Michael took his science test today and got a 14/16. What grade did he receive in total A/B/C??

Please help me please

lbvjy [14]

60 because area of a rectangle is l*b*w. So 6*2*5 will equal 60

6 0

2 years ago

Solve the system of equations by substitution. <br>y=3x-10, 2x+3y=14<br><br>Its 5-3 MathXL btw

hjlf

2x+3(3x-10)=14

2x+9x-30=14

11x=44

x=4

y=3(4)-10

y=2

7 0

3 years ago

Point A (6,2) is translated using the vector <-5,2>. Where is the new point located?

Dominik [7]

<h3>Answer: (1,4)</h3>

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

Explanation:

The notation <-5,2> is the same as writing the translation rule $(x,y) \to (x-5,y+2)$

It says: move 5 units to the left and 2 units up

The point (6,2) moves to (1,2) when moving five units to the left. Then it ultimately arrives at (1, 4) after moving 2 units up. You could move 2 units up first and then 5 units to the left later on, and you'd still arrive at (1, 4). In this case, the order doesn't matter (some combinations of transformations this won't be the case and order will matter).

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Or you could write out the steps like so

$(x,y) \to (x-5, y+2)\\\\(6,2) \to (6-5, 2+2)\\\\(6,2) \to (1, 4)\\\\$

We see that (6,2) moves to (1, 4)

4 0

3 years ago

Please help, the question below

Licemer1 [7]

Answer:

$\sf R=\left(3, -\dfrac{5}{4}\right)$

Step-by-step explanation:

Given:

P = (1, 5)
Q = (-2, 3)
R = (a, b)
R lies on line x = 3
PR = QR

If point R lies on the line x = 3, then the x-value of point R is 3.

⇒ a = 3

<u />

<u>Distance between two points</u>

$\sf d=\sqrt{(x_2-x_1)^2+(y_2-y_1)^2}$

(where (x₁, y₁) and (x₂, y₂) are the two points)

Use the <u>distance formula</u> to derive equations for PR and QR.

Let (x₁, y₁) = P = (1, 5)

Let (x₂, y₂) = R = (3, b)

$\begin{aligned} \sf PR & =\sf \sqrt{(3-1)^2+(b-5)^2}\\ & = \sf \sqrt{4+(b-5)^2} \end{aligned}$

Let (x₁, y₁) = Q = (-2, 3)

Let (x₂, y₂) = R = (3, b)

$\begin{aligned} \sf QR & =\sf \sqrt{(3-(-2))^2+(b-3)^2}\\ & = \sf \sqrt{25+(b-3)^2} \end{aligned}$

As PR = QR, equate the derived equations and solve for b:

$\begin{aligned} \sf PR & = \sf QR \\\sf \sqrt{4+(b-5)^2} & = \sf \sqrt{25+(b-3)^2}\\\sf 4+(b-5)^2 & = \sf 25+(b-3)^2\\\sf 4+b^2-10b+25 & = \sf 25 + b^2-6b+9\\\sf b^2-10b+29 & = \sf b^2 -6b +34\\\sf -10b+29 & = \sf -6b + 34\\\sf -4b & = \sf 5\\\sf b & = -\dfrac{5}{4}\end{aligned}$

Substitute the found values of a and b to find the coordinates of R:

$\sf R=(a,b)=\left(3, -\dfrac{5}{4}\right)$

Learn more about the distance formula here:

brainly.com/question/28144723

8 0

2 years ago

Read 2 more answers

Which two quadrilaterals are reflections of each other across the y-axis?

ICE Princess25 [194]

Answer:

Step-by-step explanation:

When you reflect a point across the y-axis, the y-coordinate remains the same, but the x-coordinate is taken to be the additive inverse. The reflection of point (x, y) across the y-axis is (-x, y).

Hope this helped.

7 0

3 years ago