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3 years ago

Eight less than the product of two and a number X

Mathematics

2 answers:

Alekssandra [29.7K]3 years ago

6 0

Answer:

2x - 8

Step-by-step explanation:to find the product means to multiply, so it would be 2 times X or 2x. Then you subtract 8 from that.

Hence, the answer is 2x-8

Hope it helps!

ps. plz give me brainliest

kirill115 [55]3 years ago

6 0

Answer:

PRODUCT OF 2 AND X=2X

8 LESS FROM 2X=2X-8

HOPE IT HELPS

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-2(2u-3)+3u=3(u+8)<br> u=?

dsp73

If you expand this equation first
You get -4u+6+3u=3u+24

And if you simplify that you get:
-u+6=3u+24
Now you need to solve this equation to find u

First bring 3u to the other side to -u-3u (the positive changes to negative)
Now the equation is
-4u+6= 24
Take 6 of both sides
-4u=24-6= 18
-4u= 18
So u= -18/4
Which simplifies to -9/2

Answer: -9/2

7 0

3 years ago

What is the solution to the system of equations represented by these two lines?

quester [9]

The answer for this is A.(3,1).
hope this helps

5 0

3 years ago

Explain in words how to solve the following problem.Maske sure to include all steps and the answer. 3/7 divided by 1/2

Maksim231197 [3]

To divide by a fraction is to multiply by its recipricle.
3/7 × 2/1
simplify fraction
3/7 × 2
multiply
6/7

3 0

3 years ago

Read 2 more answers

I have no idea how to do this, it is due in two days. Hopefully someone sees this before then.

elena-14-01-66 [18.8K]

Hello,

$m\ \widehat{ABC}=x\\m\ \widehat{BAC}=2*x\\\\So:\\ x+2x=90^o\\x=30^o\\$

$cos(30^o)=\dfrac{\sqrt{3} }{2} \\$

In the triangle ABC,

$cos(30^o)=\frac{BC}{BA} \\\\BA=\dfrac{cos(30^o)}{BC} \\\\BA=\frac{\dfrac{\sqrt{3} }{2} }{24} =16*\sqrt{3} \\\\$

$sin(30^o)=\dfrac{1 }{2} =\dfrac{AC}{AB} \\\\AC=\dfrac{1}{2} *16\sqrt{3} =8\sqrt{3}$

In the triangle ACB,

$cos(30^o)=\dfrac{AC}{AL} \\\\AL=\dfrac{8\sqrt{3} *2}{\sqrt{3} } =16\\$

6 0

2 years ago

Which statement is true?

love history [14]

<h2>Hello!</h2>

The answer is:

The second option,

$(\sqrt[m]{x^{a} } )^{b}=\sqrt[m]{x^{ab} }$

Discarding each given option in order to find the correct one, we have:

<h2>First option,</h2>

$\sqrt[m]{x}\sqrt[m]{y}=\sqrt[2m]{xy}$

The statement is false, the correct form of the statement (according to the property of roots) is:

$\sqrt[m]{x}\sqrt[m]{y}=\sqrt[m]{xy}$

<h2>Second option,</h2>

$(\sqrt[m]{x^{a} } )^{b}=\sqrt[m]{x^{ab} }$

The statement is true, we can prove it by using the following properties of exponents:

$(a^{b})^{c}=a^{bc}$

$\sqrt[n]{x^{m} }=x^{\frac{m}{n} }$

We are given the expression:

$(\sqrt[m]{x^{a} } )^{b}$

So, applying the properties, we have:

$(\sqrt[m]{x^{a} } )^{b}=(x^{\frac{a}{m}})^{b}=x^{\frac{ab}{m}}\\\\x^{\frac{ab}{m}}=\sqrt[m]{x^{ab} }$

Hence,

$(\sqrt[m]{x^{a} } )^{b}=\sqrt[m]{x^{ab} }$

<h2>Third option,</h2>

$a\sqrt[n]{x}+b\sqrt[n]{x}=ab\sqrt[n]{x}$

The statement is false, the correct form of the statement (according to the property of roots) is:

$a\sqrt[n]{x}+b\sqrt[n]{x}=(a+b)\sqrt[n]{x}$

<h2>Fourth option,</h2>

$\frac{\sqrt[m]{x} }{\sqrt[m]{y}}=m\sqrt{xy}$

The statement is false, the correct form of the statement (according to the property of roots) is:

$\frac{\sqrt[m]{x} }{\sqrt[m]{y}}=\sqrt[m]{\frac{x}{y} }$

Hence, the answer is, the statement that is true is the second statement:

$(\sqrt[m]{x^{a} } )^{b}=\sqrt[m]{x^{ab} }$

Have a nice day!

6 0

3 years ago