All categories

2 years ago

Can someone please help me

Mathematics

2 answers:

Sholpan [36]2 years ago

8 0

Answer:

-1.5

Step-by-step explanation:

Hope this helps, ^^

Have a great day and enjoy your assignment!

Good luck!

~sunny

klasskru [66]2 years ago

4 0

Answer:

1.5 hope this helped if you have any more questions let me know so I will be able to help you again

You might be interested in

Find the degree of the monomial. -20.5

Nadya [2.5K]

Answer:

Step-by-step explanation:

-20.5 is the same as -20.5 $x^{0}$ , since anything to the zero power is equal to 1.

7 0

2 years ago

Read 2 more answers

Select the correct answer.

Elis [28]

C would be the right answer

8 0

2 years ago

Read 2 more answers

The piggy bank contains 16 dimes, 8 nickels, and 12 quarters. This morning you removed 3 dimes and 5 nickels.Which expression co

Bond [772]

16•0.10= 1.60$
8•0.05= 0.40¢
12•0.25= 3.00$

1.60$+0.40¢+3.00$= 5.00$

3•0.10= 0.30¢
5•0.05= 0.25¢

0.30¢+0.25¢= 0.55¢

5.00$-0.55¢= 4.45$

4.45$ is left in the piggy bank.

3 0

3 years ago

(07.03)

Mkey [24]

x = 2

5(2 • 2 - 6) + 20

5(2x - 6 )+ 20 = 10

10x - 30 + 20 = 10

10x - 10 = 10

+10 +10

10 / 10x = 20 / 10

x = 2

7 0

3 years ago

Will Mark Brainiest!!! Simplify the following:

xz_007 [3.2K]

Answer:

1.B

2.A

3. B

Step-by-step explanation:

1. $\frac{x+5}{x^{2} + 6x +5 }$

We have the denominator of the fraction as following:

$x^{2} + 6x + 5 \\= x^{2} + (1 + 5)x + 5\\= x*x + 1x + 5x + 5*1\\= x ( x + 1) + 5(x + 1)\\= (x + 1) (x + 5)$

As the initial one is a fraction, so that its denominator has to be different from 0.

=> ( $x^{2} +6x+5$ ) ≠ 0

⇔ (x +1) (x +5) ≠ 0

⇔ (x + 1) ≠ 0; (x +5) ≠ 0

⇔ x ≠ -1; x ≠ -5

Replace it into the initial equation, we have:

$\frac{x+5}{x^{2} + 6x +5 } = \frac{x+5}{(x+1)(x+5)}$

As (x+5) ≠ 0; we divide both numerator and denominator of the fraction by (x +5)

=> $\frac{x+5}{x^{2} + 6x +5 } = \frac{x+5}{(x+1)(x+5)} = \frac{1}{x+1}$

So that $\frac{x+5}{x^{2} + 6x +5 } = \frac{1}{x+1}$ with x ≠ 1; x ≠ -5

So that the answer is B.

2. $\frac{(\frac{x^{2} -16 }{x-1} )}{x+4}$

As the initial one is a fraction, so that its denominator has to be different from 0

=> x + 4 ≠ 0

=> x ≠ -4

As $\frac{x^{2}-16 }{x-1}$ is also a fraction, so that its denominator (x-1) has to be different from 0

=> x - 1 ≠ 0

=> x ≠ 1

We have an equation: $x^{2} - y^{2} = (x - y ) (x+y)$

=> $x^{2} - 16 = x^{2} - 4^{2} = (x -4) (x +4)$

Replace it into the initial equation, we have:

$\frac{(\frac{x^{2} -16 }{x-1} )}{x+4} \\= \frac{x^{2} -16 }{x-1} . \frac{1}{x + 4}\\= \frac{(x-4)(x+4)}{x-1}. \frac{1}{x + 4}$

As (x + 4) ≠ 0 (proven above), we can divide both numerator and the denominator of the fraction by (x +4)

=> $\frac{(x-4)(x+4)}{x-1} .\frac{1}{x+4} =\frac{x-4}{x-1}$

So that the initial equation is equal to $\frac{x-4}{x-1}$ with x ≠-4; x ≠1

=> So that the correct answer is A

3. $\frac{x}{4x + x^{2} }$

As the initial one is a fraction, so that its denominator (4x + x^2) has to be different from 0

We have:

(4x + x^2) = 4x + x.x = x ( x + 4)

So that: (4x + x^2) ≠ 0 ⇔ x ( x + 4 ) ≠ 0

⇔ $\left \{ {{x\neq 0} \atop {(x+4)\neq0 }} \right.$ ⇔ $\left \{ {{x\neq 0} \atop {x \neq -4 }} \right.$

As (4x + x^2) = x ( x + 4) , we replace this into the initial fraction and have:

$\frac{x}{4x + x^{2} } = \frac{x}{x(x+4)}$

As x ≠ 0, we can divide both numerator and denominator of the fraction by x and have:

$\frac{x}{x(x+4)} =\frac{x/x}{x(x+4)/x} = \frac{1}{x+4}$

So that $\frac{x}{4x+x^{2} } = \frac{1}{x+4}$ with x ≠ 0; x ≠ -4

=> The correct answer is B

3 0

4 years ago