50 POINTS!!!!!!!!!! Solve for EG. A) 12 B) 15 C) 21 D) 30

Mathematics

2 answers:

Sergio039 [100]3 years ago

6 0

The answer is C) 21, because 6/14 is simplified to 3/7. assuming it’s a pattern 3/7, 6/14, 12/21.

Olegator [25]3 years ago

6 0

Assuming the parallel looking lines are actually parallel, we get proportionality from the corresponding parts:

$\dfrac{FG}{IJ} = \dfrac{EF}{HI}$

$FG = IJ \dfrac{EF}{HI} = 14 (9/6) = 21$

$EG = EF + FG = 9 + 21 = 30$

Answer: D

You might be interested in

Evidence of a chemical reaction includes anything that shows -

puteri [66]

Answer:

a change in the state of matter

Step-by-step explanation:

6 0

3 years ago

Please help it's about Phythagorean Theorem! <br> I attached the image

storchak [24]

Pythagorean Theorem is:

c² = a² + b²

Let’s find the values of a and b.

Looking at the image provided in the question we have 2 number already, 70 and 120.

Let’s name the side with length 120 as a, and the side with length 70 as b.

Then we should substitute these values into the Pythagorean Theorem:

c² = 120² + 70²

c² = 14400 + 4900

c² = 19300

Then we should work out the value of c by square rooting both sides:

√c = √19300

c = 138.92444

Therefore the ball would have to be hit a total of 138.92444 units.

If you have to round, that would be 139 units.

8 0

3 years ago

Calculate the area of the triangle with the following vertices (3, -7), (6, 4), (-2, -3)

Monica [59]

Answer:

$\boxed{\mathsf{A} \triangle = \red{\dfrac{67}{2}u.a}}$

Step-by-step explanation:

Let's follow up with the solution. Considering a triangle with the vertices $\mathsf{A(x_A, y_A)}$ , $\mathsf{B(x_B, y_B)}$ and $\mathsf{C(x_C, y_C)}$ , have a look at the representation in the cartesian plan.

From this representation we can say that the area (A) of a triangle through the knowledge of <u>analytical geometry</u> is given by the determinant of the vertices divided by two, mathematically,

$\mathsf{A} \triangle = \dfrac{\left| \begin{array}{ccc} \mathsf{x_A} & \mathsf{y_A }& 1 \\ \mathsf{x_B} & \mathsf{ y_B} & 1 \\ \mathsf{ x_C} & \mathsf{ y_C} & 1 \end{array} \right|}{2}$

So, applying this knowledge we're going to have,

$\mathsf{A} \triangle = \dfrac{\left| \begin{array}{ccc} 3 & -7 & 1 \\ 6 & 4 & 1 \\ -2 & -3 & 1 \end{array} \right|}{2}$

$\mathsf{A} \triangle = \dfrac{1}{2}\left[ \left.\begin{array}{ccc} 3 & -7 & 1 \\ 6 & 4 & 1 \\ -2& -3 & 1 \end{array} \right| \begin{array}{cc} 3 & -7 \\ 6 & 4 \\ -2 & -3 \end{array} \right]$