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mojhsa [17]
3 years ago
6

What is ​ BC ​ ? Enter your answer in the box.

Mathematics
1 answer:
Anuta_ua [19.1K]3 years ago
4 0

from the equal angles it is an isosceles triangle, AB = AC.


4x + 1 = 2x + 23

4x - 2x = 23 - 1

2x = 22

x = 22 : 2

x = 11


----------------------

4 * 11 + 1 = 2 * 11 + 23

45 = 45

--------------

now we find BC

3x - 8

3 * 11 - 8 =

25 (is the value of BC)

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Split up the interval [0, 2] into 4 subintervals, so that

[0,2]=\left[0,\dfrac12\right]\cup\left[\dfrac12,1\right]\cup\left[1,\dfrac32\right]\cup\left[\dfrac32,2\right]

Each subinterval has width \dfrac{2-0}4=\dfrac12. The area of the trapezoid constructed on each subinterval is \dfrac{f(x_i)+f(x_{i+1})}4, i.e. the average of the values of x^2 at both endpoints of the subinterval times 1/2 over each subinterval [x_i,x_{i+1}].

So,

\displaystyle\int_0^2x^2\,\mathrm dx\approx\dfrac{0^2+\left(\frac12\right)^2}4+\dfrac{\left(\frac12\right)^2+1^2}4+\dfrac{1^2+\left(\frac32\right)^2}4+\dfrac{\left(\frac32\right)^2+2^2}4

=\displaystyle\sum_{i=1}^4\frac{\left(\frac{i-1}2\right)^2+\left(\frac i2\right)^2}4=\frac{11}4

4 0
3 years ago
Solve this application problem using a system of equations: The Bright College footballteam scored 80 times last season, some on
Goryan [66]

Answer:

\large \boxed{\sf \ \ 69 \ \textbf{touchdowns } 11 \ \textbf{field goals } \ }

Step-by-step explanation:

Hello,

Touchdown is 7 points, field goals is 3 points

Let's note a the number of touchdowns and b the number of field goals, we can write

   (1)    a + b = 80

   (2)    7a + 3b = 516

(2) - 3*(1) gives

7a + 3b - 3(a + b) = 516 - 3*80

<=> 7a + 3b -3a - 3b = 516 - 240 = 276

<=> 4a = 276

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4 0
3 years ago
<img src="https://tex.z-dn.net/?f=%20%20%5Csf%20%5Chuge%7B%20question%20%5Chookleftarrow%7D" id="TexFormula1" title=" \sf \huge
BabaBlast [244]

\underline{\bf{Given \:equation:-}}

\\ \sf{:}\dashrightarrow ax^2+by+c=0

\sf Let\:roots\;of\:the\: equation\:be\:\alpha\:and\beta.

\sf We\:know,

\boxed{\sf sum\:of\:roots=\alpha+\beta=\dfrac{-b}{a}}

\boxed{\sf Product\:of\:roots=\alpha\beta=\dfrac{c}{a}}

\underline{\large{\bf Identities\:used:-}}

\boxed{\sf (a+b)^2=a^2+2ab+b^2}

\boxed{\sf (√a)^2=a}

\boxed{\sf \sqrt{a}\sqrt{b}=\sqrt{ab}}

\boxed{\sf \sqrt{\sqrt{a}}=a}

\underline{\bf Final\: Solution:-}

\\ \sf{:}\dashrightarrow \sqrt{\alpha}+\sqrt{\beta}

\bull\sf Apply\: Squares

\\ \sf{:}\dashrightarrow (\sqrt{\alpha}+\sqrt{\beta})^2= (\sqrt{\alpha})^2+2\sqrt{\alpha}\sqrt{\beta}+(\sqrt{\beta})^2

\\ \sf{:}\dashrightarrow (\sqrt{\alpha}+\sqrt{\beta})^2 \alpha+\beta+2\sqrt{\alpha\beta}

\bull\sf Put\:values

\\ \sf{:}\dashrightarrow (\sqrt{\alpha}+\sqrt{\beta})^2=\dfrac{-b}{a}+2\sqrt{\dfrac{c}{a}}

\\ \sf{:}\dashrightarrow \sqrt{\alpha}+\sqrt{\beta}=\sqrt{\dfrac{-b}{a}+2\sqrt{\dfrac{c}{a}}}

\bull\sf Simplify

\\ \sf{:}\dashrightarrow \underline{\boxed{\bf {\sqrt{\boldsymbol{\alpha}}+\sqrt{\boldsymbol{\beta}}=\sqrt{\dfrac{-b}{a}}+\sqrt{2}\dfrac{c}{a}}}}

\underline{\bf More\: simplification:-}

\\ \sf{:}\dashrightarrow \sqrt{\alpha}+\sqrt{\beta}=\dfrac{\sqrt{-b}}{\sqrt{a}}+\dfrac{c\sqrt{2}}{a}

\\ \sf{:}\dashrightarrow \sqrt{\alpha}+\sqrt{\beta}=\dfrac{\sqrt{a}\sqrt{-b}+c\sqrt{2}}{a}

\underline{\Large{\bf Simplified\: Answer:-}}

\\ \sf{:}\dashrightarrow\underline{\boxed{\bf{ \sqrt{\boldsymbol{\alpha}}+\sqrt{\boldsymbol{\beta}}=\dfrac{\sqrt{-ab}+c\sqrt{2}}{a}}}}

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