Pls help I’ll give brainliest

In ABC AD and BE intersect each other at point q according to a theorem on medians q divides AD in the ratio 2:1 what are the co

victus00 [196]

C. (3.33, -3) would be the answer

8 0

3 years ago

12What mistake did the student make when solvingtheir two-step equation?(a)b) If correctly solved what should the value of be?

mixas84 [53]

Given the equation:

$\frac{x}{6}+3=-18$

(a) You can identify that the student applied the Subtraction Property of Equality by subtraction 3 from both sides of the equation:

$\frac{x}{6}+3-(3)=-18-(3)$

However, the student made a mistake when adding the numbers on the right side.

Since you have two numbers with the same sign on the right side of the equation, you must add them, not subtract them and use the same sign in the result. Then, the steps to add them are:

- Add their Absolute values (their values without the negative sign).

- Write the sum with the negative sign.

Then:

$\frac{x}{6}=-21$

(b) The correct procedure is:

1. Apply the Subtraction Property of Equality by subtracting 3 from both sides (as you did in the previous part):

$\begin{gathered} \frac{x}{6}+3-(3)=-18-(3) \\ \\ \frac{x}{6}=-21 \end{gathered}$

2. Apply the Multiplication Property of Equality by multiplying both sides of the equation by 6:

$\begin{gathered} (6)(\frac{x}{6})=(-21)(6) \\ \\ x=-126 \end{gathered}$

Hence, the answers are:

(a) The student made a mistake by adding the numbers -18 and -3:

$-18-3=-15\text{ (False)}$

(b) The value of "x" should be:

$x=-126$

4 0

1 year ago

Please help! I will mark brainliest!

Maurinko [17]

$\left(3.5\times 10^{-2}\dfrac{kg}{L}\right)\times\left(2.64\times 10^{20}\,L\right)=(3.5\times 2.64)\times 10^{(-2+20)}\,kg\\\\=\bf{9.24\times 10^{18}\,kg}$

_____

Multiply the numbers and add the exponents.

4 0

3 years ago

A rectangle is a quadrilateral with 444 right angles. A rhombus is a quadrilateral with 444 sides of equal length. What is the m

BaLLatris [955]

A shape that has both 4 right angles AND 4 sides of equal length is a square.

5 0

3 years ago

I have a Triangle and I need to find the side BC?

Rus_ich [418]

Looking at this problem in the book, I'm guessing that you've been
introduced to a little bit of trigonometry. Or at least you've seen the
definitions of the trig functions of angles.

Do you remember the definition of either the sine or the cosine of an angle ?

In a right triangle, the sine of an acute angle is (opposite side) / (hypotenuse),
and the cosine of an acute angle is (adjacent side) / (hypotenuse).

Maybe you could use one of these to solve this problem, but first you'd need to
make sure that this is a right triangle.

Let's see . . . all three angles in any triangle always add up to 180 degrees.
We know two of the angles in this triangle ... 39 and 51 degrees.
How many degrees are left over for the third angle ?
180 - (39 + 51) = 180 - (90) = 90 degrees for the third angle.
It's a right triangle ! yay ! We can use sine or cosine if we want to.

Let's use the 51° angle.
The cosine of any angle is (adjacent side) / (hypotenuse) .
'BC' is the side adjacent to the 51° angle in the picture,
and the hypotenuse is 27 .

cosine(51°) = (side BC) / 27

Multiply each side of that equation by 27 :

Side-BC = (27) times cosine(51°)

Look up the cosine of 51° in a book or on your calculator.

Cosine(51°) = 0.62932 (rounded)

<u>Side BC</u> = (27) x (0.62932) = <u>16.992</u> (rounded)
============================================

You could just as easily have used the sine of 39° .
That would be (opposite side) / (hypotenuse) ... also (side-BC) / 27 .

4 0

3 years ago