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3 years ago

How do I find the triangles similarity and each length

Mathematics

1 answer:

Kamila [148]3 years ago

7 0

Angle g would be congruent/equal to itself because of verticals angles theorem. i hope that helped a bit

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Use the x-intercept method to find all real solutions of the equation. -9x^3-72x^2+36=3x^3+x^2-3x+8

larisa86 [58]

-9x^3-72x^2+36=3x^3+x^2-3x+8 Add 9x^3 to both sides.

-72x^2 + 36 = 3x^3 + 9x^3 + x^2 - 3x + 8 Add 72x^2 to both sides

36 = 12x^3 + 73x^2 - 3x + 8 Subtract 36 from both sides.

0 = 12x^3 + 73x^2 - 3x - 28

It does factor, but it is not very nice.

(x + 6.06)(x - 6.09)(x + 0.632)

If there is any kind of error please report it in a note below.

6 0

3 years ago

help please! Darren is finding the equation in the form y = m x + b for a trend line that passes through the points (2, 18) and

andrezito [222]

Answer: d) 14

Step-by-step explanation:

Equation of a line passing through (a,b) and (c,d):

$(y-b)=\dfrac{d-b}{c-a}(x-a)$

Equation of a line passing through (2, 18) and (–3, 8):

$(y-18)=\dfrac{8-18}{-3-2}(x-2)\\\\\Rightarrow\ (y-18)=\dfrac{-10}{-5}(x-2)\\\\\Rightarrow\ (y-18)=2(x-2)\\\\\Rightarrow\ y-18=2x-4\\\\\Rightarrow\ y=2x-4+18\\\\\Rightarrow\ y=2x+14$

Comparing resulting equation $y=2x+14$ to $y = m x + b$ , we get value of b= 14.

Hence, correct option is d) 14

6 0

3 years ago

Read 2 more answers

In a lottery the top cash prize and $663 million going to three lucky winners players pick five different numbers from 1 to 58 a

user100 [1]

Answer:

Step-by-step explanation:

5 0

3 years ago

Find the area of the triangle with the given measurements. Round the solution to the nearest hundredth if necessary. B = 67°, a

kvasek [131]

Among many other formulas, the area of a triangle is

$S = \frac 1 2 a c \sin B$

where B is the angle between sides a and c.

$S=\frac 1 2(10)(20) \sin 67^\circ \approx 92.05 \textrm{ sq cm}$

Answer: 92.05 sq cm, first choice

Bonus. Here's a formula for the area S of a triangle your teacher doesn't know:

$16S^2 = 4a^2b^2 -(c^2-a^2-b^2)^2=(a^2+b^2+c^2)^2-2(a^4+b^4+c^4)$

7 0

3 years ago

A rectangle has width that is 6 meters less than the length. The area of the rectangle is 280 square meters. Find the dimensions

salantis [7]

Dimensions are length 20 meter and width 14 meter

Solution:

Let "a" be the length of rectangle

Let "b" be the width of rectangle

Given that,

A rectangle has width that is 6 meters less than the length

Width = length - 6

b = a - 6

The area of the rectangle is 280 square meters

The area of the rectangle is given by formula:

$Area = length \times width$

Substituting the values we get,

$Area = a \times (a-6)\\\\280 = a^2-6a\\\\a^2-6a -280=0$

Solve the above equation by quadratic formula

$\mathrm{For\:a\:quadratic\:equation\:of\:the\:form\:}ax^2+bx+c=0\mathrm{\:the\:solutions\:are\:}\\\\\quad x_{1,\:2}=\frac{-b\pm \sqrt{b^2-4ac}}{2a}$

$\mathrm{For\:}\quad a=1,\:b=-6,\:c=-280:\quad a_{1,\:2}=\frac{-\left(-6\right)\pm \sqrt{\left(-6\right)^2-4\cdot \:1\left(-280\right)}}{2\cdot \:1}$

$a =\frac{6 \pm \sqrt{36+1120}}{2}\\\\a = \frac{6 \pm \sqrt{1156}}{2}\\\\a = \frac{6 \pm 34}{2}\\\\Thus\ we\ have\ two\ solutions\\\\a = \frac{6+34}{2} \text{ or } a = \frac{6-34}{2}\\\\a = 20 \text{ or } a = -14$

Since, length cannot be negative, ignore a = -14

Thus solution of length is a = 20

Therefore,

width = length - 6

width = 20 - 6 = 14

Thus dimensions are length 20 meter and width 14 meter

6 0

3 years ago