1 = \frac{6 - 2\sqrt{10}}{1}= 6-2\sqrt{10}

x_2 = \frac{6 + 2\sqrt{10}}{1}= 6+2\sqrt{10}

3) p_1 = \frac{-8 - 2\sqrt{30}}{4}= -2-\frac{1}{2}\sqrt{30}

p_2 = \frac{-8 + 2\sqrt{30}}{4}= -2+\frac{1}{2}\sqrt{30}

4) y_1 = \frac{-3 - 9}{4}=-3

y_2 = \frac{-3 + 9}{4}= \frac{3}{2}

Step-by-step explanation:

The quadratic formula is given by:

x = \frac{-b \pm \sqrt{b^2 -4ac}}{2a}

We can use this formula in order to solve the following equations:

1. x^2 − 2x = 12 → a = 1, b = −2, c = −12

For this case if we apply the quadratic formula we got:

x = \frac{-(-2) \pm \sqrt{(-2)^2 -4(1)(-12)}}{2(1)}

x_1 = \frac{2 - 2\sqrt{13}}{2}= 1-\sqrt{13}

x_2 = \frac{2 + 2\sqrt{13}}{2}= 1+\sqrt{13}

2. 1/2x^2 − 6x = 2 → a = 1 / 2, b = −6, c = −2

For this case if we apply the quadratic formula we got:

x = \frac{-(-6) \pm \sqrt{(-6)^2 -4(1/2)(-2)}}{2(1/2)}

x_1 = \frac{6 - 2\sqrt{10}}{1}= 6-2\sqrt{10}

x_2 = \frac{6 + 2\sqrt{10}}{1}= 6+2\sqrt{10}

3. 2p^2 + 8p = 7 → a = 2, b = 8, c = −7

For this case if we apply the quadratic formula we got:

p = \frac{-(8) \pm \sqrt{(8)^2 -4(2)(-7)}}{2(2)}

p_1 = \frac{-8 - 2\sqrt{30}}{4}= -2-\frac{1}{2}\sqrt{30}

p_2 = \frac{-8 + 2\sqrt{30}}{4}= -2+\frac{1}{2}\sqrt{30}

4. 2y^2 + 3y − 5 = 4 → a = 2, b = 3, c = −9

For this case if we apply the quadratic formula we got:

y = \frac{-(3) \pm \sqrt{(3)^2 -4(2)(-9)}}{2(2)}

y_1 = \frac{-3 - 9}{4}=-3

y_2 = \frac{-3 + 9}{4}= \frac{3}{2}

8 0
4 years ago
Am i right for 6 and 7 also help with 8, 9, 10 and 11 plz
marysya [2.9K]
Your right on 6 and 7 but I don't understand 10 @ 11
3 0
3 years ago
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