481.65 = 39(x+2.35) solve this problen

Consider ARST and ARYX If the triangles are similar, which must be true?

Sergeeva-Olga [200]

The second choice
The second one is the answer because it is the only one that is consistently matching the sides of the similar triangles with one another

6 0

4 years ago

Jordan bought 9/16 of a proud of sunflower seeds and divided and them evenly into 9 snack bags for his lunch. What is the weight

stepladder [879]

Answer:

The weight of sunflower seeds in each snack bag is 1/16 of a pound

Step-by-step explanation:

Here, we want to know the weight of sunflower seeds in each snack bag.

The total weight bought is 9/16 pound and the division is into 9 bags

So the weight of the individual bags will be ;

9/16 divided by 9

= 9/16 * 1/9 = 1/16

5 0

3 years ago

lyudmila [28]

${\large{\textsf{\textbf{\underline{\underline{Given :}}}}}}$

★ ∆ ABC is similar to ∆DEF

★ Area of triangle ABC = 64cm²

★ Area of triangle DEF = 121cm²

★ Side EF = 15.4 cm

${\large{\textsf{\textbf{\underline{\underline{To \: Find :}}}}}}$

★ Side BC

${\large{\textsf{\textbf{\underline{\underline{Solution :}}}}}}$

Since, ∆ ABC is similar to ∆DEF

[ Whenever two traingles are similar, the ratio of the areas of two similar triangles is equal to the square of the ratio of their corresponding sides. ]

$\therefore \tt \boxed{ \tt \dfrac{area( \triangle \: ABC )}{area( \triangle \: DEF)} = { \bigg(\frac{BC}{EF} \bigg)}^{2} }$

❍ Putting the values, [Given by the question]

• Area of triangle ABC = 64cm²

• Area of triangle DEF = 121cm²

• Side EF = 15.4 cm

$\implies \tt \dfrac{64 \: {cm}^{2} }{12 \: {cm}^{2} } = { \bigg( \dfrac{BC}{15.4 \: cm} \bigg) }^{2}$

❍ By solving we get,

$\implies \tt \sqrt{\dfrac{{64 \: cm}^{2} }{ 121 \: {cm}^{2} }} = \bigg( \dfrac{BC}{15.4 \: cm} \bigg)$

$\implies \tt \sqrt{\dfrac{{(8 \: cm)}^{2} }{ {(11 \: cm)}^{2} }} = \bigg( \dfrac{BC}{15.4 \: cm} \bigg)$

$\implies \tt \dfrac{8 \: cm}{11 \: cm} = \dfrac{BC}{15.4 \: cm}$

$\implies \tt \dfrac{8 \: cm \times 15.4 \: cm}{11 \: cm} = BC$

$\implies \tt \dfrac{123.2 }{11 } cm = BC$

$\implies \tt \purple{ 11.2 \: cm} = BC$

Hence, BC = 11.2 cm.

${\large{\textsf{\textbf{\underline{\underline{Note :}}}}}}$

★ Figure in attachment.

$\rule{280pt}{2pt}$

4 0

3 years ago

Given: AB ∥ DC , m CB =62°, m∠DAB=104° Find: m∠DEA, m∠ADB

maria [59]

Answer:

The values of the angles are;

m∠DEA = 62°, m∠ADB = 45°

Step-by-step explanation:

Specify an arc or an angle three letters

Angle opposite an arc on the circumference

m DA ≅ m CB = 62° (Arc between parallel lines are congruent)

∠CAB = 1/2 × m CB = 1/2 × 62° = 31° (Angle at the center = 2 × Angle st the circumference)

∠DBA = 31° (Angle at the center m DA = 2 × Angle st the circumference)

m∠DAB = 104° (Given)

∠ADB = 180° - m∠DAB - ∠DBA = 180° - 104° - 31° = 45° (Interior angles of triangle ΔADB

m∠ADB = 45°

∠AEB = 180 - ∠CAB - ∠DBA = 180° - 31° - 31° = 118°

∠AEB ≅ ∠COD (Vertically opposite angles)

∠DEA ≅ ∠CEB (Vertically opposite angles)

∠AEB + ∠COD + ∠DEA + ∠CEB = 360° (Sum of angles at a point)

118° + 118° + ∠DEA + ∠CEB = 360°

∠DEA + ∠CEB = 360° - 118° - 118° = 124°

Given that ∠DEA = ∠CEB we have;

2 × ∠DEA = 124°

∠DEA = 124°/2 = 62°

m∠DEA = 62°.

3 0

3 years ago

Gloria earns 1.5 times her normal hourly pay for each hour that she works over 40 hours in a week. Her normal pay is $9.70 per h

ivann1987 [24]

"for each hour that she works over 40hrs in a week" key words "for each hour" in that section of the sentence. so we take 489.85 divide that by 9.7 = 50 and then we divide 489.85 by 5 so that we can get 1/5 of 50, then we take 9.7 times 1.5 = 14.55 and divide that by 97.97

489.85/5 = 97.97/14.55 = 6.73 + 40hrs = 46.73hrs

Not sure why it's a decimal.. but I'm 100% sure that's how you solve it

3 0

3 years ago