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yan [13]
3 years ago
9

Pls help 10 ptsssssssssssss NO LINKS

Mathematics
2 answers:
Hunter-Best [27]3 years ago
3 0

Answer:

the answer is 14.

Step-by-step explanation:

frutty [35]3 years ago
3 0

Answer: If m=2, then 28 divided by 2 equals 14. Plug m into the equation.

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Tesfa is x years old his father is 4 times as old as tesfa his mother is 7 years younger than his father If their ages add up to
Sergio039 [100]

Step-by-step explanation:

his father's age= 4x

his mother's age= 4x- 7

his age= x

x+4x+4x-7=101

x+ 4x+ 4x= 101+ 7

9x= 108

\frac{9x}{9}= \frac{108}{9}

x= 12

Tesfa is 12 years old

4 0
3 years ago
Read 2 more answers
Help ? I need steps on how to do this
Furkat [3]

<em>Answer</em><em> </em><em>:</em><em>-</em><em> </em>

<em>the</em><em> </em><em>quoti</em><em>ent</em><em> </em><em>is</em><em> </em><em>(</em><em> </em><em>4</em><em>x</em><em>²</em><em> </em><em>-</em><em> </em><em>5</em><em>x</em><em> </em><em>+</em><em> </em><em>7</em><em> </em><em>)</em>

Step-by-step explanation:

[ Refer to the attachment for steps ]

  • We have to eliminate the highest degree coefficient in each step.

  • And as in division of normal numbers we subtract the things here we do the same ,

but while subtracting we have to take care about the signs !

  • The negative sign changes the negative sign into positive sign and positive sign into negative sign.

  • Whereas , a positive sign don't changes the sign.

8 0
3 years ago
Un apostador pierde en su primer juego el 30% de su dinero, en el segundo juego pierde el 50% de lo que perdió; finalmente en el
sergiy2304 [10]

Responder:

26,62

Explicación paso a paso:

Sea x el dinero original que tenía el jugador:

si un jugador pierde en su primer juego el 30% de su dinero, la cantidad perdida será;

= 30/100 \ of \ x\\= 0.3x

Si en el segundo juego pierde el 50% de lo que perdió, entonces la cantidad perdida en el segundo juego será:

= 50/100 \ of \ 0.3x\\= 0.5 \times 0.3x\\= 0.15x

Si en el tercer juego pierde el 40% de todo lo que ha perdido, la cantidad perdida en el tercer juego será:

=\frac{40}{100} \ of \ (0.3x+0.15x) \\= 0.4(0.45x)\\= 0.2025x

Si la cantidad que le queda para seguir apostando es de 37 soles, entonces para calcular la cantidad original que tiene, sumaremos toda la cantidad perdida y la cantidad restante y equipararemos la cantidad original x como se muestra:

0,3x + 0,15x + 0,2025x + 37 = x

0,6525x + 37 = x

x-0,6525x = 37

0,3475x = 37

x = 37 / 0,3475

x = 106,48

La cantidad que tenía originalmente era de 106,48

75% de 106,48

= 75/100 * 106,48

= 0,75 * 106,48

= 79,86

Tomando la diferencia entre su monto original y su 75% será:

= 106.48-79.86\\= 26.62

3 0
3 years ago
A bike accelerates at 3m/s2. How fast will it be traveling after 8 seconds?
Likurg_2 [28]

Answer:5 x (25 + 125 )/ 2 = 750/2 m = 375 m.


Step-by-step explanation:


3 0
3 years ago
1. Write the equation of the piece-wise function graphed below​.
DaniilM [7]

Problem 4

<h3>Answer:</h3>

f(x) = \begin{cases}2x+4 \text{ if } x < 1\\x-3 \text{ if } x \ge 1\end{cases}

------------------

Work Shown:

The left line goes through (-2,0) and (0,4)

The slope of this line is

m = (y2-y1)/(x2-x1)

m = (4-0)/(0-(-2))

m = (4-0)/(0+2)

m = 4/2

m = 2

The y intercept is b = 4

Since m = 2 and b = 4, this means y = mx+b turns into y = 2x+4. This portion is only done when x < 1. Note the open circle at the endpoint of this portion. So we do not include x = 1 as part of this piece.

---

The line on the right side goes through (1,-2) and (2,-1)

Slope

m = (y2-y1)/(x2-x1)

m = (-1-(-2))/(2-1)

m = (-1+2)/(2-1)

m = 1/1

m = 1

The y intercept is b = -3. You can see this if you extend the line until it crosses the y axis.

Alternatively, plug in (x,y) = (1,-2) and m = 1 into y = mx+b to find that b = -3

So y = mx+b turns into y = 1x+(-3) or just y = x-3

We combine both parts to end up with f(x) = \begin{cases}2x+4 \text{ if } x < 1\\x-3 \text{ if } x \ge 1\end{cases}

This is only graphed when x \ge 1 (note the closed or filled in circle for the endpoint of this portion).

===================================================

Problem 5

Answer:

<h3>f(x) = \frac{1}{2}|x+3| is the absolute value function</h3><h3>while this is the piecewise function</h3>

f(x) = \begin{cases}-\frac{1}{2}(x+3) \text{ if } x < -3\\\frac{1}{2}(x+3) \text{ if } x \ge -3\end{cases}\\

------------------

Work Shown:

y = |x| .... parent function

y = |x+3| ... shift 3 units to the left

y = (1/2)*|x+3| .... vertically compress by factor of 1/2

f(x) = (1/2)*|x+3|

------

Break that down into a piecewise function

when x < -3, then y = -(1/2)(x+3)

when x \ge -3, then y = (1/2)(x+3)

I'm using the rule that y = |x| turns into y = -x when x < 0 and y = x when x \ge 0

So that is how we get f(x) = \begin{cases}-\frac{1}{2}(x+3) \text{ if } x < -3\\\frac{1}{2}(x+3) \text{ if } x \ge -3\end{cases}\\as the piecewise function.

8 0
3 years ago
Read 2 more answers
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