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3 years ago

What is the following product? (√12+√6)(√6-√10)

Mathematics

2 answers:

iren [92.7K]3 years ago

8 0

Answer is A, (√12+√6)(√6-√10)

Tatiana [17]3 years ago

3 0

Answer:

The correct option is A.

Step-by-step explanation:

The given expression is

$(\sqrt{12}+\sqrt{6})(\sqrt{6}-\sqrt{10})$

Using distributive property of multiplication, we get

$\sqrt{12}(\sqrt{6}-\sqrt{10})+\sqrt{6}(\sqrt{6}-\sqrt{10})$

Using distributive property of multiplication, we get

$\sqrt{12}(\sqrt{6})+\sqrt{12}(-\sqrt{10})+\sqrt{6}(\sqrt{6})+\sqrt{6}(-\sqrt{10})$

$\sqrt{72}-\sqrt{120}+\sqrt{36}-\sqrt{60}$

$6\sqrt{2}-2\sqrt{30}+6-2\sqrt{15}$

The simplified form of the given expression is $6\sqrt{2}-2\sqrt{30}+6-2\sqrt{15}$ . Therefore the correct option is A.

You might be interested in

Devin needs to fill a square pyramid that has a base of 6cm and height of 10cm,he needs to melt cubes that are 2cm by 2cm by 2cm

aleksandrvk [35]

Answer:

Three cubes

Step-by-step explanation:

Lets first write down what we know.

The dimensions of the pyramid are a base of 6cm and a height of 10cm.

The cubes are 2cm by 2cm by 2cm.

We need to find the volume of both to solve.

First, lets solve for the volume of the pyramid. The equation is Volume=base*height/3

Lets put in the values we know.

volume=6*10/3

Multiply.

volume=60/3

Divide.

volume=20cm

Now that we know the volume of the pyramid is 20cm, all we have to do is find the volume of the cubes, and then divide 20cm by their volume.

To find the volume of the cubes, we just need to multiply length, width, and height together.

2cm*2cm*2cm=8cm

Now divide 20cm by 8cm.

20/8=2.5 rounding up to 3.

Devin will need to melt 3 cubes to fill the square pyramid.

Hope this helps!

3 0

3 years ago

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zhenek [66]

Answer: $\bold{y=\dfrac{1\pm \sqrt{4x+13}}{2}}$

Step-by-step explanation:

Inverse is when you swap the x's and y's and solve for y:

$y+\dfrac{13}{4}=\bigg(x-\dfrac{1}{2}\bigg)^2\\\\\\x+\dfrac{13}{4}=\bigg(y-\dfrac{1}{2}\bigg)^2\rightarrow \text{(swapped the x's and y's)}\\\\\\\sqrt{x+\dfrac{13}{4}}=\sqrt{\bigg(y-\dfrac{1}{2}\bigg)^2}\rightarrow \text{(took square root of both sides)}\\\\\\\sqrt{\dfrac{4x+13}{4}}=\pm \bigg(y-\dfrac{1}{2}\bigg)}\rightarrow \text{(created common denominator in radical)}\\\\\\ \dfrac{\sqrt{4x+13}}{2}=\pm \bigg(y-\dfrac{1}{2}\bigg)}\rightarrow \text{(simplified radical)}$

$\pm \dfrac{\sqrt{4x+13}}{2}=y-\dfrac{1}{2}}\rightarrow \text{(divided both sides by}\ \pm )\\\\\\\dfrac{1}{2}\pm \dfrac{\sqrt{4x+13}}{2}=y\quad \rightarrow \text{(added}\ \dfrac{1}{2}\ \text{to both sides)}\\\\\\\dfrac{1\pm \sqrt{4x+13}}{2}=y\quad \rightarrow \text{(combined numerators)}$