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olasank [31]
3 years ago
15

What is the image of (-5, 6) after a reflection over the line y = -2

Mathematics
1 answer:
prohojiy [21]3 years ago
6 0

Answer:

I'm pretty sure it would be (5,6)

Step-by-step explanation:

Reflecting a point over the Y-axis would change the x-coordinate, but not the y-coordinate

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Find the area of the circle to the nearest whole number, if necessary!
astra-53 [7]

Answer:

The circle has an area of about 1385 square mm.

Step-by-step explanation:

Let's recall that circles have an area that can be found with the following formula:

A=\pi r^{2}

where r is the radius of the circle.

Now, focus your eyes on the circle. We are shown that the diameter of this circle is 42 mm, but we only want the radius. Since the radius is half the diameter, the radius is 21 mm. Now, we can solve for the area of the circle.

A=\pi (21)^2\\A=441\pi \\A=1385.44236...

So, to the nearest whole number, the area of the circle is 1385 square mm.

5 0
2 years ago
What is the solution to the inequality below?
Svetlanka [38]

Answer:

B. x < -8 or x > 8

Step-by-step explanation:

You can use process of elimination to solve this problem by going through every solution and testing them out, but let's jump right to B.

Process:

You know that since the inequality states that x^2 has to be greater than 64, x  has to be more than 8, or less than -8.

This is because 8^2 = 64, and -8^2 = 64, and the inequality requires the answer to be more than 64.

Looking at B., you can see that if x is < -8, the square of, for example, -9, would be 81. This is greater than 64, so this works!

Now, B. also has an alternative. The 'or' is a major clue to which is the correct answer, since the square root of any number can be positive or negative. (-8^2 = 8^2)

The 'or' states that x must be greater than 8. So, for example, if we take the square of 10, we get 100, and that is also greater than 64.

We've proven that this solution is accurate for both parts, so it is definitely the one we want!

Hope this helps!

4 0
3 years ago
Conan installs windows for a living and has propped his ladder against a customer's
Luden [163]

Answer:

Step-by-step explanation:

By the Pythagorean Theorem

h^2=x^2+y^2

Which means that the hypotenuse (longest side of a right triangle) squared is equal to the sum of its squared sides. In this case we are given the side lengths of 8 and 15 feet.

h^2=8^2+15^2

h^2=64+225

h^2=289

h=17 ft

The ladder is 17 feet long.

3 0
3 years ago
Multiply.
Nady [450]

\boxed{3\sqrt{22}\sqrt{58}\sqrt{18}=18\sqrt{638}}

<h2>Explanation:</h2>

Here we have the following expression:

3\sqrt{22}\sqrt{58}\sqrt{18}

So we need to simplify that radical expression. By property of radicals we know that:

\sqrt[n]{a}\sqrt[n]{b}=\sqrt[n]{ab}

So:

3\sqrt{22}\sqrt{58}\sqrt{18}=3\sqrt{22\times 58 \times 18}=3\sqrt{22968}

The prime factorization of 22968 is:

22968=2^3\cdot 3^2\cdot11\cdot 29

Hence:

3\sqrt{22968}=3\sqrt{2^3\cdot 3^2\cdot11\cdot 29}=3\sqrt{2^2\cdot 3^2\cdot 2\cdot 11\cdot 29}

By property:

\sqrt[n]{a^n}=a

So:

3\sqrt{2^2\cdot 3^2\cdot 2\cdot 11\cdot 29} \\ \\ 3(2)(3)\sqrt{2\cdot 11\cdot 29}=18\sqrt{638}

Finally:

\boxed{3\sqrt{22}\sqrt{58}\sqrt{18}=18\sqrt{638}}

<h2>Learn more:</h2>

Radical expressions: brainly.com/question/13452541

#LearnWithBrainly

3 0
3 years ago
What number is 4 times as many as 25
Oduvanchick [21]
This is a standard relationship/systems of equations question.  Here is how you attack it.  Firstly, set up equations to represent the relationships. Ed = 4*Kim That shows that Ed has four times as many pins as Kim.  Next, we see that the two have 25 together.  So: Ed + Kim = 25 Now we have our two equations.  In order to solve, we need to get one of the equations down to one variable.  We can achieve this via substitution.  The first equation tells us that we can substitute 4*Kim for Ed (in the second equation).  So, let's do just that: 4*Kim + Kim = 255*Kim = 25Kim = 5 Using the first equation again, we can solve for Ed: Ed = 4*KimEd = 4*(5)<span>Ed = 20</span>
6 0
3 years ago
Read 2 more answers
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