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3 years ago

Solve the system using substitution x+5y=0 3y+2x=-21

Mathematics

2 answers:

Crazy boy [7]3 years ago

5 0

Answer:

$\left \{ {{x = -15} \atop {y=3}} \right.$

Step-by-step explanation:

$\left \{ {{x + 5y = 0} (1) \atop {3y + 2x = -21}(2)} \right. \left \{ {{x = -5y} \atop {3y + 2x = -21}} \right. \left \{ {{x = -5y} \atop {3y + 2*(-5y) = -21}} \right. \left \{ {{x = -5y} \atop {3y - 10y = -21}} \right. \\ \left \{ {{x = -5y } \atop {-7y = -21}} \right. \left \{ {{x = -5y} \atop {y= 3 }} \right. \left \{ {{x = -5 * 3} \atop {y=3}} \right.=> \left \{ {{x=-15} \atop {y=3}} \right.$

pochemuha3 years ago

4 0

Ok done. Thank to me :>

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Subtract. 5x^2-5x+1-(2x^2+9x-6)

Talja [164]

<h2>Hello!</h2>

The answer is:

$3x^{2} -14x+7$

To solve the problem we need to add/subtract like terms. We need to remember that like terms are the terms that share the same variable and the same exponent.

For example, we have:

$x^{2} +2x+3x=x^{2} +(2x+3x)=x^{2}+5x$

We have that we were able to add just the terms that were sharing the same variable and exponenr (x for this case).

So, we are given the expression:

$5x^2-5x+1-(2x^2+9x-6)=5x^{2}-2x^{2}-5x-9x+1-(-6)\\\\(5x^{2}-2x^{2})+(-5x-9x)+(1-(-6))=3x^{2}-14x+7$

Hence, the answer is:

$3x^{2} -14x+7$

Have a nice day!

4 0

3 years ago

A ball is thrown into the air by a baby alien on a planet in the system of Alpha Centauri with a velocity of 30 ft/s. Its height

Crank

Answer:

a) h = 0.1: $\bar v = -11\,\frac{ft}{s}$ , h = 0.01: $\bar v = -10.1\,\frac{ft}{s}$ , h = 0.001: $\bar v = -10\,\frac{ft}{s}$ , b) The instantaneous velocity of the ball when $t = 2\,s$ is $-10$ feet per second.

Step-by-step explanation:

a) We know that $y = 30\cdot t -10\cdot t^{2}$ describes the position of the ball, measured in feet, in time, measured in seconds, and the average velocity ( $\bar v$ ), measured in feet per second, can be done by means of the following definition:

$\bar v = \frac{y(2+h)-y(2)}{h}$

Where:

$y(2)$ - Position of the ball evaluated at $t = 2\,s$ , measured in feet.

$y(2+h)$ - Position of the ball evaluated at $t =(2+h)\,s$ , measured in feet.

$h$ - Change interval, measured in seconds.

Now, we obtained different average velocities by means of different change intervals:

$h = 0.1\,s$

$y(2) = 30\cdot (2) - 10\cdot (2)^{2}$

$y (2) = 20\,ft$

$y(2.1) = 30\cdot (2.1)-10\cdot (2.1)^{2}$

$y(2.1) = 18.9\,ft$

$\bar v = \frac{18.9\,ft-20\,ft}{0.1\,s}$

$\bar v = -11\,\frac{ft}{s}$

$h = 0.01\,s$

$y(2) = 30\cdot (2) - 10\cdot (2)^{2}$

$y (2) = 20\,ft$

$y(2.01) = 30\cdot (2.01)-10\cdot (2.01)^{2}$

$y(2.01) = 19.899\,ft$

$\bar v = \frac{19.899\,ft-20\,ft}{0.01\,s}$

$\bar v = -10.1\,\frac{ft}{s}$

$h = 0.001\,s$

$y(2) = 30\cdot (2) - 10\cdot (2)^{2}$

$y (2) = 20\,ft$

$y(2.001) = 30\cdot (2.001)-10\cdot (2.001)^{2}$

$y(2.001) = 19.99\,ft$

$\bar v = \frac{19.99\,ft-20\,ft}{0.001\,s}$

$\bar v = -10\,\frac{ft}{s}$

b) The instantaneous velocity when $t = 2\,s$ can be obtained by using the following limit:

$v(t) = \lim_{h \to 0} \frac{x(t+h)-x(t)}{h}$

$v(t) = \lim_{h \to 0} \frac{30\cdot (t+h)-10\cdot (t+h)^{2}-30\cdot t +10\cdot t^{2}}{h}$

$v(t) = \lim_{h \to 0} \frac{30\cdot t +30\cdot h -10\cdot (t^{2}+2\cdot t\cdot h +h^{2})-30\cdot t +10\cdot t^{2}}{h}$

$v(t) = \lim_{h \to 0} \frac{30\cdot t +30\cdot h-10\cdot t^{2}-20\cdot t \cdot h-10\cdot h^{2}-30\cdot t +10\cdot t^{2}}{h}$

$v(t) = \lim_{h \to 0} \frac{30\cdot h-20\cdot t\cdot h-10\cdot h^{2}}{h}$

$v(t) = \lim_{h \to 0} 30-20\cdot t-10\cdot h$

$v(t) = 30\cdot \lim_{h \to 0} 1 - 20\cdot t \cdot \lim_{h \to 0} 1 - 10\cdot \lim_{h \to 0} h$

$v(t) = 30-20\cdot t$

And we finally evaluate the instantaneous velocity at $t = 2\,s$ :

$v(2) = 30-20\cdot (2)$

$v(2) = -10\,\frac{ft}{s}$

The instantaneous velocity of the ball when $t = 2\,s$ is $-10$ feet per second.

8 0

3 years ago

Plz help! 20 points!

Colt1911 [192]

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5 0

4 years ago

Jessie's daughter went to Target and bought $1 or $2 items from the One Spot area. She wants to spend less than $20 at Target

jenyasd209 [6]

She can buy 20 $1 items or 10 $2 items or a mixture of both.

6 0

4 years ago

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Explain how to use friendly percents to find 35% of 80<br>

tresset_1 [31]

You first need to change the percent into a decimal then multiply by 80 to get your answer
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2 years ago