ABC and QRS are complementary angles. ABC=52, what is QRS?

Which equation is an identity? (Can someone explain to me how they got their answer, I don't get this.)

Greeley [361]

5w + 8 - w = 6w - 2(w -4).

Let's reduce the equation on the left:

4w + 8; (1)

and now let's reduce the equation on the right

6w - 2w +8 = 4w +8 (2).

We notice that (1) ≈ (2), and this is the only IDENTITY

7 0

3 years ago

In Exercises 9-14, decide whether enough information

Degger [83]

The triangles that can be proved to be congruent by the SAS Congruence Theorem are:

10. ΔLMN and ΔNQP

13. ΔEFH and ΔGHF

The triangles that cannot be proved to be congruent by the SAS Congruence Theorem due to insufficient information are:

9. ΔABD and ΔCDB

12. ΔQRV and ΔTSU

14. ΔKLM and ΔMNK

11. ΔYXZ and ΔWXZ

<h3>What is the SAS Congruence Theorem?</h3>

If two triangles have two pairs of corresponding sides that are congruent, and a pair of corresponding included angle that are congruent, both triangles can be proven to be congruent triangles by the SAS Congruence Theorem.

Thus:

9. ΔABD and ΔCDB have:

one pair of congruent non-included angles - ∠ABD ≅ ∠CDB
two pairs of congruent sides - AD ≅ BC, and BD ≅ BD

ΔABD and ΔCDB lack a pair of included angles, therefore, the information given is not enough to prove the triangles are congruent by the SAS Congruence Theorem.

10. ΔLMN and ΔNQP have:

one pair of congruent included angles - ∠LMN ≅ ∠NQP
two pairs of congruent sides - LM ≅ NQ, and MN ≅ QP

ΔLMN and ΔNQP, therefore, have enough information to prove that they are congruent triangles by the SAS Congruence Theorem.

11. ΔYXZ and ΔWXZ have:

one pair of congruent non-included angles - ∠YXZ ≅ ∠WXZ
two pairs of congruent sides - XZ ≅ XZ, and XW ≅ YZ

ΔYXZ and ΔWXZ lack a pair of included angles, therefore, the information given is not enough to prove the triangles are congruent by the SAS Congruence Theorem.

12. ΔQRV and ΔTSU do not have enough information to prove they are congruent by the SAS Congruence Theorem.

13. ΔEFH and ΔGHF have two pairs of congruent sides and a pair of congruent included angles, therefore, there is enough information to prove they are congruent by the SAS Congruence Theorem.

14. ΔKLM and ΔMNK do not have enough information to prove they are congruent by the SAS Congruence Theorem.

Therefore, the triangles that can be proved to be congruent by the SAS Congruence Theorem are:

10. ΔLMN and ΔNQP

13. ΔEFH and ΔGHF

The triangles that cannot be proved to be congruent by the SAS Congruence Theorem due to insufficient information are:

9. ΔABD and ΔCDB

12. ΔQRV and ΔTSU

14. ΔKLM and ΔMNK

11. ΔYXZ and ΔWXZ

Learn more about SAS Congruence Theorem on:

brainly.com/question/13408604

3 0

3 years ago

Find the areas of the figures below. Show all work on for each problem.

Marysya12 [62]

Answer: A = ≈28.55 cm²

Step-by-step explanation:

It is drawn as if it should be a triangle minus a rectangle with an area of

½(10)(13) - (7)(6) = 23 cm²

However it is not to scale.

The angle on the right side is

θ = arctan10/13 = 37.56859...°

Meaning the right triangle has a height of

h = 4tan37.56859 = 3.07692 cm

So the right triangle has an area of

Ar = ½(3.07692)(4) = 6.15384... cm²

The left hand side consists of a triangle over a rectangle

The height of the triangle is 4, the base is is 4/tan37.56859 = 5.2 cm

so the left area is

(2)(6) + ½(5.2)(4) = 22.4 cm²

So the sum of the areas of the two enclosed shapes is

A = 22.4 + 6.15384... = 28.55384... ≈28.55 cm²

5 0

3 years ago

Help with special right triangle please

Ahat [919]

Answer:

Step-by-step explanation:

Looking at the right angle triangle that contains x and y, we would apply trigonometric ratio to determine x an y. Taking 60 degrees as the reference angle,

Hypotenuse = y

Adjacent side = x

Opposite side = 36

Therefore,

Sin # = opposite side / hypotenuse

Sin 60 = 36/y

y = 36/Sin60 = 36/0.8660 = 41.57

Cos # = adjacent side/hypotenuse

Cos 60 = x/41.57

x = 41.57Cos 60 = 41.57 × 0.5 = 20.785

Sin45 = z/36

z = 36Sin45 = 36 × 0.7071 = 25.4556

7 0

3 years ago

Please I need help with this question!

likoan [24]

Answer:

I. 60%

II. 75.4 kg

Step-by-step explanation:

We will use the z-scores and the standard normal distribution to answer this questions.

We have a normal distribution with mean 69 kg and variance 25 kg^2 (therefore, standard deviation of 5 kg).

I. What percentage of adult male in Boston weigh more than 72 kg?

We calculate the z-score for 72 kg and then calculate the associated probability:

$z=\dfrac{X-\mu}{\sigma}=\dfrac{72-69}{5}=\dfrac{3}{5}=0.6\\\\\\P(X>72)=P(z>0.6)=0.274$

II. What must an adult male weigh in order to be among the heaviest 10% of the population?

We have to calculate tha z-score that satisfies:

$P(z>z^*)=0.1$

This happens for z=1.28 (see attachment).

Then, we can calculate the weight using this transformation:

$X=\mu+z^*\cdot\sigma=69+1.28\cdot 5=69+6.4=75.4$

5 0

3 years ago