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3 years ago

Simplify the following

Mathematics

2 answers:

-Dominant- [34]3 years ago

7 0

Answer:

1, B

Step-by-step explanation:

When bases are same powers are to be added

=x^(1-2)/x^(-1+0)

=x^(-1)/x^(-1)

eimsori [14]3 years ago

7 0

Step-by-step explanation:

$\\ \frac{x. {x}^{ - 2} }{ {x}^{ - 1}. {x}^{0} } = \frac{{x}^{ - 2 + 1} }{ {x}^{ - 1}.1 } = \frac{{x}^{ - 1} }{ {x}^{ - 1} } = 1$

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Si al numerador de una fraccion se añade 5, el valor de la fraccion es 2, y si al numerador se resta 2, al valor de la fraccion

konstantin123 [22]

Answer:

Respuesta.- La fracción es 9/7

Step-by-step explanation:

1) Llamando X e Y a la fracción, tenemos: X/Y

Si al numerador se le añade 5, es decir X+5, el valor de la fracción es 2, luego:

(X+5)/Y = 2

Y si al numerador se le resta 2, es decir X-2, el valor de la fracción es 1, luego:

(X-2)/Y = 1

Juntando amabas ecuaciones:

(X+5)/Y= 2

(X-2)/Y = 1

Despejando Y en ambas:

Y = (X+5)/2

Y = (X-2)..........(1)

Igualandolas:

(X+5)/2= (X-2)

De donde:

X+5 = 2X-4

X = 9

Y finalmente, sustituyendo el valor de X = 9 en la ecuaión (1), tendremos:

Y = (X-2)

Y = (9-2)

Y = 7

Respuesta.- La fracción es 9/7

6 0

4 years ago

I need step by step help, thank you:)

prohojiy [21]

The area of a triangle is 1/2 of base times height
for ΔPQS, the base is PS=6, the area is 16.5, so the height is 16.5×2÷=5.5
the height of the triangles is also the height of the trapezoid.

7 0

3 years ago

Determine the value of x for which rls if<br> A)10<br> B)0.75<br> C)1.25<br> D)20

lesantik [10]

Answer:

x = 1.25

Step-by-step explanation:

<1 and <2 are corresponding angles and if the lines are parallel, corresponding angles are equal

<1 = <2

20-8x = 30-16x

Add 16x to each side

20-8x+16x = 30-16x+16x

20 +8x = 30

Subtract 20 from each side

20+8x-20 = 30-20

8x= 10

Divide by 8

8x/8 = 10/8

x = 5/4

x = 1.25

7 0

3 years ago

Read 2 more answers

What is 92.9 rounded to the nersst whole number

sammy [17]

The nearest whole number would be 93 because you round up if the decimal place is greater than 5 and down if it's less.

5 0

3 years ago

PLEASE HELP

sweet [91]

To model and solve our situation we are going to use the equation: $s= \frac{d}{t}$
where
$s$ is speed
$d$ is distance
$t$ is time

1. We know that the distance between the cities is 2400 miles, so $d=2400$ . We also know that the speed of the plane is 450 mi/h. Since we don't know the speed of the air, $S_{a}=?$ . We don't know how much the westward trip takes, so $t_{w}=?$ , and we also don't know how much the eastward trip takes, so $t_{e}=?$ .

Going westward. Here the plane is flying against the air, so we need to subtract the speed of the air from the speed of the plane:
$450-S_{a}= \frac{2400}{t_{w} }$
Going eastward. Here the plane is flying with the the air, so we need to add the speed of the air to the speed of the plane:
$450+S_{a}= \frac{2400}{t_{e} }$

2. We know for our problem that the round trip takes 11 hours; so the total time of the trip is 11, $t_{t}=11$ . Notice that we also know that the total time of the trip equals time of the tip going westward plus time of the trip going eastward, so $t_{t}=t_{w}+t_{e}$ . Since we know that the total trip takes 11 hours, we can replace that value in our total time equation and solve for $t_{w}$ :
$11=t_{w}+t_{e}$
$t_{w}=11-t_{e}$

Now we can replace $t_{w}$ in our going westward equation to model our round trip with a system of equations:
$450-S_{a}= \frac{2400}{t_{w}}$
$450-S_{a}= \frac{2400}{11-t_{e} }$ equation (1)
$450+S_{a}= \frac{2400}{t_{e}}$ equation (2)

3. To solve our system of equations, we are going to solve for $t_{e}$ in equations (1) (2):

From equation (1)
$450-S_{a}= \frac{2400}{11-t_{e} }$
$11-t_{e}= \frac{2400}{450-S_{a} }$
$-t_{e}= \frac{2400}{450-S_{a} } -11$
$t_{e}=11- \frac{2400}{450-S_{a} }$
$t_{e}= \frac{4950-11S_{a} -2400}{450-S_{a} }$
$t_{e}= \frac{2550-11S_{a} }{450-S_{a} }$ equation (3)

From equation (2):
$450+S_{a}= \frac{2400}{t_{e} }$
$t_{e}= \frac{2400}{450+S_{a} }$ equation (4)

Replacing (4) in (3)
$\frac{2400}{450+S_{a}} = \frac{2550-11S_{a}}{450-S_{a} }$
Now, we can solve for $S_{a}$ to find the speed of the wind:
$2400(450-S_{a})=(450+S_{a})(2550-11S_{a})$
$1080000-2400S_{a}=1147500-4950S_{a}+2550S_{a}-11(S_{a})^{2}$
$11(S_{a})^{2}-67500=0$
$11(S_{a})^{2}=67500$
$(S_{a})^{2}= \frac{67500}{11}$
$S_{a}=+/- \sqrt{ \frac{67500}{11} }$
Since speed cannot be negative, the solution of our equation is:
$S_{a}= \sqrt{ \frac{67500}{11} }$
$S_{a}=78.33$

We can conclude that the speed of the wind is 78 mph.

3 0

4 years ago