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3 years ago

The gray figure has been transformed in different ways.

Mathematics

1 answer:

yanalaym [24]3 years ago

3 0

Answer:

False ,false,true, true and true

Step-by-step explanation:

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-0.4a + 3 = 7 <br> a=? <br> please help

AleksandrR [38]

Answer:

a = -10

Step-by-step explanation:

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3 years ago

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Question :

Elza [17]

Answer:

C) 405.3 ft

Step-by-step explanation:

A triangle is a polygon with three angles and three sides. Types of triangle are scalene, isosceles, equilateral, right angled triangle.

Given a triangle with angles A, B, C and the corresponding sides directly opposite to the angles as a, b, c. The sine rule states that:

$\frac{a}{sin(A)}= \frac{b}{sin(B)} =\frac{c}{sin(C)}$

In the triangle formed by point C, ship B and ship A, we have ∠A = 82°, ∠B = 78°, c = AB = 140 ft. Hence:

∠A + ∠B + ∠C = 180° (sum of angles in a triangle)

82 + 78 + ∠C = 180

∠C + 160 = 180

∠C = 20°

Using sine rule:

$\frac{c}{sin(C)}=\frac{a}{sin(A)}\\\\\frac{140}{sin(20)}=\frac{a}{sin(82)}\\\\a= \frac{140*sin (82)}{sin(20)}\\\\a=405.3\ ft$

a = distance from Ship B to the signal fire at point C

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2 years ago

Expressions that are equal to 4 times the square root of 5

Morgarella [4.7K]

Answer:

Simplify 4/ (square root of 5) 4 √5 4 5 Multiply 4 √5 4 5 by √5 √5 5 5. 4 √5 ⋅ √5 √5 4 5 ⋅ 5 5

Step-by-step explanation:

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2 years ago

How do I do this.?.?.

viktelen [127]

108 = ( x)(48-x)
108 = 48x - x^2
48x -x^2 -108 = 0

I'm assuming you're doing solve the square right now, so you should be able to do the rest.

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3 years ago

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Please answer correctly !!!!!!!!! Will<br> Mark brainliest !!!!!!!!!!!!!

il63 [147K]

Answer:

$x=-\frac{-20+\sqrt{-20w+3600}}{10},\:x=-\frac{-20-\sqrt{-20w+3600}}{10}$

Step-by-step explanation:

$w=-5\left(x-8\right)\left(x+4\right)\\\mathrm{Expand\:}-5\left(x-8\right)\left(x+4\right):\quad -5x^2+20x+160\\w=-5x^2+20x+160\\Switch\:sides\\-5x^2+20x+160=w\\\mathrm{Subtract\:}w\mathrm{\:from\:both\:sides}\\-5x^2+20x+160-w=w-w\\Simplify\\-5x^2+20x+160-w=0\\Solve\:with\:the\:quadratic\:formula\\\mathrm{Quadratic\:Equation\:Formula:}\\\mathrm{For\:a\:quadratic\:equation\:of\:the\:form\:}ax^2+bx+c=0\mathrm{\:the\:solutions\:are\:}\\x_{1,\:2}=\frac{-b\pm \sqrt{b^2-4ac}}{2a}$

$\mathrm{For\:}\quad a=-5,\:b=20,\:c=160-w:\quad x_{1,\:2}=\frac{-20\pm \sqrt{20^2-4\left(-5\right)\left(160-w\right)}}{2\left(-5\right)}\\x=\frac{-20+\sqrt{20^2-4\left(-5\right)\left(160-w\right)}}{2\left(-5\right)}:\quad -\frac{-20+\sqrt{-20w+3600}}{10}\\x=\frac{-20-\sqrt{20^2-4\left(-5\right)\left(160-w\right)}}{2\left(-5\right)}:\quad -\frac{-20-\sqrt{-20w+3600}}{10}\\The\:solutions\:to\:the\:quadratic\:equation\:are\\x=-\frac{-20+\sqrt{-20w+3600}}{10},\:x=-\frac{-20-\sqrt{-20w+3600}}{10}$

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3 years ago