Darius solved the equation 5 ⋅ d = 15 . his work is shown.

Mathematics

1 answer:

Dafna11 [192]3 years ago

8 0

Answer:

He multiplied both sides by 5.

He should have divided instead.

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Complete this proof. Given: EC AC, DB AC, ∠A = ∠F Prove: ΔMDF ∼ ΔNCA

Gelneren [198K]

Answer:

Proved

Step-by-step explanation:

Given: EC || AC, DB || AC, ∠A = ∠F

Prove: ΔMDF ∼ ΔNCA

Solution

See diagram attached to the solution to better understand the following workings.

Redrawing ΔMDF or rotating to be facing the same direction.

EC is parallel to AC

DB parallel to AC

Using similar triangle theorem:

If ΔMDF ∼ ΔNCA

Ratio of Corresponding sides would be equal

(adjacent of ΔMDF)/(adjacent of ΔNCA) = (Opposite of ΔMDF)/(opposite of ΔNCA) = (hypotenuse of ΔMDF)/(hypotenuse of ΔNCA)

DF/ CA = MD/NC = FM/AN

∠A = ∠F

∠M = ∠N

∠D = ∠C

Since the ratio of Corresponding sides and angle are equal, ΔMDF is similar to ΔNCA.

ΔMDF ∼ ΔNCA

6 0

3 years ago

Read 2 more answers

Evaluate the iterated integral 2 0 2 x sin(y2) dy dx. SOLUTION If we try to evaluate the integral as it stands, we are faced wit

nignag [31]

Answer:

Step-by-step explanation:

Given that:

$\int^2_0 \int^2_x \ sin (y^2) \ dy dx \\ \\ \text{Using backward equation; we have:} \\ \\ \int^2_0\int^2_0 sin(y^2) \ dy \ dx = \int \int_o \ sin(y^2) \ dA \\ \\ where; \\ \\ D= \Big\{ (x,y) | }0 \le x \le 2, x \le y \le 2 \Big\}$

$\text{Sketching this region; the alternative description of D is:} \\ D= \Big\{ (x,y) | }0 \le y \le 2, 0 \le x \le y \Big\}$

$\text{Now, above equation gives room for double integral in reverse order;}$

$\int^2_0 \int^2_0 \ sin (y^2) dy dx = \int \int _o \ sin (y^2) \ dA \\ \\ = \int^2_o \int^y_o \ sin (y^2) \ dx \ dy \\ \\ = \int^2_o \Big [x sin (y^2) \Big] ^{x=y}_{x=o} \ dy \\ \\= \int^2_0 ( y -0) \ sin (y^2) \ dy \\ \\ = \int^2_0 y \ sin (y^2) \ dy \\ \\ y^2 = U \\ \\ 2y \ dy = du \\ \\ = \dfrac{1}{2} \int ^2 _ 0 \ sin (U) \ du \\ \\ = - \dfrac{1}{2} \Big [cos \ U \Big]^2_o \\ \\ = - \dfrac{1}{2} \Big [cos \ (y^2) \Big]^2_o \\ \\ = - \dfrac{1}{2} cos (4) + \dfrac{1}{2} cos (0) \\ \\$