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egoroff_w [7]
3 years ago
15

Find the distance between the parallel lines y=x+8 and y=x+2

Mathematics
1 answer:
bulgar [2K]3 years ago
6 0

Answer:

Step-by-step explanation:

6

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Indicate whether the measures 11, 12, and 20 can be the side lengths of a triangle. If they can, classify the triangle.
IrinaVladis [17]

Answer:

Its obtuse

Step-by-step explanation:

one angel is above 90 degrees.

4 0
2 years ago
The age of father is a two digit number. If 5 is added to the double of the digit in the tenth place of father's age will be the
yuradex [85]

Answer:

The father is 50 and his son is 15.

Step-by-step explanation:

Double the tens digit of the father's age = 10

Plus 5 = 15 (the age of the son)

50 + 15 = 65 (the sum of their ages)

The reverse of the digits of the son's age is 51.

Subtract 1 = 50 (the age of the father)

7 0
3 years ago
What is the product of <br><br> 2x+3 and 4x^2-5x+6 .
stira [4]

Answer:

Explanation:

We have:

(

2

x

+

3

)

(

4

x

2

−

5

x

+

6

)

Now let's distribute this piece by piece:

(

2

x

)

(

4

x

2

)

=

8

x

3

(

2

x

)

(

−

5

x

)

=

−

10

x

2

(

2

x

)

(

6

)

=

12

x

(

3

)

(

4

x

2

)

=

12

x

2

(

3

)

(

−

5

x

)

=

−

15

x

(

3

)

(

6

)

=

18

And now we add them all up (I'm going to group terms in the adding):

8

x

3

−

10

x

2

+

12

x

2

+

12

x

−

15

x

+

18

And now simplify:

8

x

3

+

2

x

2

−

3

x

+

18Step-by-step explanation:

3 0
2 years ago
Read 2 more answers
Use the product property of roots to choose the expression equivalent to 3√5x*3√25x^2?
nignag [31]

Answer: \sqrt[3]{125x^3}.


Step-by-step explanation: Given radical expression

\sqrt[3]{5x} \times \sqrt[3]{25x^2}.

According to the product property of roots.

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

On applying above rule, we get

\sqrt[3]{5x} \times \sqrt[3]{25x^2} = \sqrt[3]{5x \times 25x^2}

5 × 25 = 125 and

x \times x^2 = x^3

Therefore,

\sqrt[3]{5x \times 25x^2}= \sqrt[3]{125x^3}

<h3>So, the correct option would be second option \sqrt[3]{125x^3}.</h3>

3 0
3 years ago
Read 2 more answers
Prove that :( 1 + 1/<img src="https://tex.z-dn.net/?f=tan%5E%7B2%7DA" id="TexFormula1" title="tan^{2}A" alt="tan^{2}A" align="ab
Aliun [14]

Answer:

See explanation

Step-by-step explanation:

Simplify left and right parts separately.

<u>Left part:</u>

\left(1+\dfrac{1}{\tan^2A}\right)\left(1+\dfrac{1}{\cot ^2A}\right)\\ \\=\left(1+\dfrac{1}{\frac{\sin^2A}{\cos^2A}}\right)\left(1+\dfrac{1}{\frac{\cos^2A}{\sin^2A}}\right)\\ \\=\left(1+\dfrac{\cos^2A}{\sin^2A}\right)\left(1+\dfrac{\sin^2A}{\cos^2A}\right)\\ \\=\dfrac{\sin^2A+\cos^2A}{\sin^2A}\cdot \dfrac{\cos^2A+\sin^A}{\cos^2A}\\ \\=\dfrac{1}{\sin^2A}\cdot \dfrac{1}{\cos^2A}\\ \\=\dfrac{1}{\sin^2A\cos^2A}

<u>Right part:</u>

\dfrac{1}{\sin^2A-\sin^4A}\\ \\=\dfrac{1}{\sin^2A(1-\sin^2A)}\\ \\=\dfrac{1}{\sin^2A\cos^2A}

Since simplified left and right parts are the same, then the equality is true.

3 0
3 years ago
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