18 19 O 18-1/4 O 1/4" 1-1/4

topjm [15]

Ah yes, good ol’ Pythagorean’s theorem.

The hypotenuse is the opposite of the 90 degree angle, the longest side of the triangle.

The legs are opposite of the acute angles

Formula: a^2 + b^2 = c^2

2:

Let’s plug in some values

(6)^2 + (3)^2 = c^2

C = 6.7

The side will be 6.7 units

3)

Let’s plug in some values

(13.3)^2 = (9.7)^2 + (b)^2

(13.3)^2 - (9.7)^2 = b^2

176.89 - 94.09 = b^2

b = 9.1

The side will be 9.1 units

3 0

2 years ago

Solve for x: 4/x+4/x2-9=3/x-3

anzhelika [568]

Answer:

x =1/12(1-√(97) )

Step-by-step explanation:

4 0

2 years ago

Monique's son just turned 2 years old and is 34 inches tall. Monique heard that the average boy will grow approximately 2 5/8 in

tatuchka [14]

Answer:

The equation representing how old Monique son is $\mathbf{a = 2 + \dfrac{8}{21}(q-34)}$

Step-by-step explanation:

From the given information:

A linear function can be used to represent the constant growth rate of Monique Son.

i.e.

$q(t) = \hat q \times t + q_o$

where;

$q_o$ = initial height of Monique's son

$\hat q$ = growth rate (in)

t = time

So, the average boy grows approximately 2 5/8 inches in a year.

i.e.

$\hat q = 2 \dfrac{5}{8} \ in/yr$

$\hat q = \dfrac{21}{8} \ in/yr$

Then; from the equation $q(t) = \hat q \times t + q_o$

$34 = \dfrac{21}{8} \times 0 + q_o$

$q_o = 34\ inches$

The height of the son as a function of the age can now be expressed as:

$q(t) = \dfrac{21}{8} \times t + 34$

Then:

Making t the subject;

$q - 34 = \dfrac{21}{8} \times t$

$t = \dfrac{8}{21}(q-34)$

and the age of the son i.e. ( a (in years)) is:

a = 2 + t

So;

$\mathbf{a = 2 + \dfrac{8}{21}(q-34)}$

SO;

if q (growth rate) = 50 inches tall

Then;

$\mathbf{a = 2 + \dfrac{8}{21}(50-34)}$

$\mathbf{a = 2 + \dfrac{8}{21}(16)}$

a = 2 + 6.095

a = 8.095 years

a ≅ 8 years

i.e.

Monique son will be 8 years at the time Monique is 50 inches tall.

8 0

2 years ago

Segments

Ipatiy [6.2K]

There are several ways two triangles can be congruent.

$\mathbf{AC = BD}$ congruent by SAS
$\mathbf{\angle ABC \cong \angle BAD}$ congruent by corresponding theorem

In $\mathbf{\triangle AOL}$ and $\mathbf{\triangle BOK}$ (see attachment), we have the following observations

$1.\ \mathbf{AO = DO}$ --- Because O is the midpoint of line segment AD

$2.\ \mathbf{BO = CO}$ --- Because O is the midpoint of line segment BC

$3.\ \mathbf{\angle AOB =\angle COD}$ ---- Because vertical angles are congruent

$4.\ \mathbf{\angle AOC =\angle BOD}$ ---- Because vertical angles are congruent

Using the SAS (side-angle-side) postulate, we have:

$\mathbf{AC = BD}$

Using corresponding theorem,

$\mathbf{\angle ABC \cong \angle BAD}$ ---- i.e. both triangles are congruent

The above congruence equation is true because:

2 sides of both triangles are congruent
1 angle each of both triangles is equal
Corresponding angles are equal

See attachment

Read more about congruence triangles at:

brainly.com/question/20517835

3 0

2 years ago

Mrs. Goldstein is building a new house. She creates a scale drawing of the house on a piece of paper. The scale is 1cm = 5.5ft.

enyata [817]

Answer:11343.75 feet

Step-by-step explanation:

so you do 15 x 5.5 and then 25 x 5.5 then you times the answers of the two and you get your answer

this should be right.

Hoped this helped

4 0

2 years ago