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3 years ago

Solve by substitution-5x+4y=22 x - 3y = 0A) No solution.

1 answer:

5 0

We want to use substitution to solve
-5x + 4y = 22 (1)
x - 3y = 0 (2)

From (2), obtain
x = 3y (3)
Substitute (3) into (1).
-5(3y) + 4y = 22
-15y + 4y = 22
-11y = 22
y = -2
From (3), obtain
x = 3y = 3*(-2) = -6

Answer: e. (-6, -2)

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If a polynomial has one root in the form a+ vb, it has a second root in the form of a__ vb.

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(A)using geometry vocabulary, describe a sequence of transformations that maps figure P (-1,2)(-1,4) (-4,2) (-4,4) onto figure Q

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Before we proceed on determining the transformation happening on this problem, it's better to see first the location of the figure by drawing it in a cartesian coordinate plane. We have

If we observe the figures and the coordinates of the plot, we can see that there is a difference of 1 on the x coordinates of P and y coordinates of Q. Therefore, the first transformation that we consider here is the movement of figure P by 1 unit to the left. We have

$\begin{gathered} P_1=(-1-1,2_{})=(-2,2) \\ P_2=(-1-1,4)=(-2,4) \\ P_3=(-4-1,2)=(-5,2) \\ P_4=(-4-1,4)=(-5,4) \end{gathered}$

This transformation changes the location of figure P into

The next transformation will be the rotation of the red dotted figure on the figure above by 90 degrees counterclockwise. With this transformation, the coordinates will transform as

$P_{ccw,90}=(-y,x)$

Hence, for the rotation, we have the new coordinates.

$\begin{gathered} P_1^{\prime}=(-2,-2) \\ P_2^{\prime}=(-4,-2) \\ P_3^{\prime}=(-2,-5) \\ P_4^{\prime}=(-4,-5) \end{gathered}$

The transformed image, which is represented as NMPO, will now be at

For the last transformation, we will be reflecting the figure NMPO over the y axis. This changes the coordinates as

$P_{\text{rotation,y}-\text{axis}}=(-x,y)$

We now have the new coordinates:

$\begin{gathered} P^{\doubleprime}_1=(2,-2)=Q_1_{}_{} \\ P_2^{\doubleprime}=(4,-2)=Q_3 \\ P_3^{\doubleprime}=(2,-5)=Q_2 \\ P_4^{\doubleprime}_{}=(4,-5)=Q_4_{} \end{gathered}$

As you can see, they have the same coordinates as figure Q.

The mapping rules for the sequence described above are as follows:

First transformation (moving one unit to the left (x-1,y))

$\begin{gathered} P_1(-1,2)\rightarrow P_1(-1-1,2)\rightarrow P_1(-2,2) \\ P_2(-1,4)\rightarrow P_1(-1-1,4)\rightarrow P_2(-2,4) \\ P_3(-4,2)\rightarrow P_1(-4-1,2)\rightarrow P_3(-5,2) \\ P_4(-4,4)\rightarrow P_1(-4-1,4)\rightarrow P_4(-5,4) \end{gathered}$

Second transformation (rotation counter clockwise (-y,x))

$\begin{gathered} P_1(-2,2)\rightarrow P^{\prime}_1(-2,-2)_{} \\ P_2(-2,4)\rightarrow P^{\prime}_2(-4,-2) \\ P_3(-5,2)\rightarrow P^{\prime}_3(-2,-5)_{} \\ P_4(-5,4)\rightarrow P^{\prime}_4(-4,-5)_{} \end{gathered}$

Third Transformation (reflection over y-axis (-x,y))

$\begin{gathered} P^{\prime}_1(-2,-2)\rightarrow P^{\doubleprime}_1(-(-2),-2)\rightarrow P^{\doubleprime}_1=(2,-2)=Q_1 \\ P^{\prime}_2(-4,-2)\rightarrow P^{\doubleprime}_1(-(-4),-2)\rightarrow P^{\doubleprime}_1=(4,-2)=Q_3 \\ P^{\prime}_3(-2,-5)\rightarrow P^{\doubleprime}_1(-(-2),-5)\rightarrow P^{\doubleprime}_1=(2,-5)=Q_2 \\ P^{\prime}_4(-4,-5)\rightarrow P^{\doubleprime}_1(-(-4),-5)\rightarrow P^{\doubleprime}_1=(4,-5)=Q_4 \end{gathered}$

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1 year ago

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Answer:

Step-by-step explanation:

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3 years ago

Civil an airport, a factory, and a shopping center are at the vertices of a right triangle formed by three highways. the airport

vekshin1

Answer:

The shortest possible length for the service road is 2.88 miles.

Step-by-step explanation:

According to the below diagram, $A, B$ and $C$ are the positions of airport, shopping center and factory respectively.

Given that, $AB= 3.6 miles, BC= 4.8 miles$ and $AC= 6.0 miles$

In right triangle $ABC$

$tan(\angle ACB)=\frac{AB}{BC} \\ \\ tan(\angle ACB)= \frac{3.6}{4.8}=0.75\\ \\ \angle ACB= tan^-^1(0.75)=36.8698.... degree$

The shortest possible length for the service road from the shopping center to the highway that connects the airport and factory is $BD$ .

That means, $\triangle BCD$ is also a right triangle in which $\angle BDC=90\°$ , Hypotenuse $(BC)= 4.8$ miles and $BD$ is the opposite side in respect of $\angle DCB$ or $\angle ACB$ .

Now in right triangle $BCD$

$Sin(\angle ACB)=\frac{BD}{BC}\\ \\ Sin(36.8698...)=\frac{BD}{4.8}\\ \\ BD=4.8*Sin(36.8698...)=2.88$

So, the shortest possible length for the service road is 2.88 miles.

4 0

3 years ago