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3 years ago

Which pair of expressions is equivalent?

1 answer:

4 0

A) 2x + 10 equals 2x + 10
B) 7x + 3 does NOT equal 7x + 15
C) 4y - 8 doe NOT equal 4y - 2
D) 2y + 12 does NOT equal 2y - 12

Thus, A is the appropriate answer.

Hope that helped!

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Consider the triangle. 45° 80° Find the measure of the unknown third angle.

Elina [12.6K]

Answer:

$55 \degree$

Step-by-step explanation:

Let the unknown angle be x

As the interior angles in a triangle are added up to 180°

$x + 45 + 80 = 180 \\ x + 125 = 180 \\ x = 180 - 125 \\ x = 55 \degree$

Hope this helps you.

Let me know if you have any other questions:-)

6 0

3 years ago

Read 2 more answers

Given that k is a positive integer. Find, in terms of k, an expression for Sn, which is the sum of the integers from 2k to 4k in

Elden [556K]

Answer:

$S_n=3k\cdot (2k+1)=6k^2 +3k$

Step-by-step explanation:

k is a positive integer.

Consider an arithmetic sequence:

$a_1=2k\\ \\a_n=4k\\ \\d=1$

First, find n:

$a_n=a_1+(n-1)\cdot d\\ \\4k=2k+(n-1)\cdot 1\\ \\2k=n-1\\ \\n=2k+1$

Now, find the sum of these 2k+1 terms:

$S_n=\dfrac{a_1+a_n}{2}\cdot n\\ \\S_n=\dfrac{2k+4k}{2}\cdot (2k+1)=\dfrac{6k}{2}\cdot (2k+1)=3k\cdot (2k+1)=6k^2 +3k$

7 0

3 years ago

The sides of a square are five to the power of two fifths inches long. What is the area of the square?

Feliz [49]

I believe the correct answer from the choices listed above is the first option. If the sides of a square are five to the power of two fifths inches long, then the are of the square would be five to the power of four fifths square inches. Hope this answers the question.

3 0

3 years ago

Please help? , no decimals please,please explain to see picture.

dalvyx [7]

Answer:

$8x\sqrt{3xy$

Step-by-step explanation:

7 0

2 years ago

Help please and thank you

Dominik [7]

Answer:

Step-by-step explanation:

We assume the rotation R is counterclockwise 60°.

The exponent on R is the number of times it is applied. That is, R² = R(R(figure)). So, the composition is equivalent to R^(2-4) = R^-2.

When the exponent of R is negative, it is essentially the inverse function. That is, applying the function R to the result will give the figure you started with. Equivalently, it is rotation in the other direction.

$(\mathcal{R}^2\circ\mathcal{R}^{-4})(\text{figure})=\mathcal{R}^{-2}(\text{figure})=\text{figure rotated $120^{\circ}$ CW}$

The point 120° clockwise from B is D.

The desired image point is D.

3 0

3 years ago