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2 years ago

Please Help me with the problem please!

Mathematics

1 answer:

evablogger [386]2 years ago

4 0

Answer:

35%

total number of 11th graders = 32+18 = 50

calculating percentage:

18÷50×100

= 36%

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Equation equivalent to 5x(3x+4)

Vika [28.1K]

use the distributive property

5x(3x+4)

15x^2 +20x

7 0

2 years ago

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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.

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

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1 year ago

25 of blank is 60 enter the unknown value that makes it true

Svetradugi [14.3K]

Answer:

240

Step-by-step explanation:

25% of 240 is 60

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1 year ago

*I WILL MARK YOU BRAINLIEST!!!!!*

AlekseyPX

<h2> Answer:</h2>

-3

Explanation:

Rate of change in the interval

X=0 to X=3 is given by,

 $r = \frac{f(3) - f(0)}{3 - 0} \\ here \\ f(3) = 1 \\ f(0) = 10 \\ r = \frac{1 - 10}{3} \\ \: \: \: = \frac{ - 9}{ 3} \\ \: \: \: \: = - 3 \\ hope \: it \: helps.....$ 

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2 years ago

4/9+4(x-1)=28/9-(x-7x)+1

kobusy [5.1K]

X= -23/6 or x= -3.83

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2 years ago

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Please Help me with the problem please!​

Please Help me with the problem please!