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2 years ago

HELP ASAP GIVING BRAINLIEST!!! There are other questions, feel free to answer.

Mathematics

2 answers:

N76 [4]2 years ago

7 0

Hi! I hope this helps.. i think this is right, not positive!!

133.3

sweet-ann [11.9K]2 years ago

4 0

Answer:

400

Step-by-step explanation:

it is 400 hundred because if you do 900-500=400 so this would be the change in elevation.

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A portable music player has 2 gigabytes of storage and can hold about 500 songs. A similar but larger media player has 80 gigaby

anyanavicka [17]

Answer:

20,000.

Step-by-step explanation:

The first music player has 500 songs stored by 2 GB. So 500/2 is 250. You take 250 and you multiply it by 80 and you get 20,000.

5 0

3 years ago

Help i have no idea how to solve this

ASHA 777 [7]

Answer:

AB =24

Step-by-step explanation:

So, since all the sides are equal, you can just take two of the sides and set em equal to each other to find x and then, solve for AB

4x+8 =6x

8= 6x-4x

8=2x

8/2 =x

x= 4

Now, find AB, which is 4x+8

4(4) + 8

16+8 = 24

Hope this helps!

Please mark brainliest if you think I helped! Would really appreciate!

5 0

3 years ago

GIVING OUT BRAINLIEST TO WHOEVER GETS ALL OF THEM RIGHT

Thepotemich [5.8K]

Answer:

4) $\frac{x}{7\cdot x +x^{2}}$ is equivalent to $\frac{1}{7+x}$ for all $x \ne -7$ . (Answer: A)

5) $\frac{-14\cdot x^{3}}{x^{3}-5\cdot x^{4}}$ is equivalent to $-\frac{14}{1-5\cdot x}$ for all $x \ne \frac{1}{5}$ . (Answer: B)

6) $\frac{x+7}{x^{2}+4\cdot x - 21}$ is equivalent to $\frac{1}{x-3}$ for all $x \ne 3$ . (Answer: None)

7) $\frac{x^{2}+3\cdot x -4}{x+4}$ is equivalent to $x - 1$ . (Answer: None)

8) $\frac{2}{3\cdot a}\cdot \frac{2}{a^{2}}$ is equivalent to $\frac{4}{3\cdot a^{3}}$ for all $a\ne 0$ . (Answer: A)

Step-by-step explanation:

We proceed to simplify each expression below:

4) $\frac{x}{7\cdot x +x^{2}}$

(i) $\frac{x}{7\cdot x +x^{2}}$ Given

(ii) $\frac{x}{x\cdot (7+x)}$ Distributive property

(iii) $\frac{1}{7+x} \cdot \frac{x}{x}$ Distributive property

(iv) $\frac{1}{7+x}$ Existence of multiplicative inverse/Modulative property/Result

Rational functions are undefined when denominator equals 0. That is:

$7+x = 0$

$x = -7$

Hence, we conclude that $\frac{x}{7\cdot x +x^{2}}$ is equivalent to $\frac{1}{7+x}$ for all $x \ne -7$ . (Answer: A)

5) $\frac{-14\cdot x^{3}}{x^{3}-5\cdot x^{4}}$

(i) $\frac{-14\cdot x^{3}}{x^{3}-5\cdot x^{4}}$ Given

(ii) $\frac{x^{3}\cdot (-14)}{x^{3}\cdot (1-5\cdot x)}$ Distributive property

(iii) $\frac{x^{3}}{x^{3}} \cdot \left(-\frac{14}{1-5\cdot x} \right)$ Distributive property

(iv) $-\frac{14}{1-5\cdot x}$ Commutative property/Existence of multiplicative inverse/Modulative property/Result

Rational functions are undefined when denominator equals 0. That is:

$1-5\cdot x = 0$

$5\cdot x = 1$

$x = \frac{1}{5}$

Hence, we conclude that $\frac{-14\cdot x^{3}}{x^{3}-5\cdot x^{4}}$ is equivalent to $-\frac{14}{1-5\cdot x}$ for all $x \ne \frac{1}{5}$ . (Answer: B)

6) $\frac{x+7}{x^{2}+4\cdot x - 21}$

(i) $\frac{x+7}{x^{2}+4\cdot x - 21}$ Given

(ii) $\frac{x+7}{(x+7)\cdot (x-3)}$ $x^{2} -(r_{1}+r_{2})\cdot x +r_{1}\cdot r_{2} = (x-r_{1})\cdot (x-r_{2})$

(iii) $\frac{1}{x-3}\cdot \frac{x+7}{x+7}$ Commutative and distributive properties.

(iv) $\frac{1}{x-3}$ Existence of multiplicative inverse/Modulative property/Result

Rational functions are undefined when denominator equals 0. That is:

$x-3 = 0$

$x = 3$

Hence, we conclude that $\frac{x+7}{x^{2}+4\cdot x - 21}$ is equivalent to $\frac{1}{x-3}$ for all $x \ne 3$ . (Answer: None)

7) $\frac{x^{2}+3\cdot x -4}{x+4}$

(i) $\frac{x^{2}+3\cdot x -4}{x+4}$ Given

(ii) $\frac{(x+4)\cdot (x-1)}{x+4}$ $x^{2} -(r_{1}+r_{2})\cdot x +r_{1}\cdot r_{2} = (x-r_{1})\cdot (x-r_{2})$

(iii) $(x-1)\cdot \left(\frac{x+4}{x+4} \right)$ Commutative and distributive properties.

(iv) $x - 1$ Existence of additive inverse/Modulative property/Result

Polynomic function are defined for all value of $x$ .

$\frac{x^{2}+3\cdot x -4}{x+4}$ is equivalent to $x - 1$ . (Answer: None)

8) $\frac{2}{3\cdot a}\cdot \frac{2}{a^{2}}$

(i) $\frac{2}{3\cdot a}\cdot \frac{2}{a^{2}}$

(ii) $\frac{4}{3\cdot a^{3}}$ $\frac{a}{b}\cdot \frac{c}{d} = \frac{a\cdot b}{c\cdot d}$ /Result

Rational functions are undefined when denominator equals 0. That is:

$3\cdot a^{3} = 0$

$a = 0$

Hence, $\frac{2}{3\cdot a}\cdot \frac{2}{a^{2}}$ is equivalent to $\frac{4}{3\cdot a^{3}}$ for all $a\ne 0$ . (Answer: A)

6 0

3 years ago

Determine whether this pair of lines is parellel, perpendicular, or neither 6+2x=3y 3x+2y=9 Choose the correct answer below. A.

andriy [413]

Answer:

Option A. These two lines are perpendicular

Step-by-step explanation:

we know that

If two lines are parallel, then their slopes are equal

If two lines are perpendicular, then their slopes are opposite reciprocal (the product of their slopes is equal to -1)

step 1

Convert the given equation A in slope intercept form

$y=mx+b$

where

m is the slope

b is the y-intercept

we have

$6+2x=3y$ ----> equation A

Solve for y

That means----> isolate the variable y

Divide by 3 both sides

$y=\frac{2}{3}x+2$

$m_A=\frac{2}{3}$

step 2

Convert the given equation B in slope intercept form

$y=mx+b$

where

m is the slope

b is the y-intercept

we have

$3x+2y=9$ ----> equation B

Solve for y

That means----> isolate the variable y

subtract 3x both sides

$2y=-3x+9$

Divide by 2 both sides

$y=-\frac{3}{2}x+4.6$

$m_B=-\frac{3}{2}$

step 3

Compare the slopes

$m_A=\frac{2}{3}$

$m_B=-\frac{3}{2}$

The slopes are opposite reciprocal (the product is equal to -1)

therefore

These two lines are perpendicular

3 0

4 years ago

The teacher is sharpening pencils before a test. He needs to sharpen 40 pencils in 2

aivan3 [116]

Answer:

3 seconds

Step-by-step explanation:

2 minutes equals 120 seconds

120 divided by 40 is 3

3 0

3 years ago