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2 years ago

>: If the scale factor is a 1. What happens to the shape? How can you prove this?

Mathematics

1 answer:

maria [59]2 years ago

3 0

The shape would be exactly the same because any number multiplied by 1 equals itself.

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Three varieties of coffee Coffee A, Coffee B, and Coffee C are combined and roasted, yielding a 55-lb batch of coffee beans.

Klio2033 [76]

Coffee A is 12.241 pounds, Coffee B is 18.269 pounds and Coffee C is 24.482 pounds.

<h3>What is Algebra?</h3>

A branch of mathematics known as algebra deals with symbols and the mathematical operations performed on them.

Variables are the name given to these symbols because they lack set values.

In order to determine the values, these symbols are also subjected to various addition, subtraction, multiplication, and division arithmetic operations.

Given:

Let a pounds of coffee A needed

Let b pounds of coffee B needed

Let c pounds of coffee C needed

c= 2a.................(1)

a+ b+c= 55..........(2)

and, 14.99 a+ 13.99 b + 10.39 c /55 = 12.61

1499a + 1399 b + 1039 c = 69,355...............(3)

Subtract (2) from (3), we get

1499 a+ 1399 b + 1039 c= 69355

- 1039 a - 1039 b - 1039 c = 57145

_____________________________

460 a + 360 b = 12210

46a+ 36 b = 1221

b= 1221/ 36 - 46 a/ 36

b= 407 / 12 - 23 a / 18

Put b in Equation (2)

a + 407 / 12 - 23 a / 18 + 2a= 55

-5a / 18 + 2a + 407/ 12 = 55

31a/ 18 = 253/ 12

a= 4554 / 372

a = 2277/ 186

a= 759/ 62

a= 12.241

and, c=2a = 24.482

b= 33.91 - 15.641

b= 18.269

Learn more about Algebra here:

brainly.com/question/24875240

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7 0

8 months ago

Let L be the line with parametric equations x=5+t,y=6,z=−2−3t. Find the vector equation for a line that passes through the point

scZoUnD [109]

Answer:

The required equations are

$(-5 \hat i + 7 \hat j + 8 \hat k )+\lambda \left((10+\frac {3}{\sqrt {10}})\hat i -\hat j +(6- \frac {9}{\sqrt {10}})\hat k\right)=0$ and

$(-5 \hat i + 7 \hat j + 8 \hat k )+\lambda \left((10-\frac {3}{\sqrt {10}})\hat i -\hat j +(6+ \frac {9}{\sqrt {10}})\hat k\right)=0$ .

Step-by-step explanation:

The given parametric equation of the line, $L$ , is $x=5+t, y=6, z=-2-3t,$ so, an arbitrary point on the line is $R(x,y,z)=R(5+t, 6, -2-3t)$

The vector equation of the line passing through the points $P(-5,7,-8)$ and $R(5+t, 6, -2-3t)$ is

$\vec P + \lambda \vec{(PR)}=0$

$\Rightarrow (-5 \hat i + 7 \hat j - 8 \hat k )+\lambda \left((5+t-(-5))\hat i + (6-7)\hat j +(-2-3t-8)\hat k\right)=0$

$\Rightarrow (-5 \hat i + 7 \hat j - 8 \hat k )+\lambda \left((10+t)\hat i -\hat j +(6-3t)\hat k\right)=0\;\cdots (i)$

The given equation can also be written as

$\frac {x-5}{1}=\frac {v-6}{0}=\frac{z+2}{-3}=t \; \cdots (ii)$

The standard equation of the line passes through the point $P_0(x_0,y_0,z_0)$ and having direction $\vec v= a_1 \hat i +a_2 \hat j +a_3 \hat k$ is

$\frac {x-x_0}{a_1}=\frac {y-y_0}{a_2}=\frac{z-z_0}{a_3}=t \;\cdots (iii)$

Here, The value of the parameter, $t$ , of any point $R$ at a distance $d$ from the point, $P_0$ , can be determined by

$|t \vec v|=d\;\cdots (iv)$

Comparing equations $(ii)$ and $(iii)$

The line is passing through the point $P_0 (5,6,-2)$ having direction $\vec v=\hat i -3 \hat k$ .

Note that the point $Q(5,6,-2)$ is the same as $P_0$ obtained above.

Now, the value of the parameter, $t$ , for point $R$ at distance $d=3$ from the point $Q(5,6,-2)$ can be determined by equation $(iv)$ , we have

$|t(\hat i -3 \hat k)|=3$

$\Rightarrow t^2|(\hat i -3 \hat k)|^2=9$

$\Rightarrow 10t^2=9$

$\Rightarrow t^2=\frac {9}{10}$

$\Rightarrow t=\pm \frac {3}{\sqrt {10}}$

Put the value of $t$ in equation $(i)$ , the required equations are as follows:

For $t= \frac {3}{\sqrt {10}}$

$(-5 \hat i + 7 \hat j - 8 \hat k )+\lambda \left((10+\frac {3}{\sqrt {10}})\hat i -\hat j +\left(6-3\times \frac {3}{\sqrt {10}})\hat k\right)=0$

$\Rightarrow (-5 \hat i + 7 \hat j - 8 \hat k )+\lambda \left((10+\frac {3}{\sqrt {10}})\hat i -\hat j +(6- \frac {9}{\sqrt {10}})\hat k\right)=0$

and for $t= -\frac {3}{\sqrt {10}}$ ,

$(-5 \hat i + 7 \hat j - 8 \hat k )+\lambda \left((10+\left (-\frac {3}{\sqrt {10}}\right))\hat i -\hat j +(6-3\times \left(-\frac {3}{\sqrt {10}}\right)\hat k\right)=0$

$\Rightarrow (-5 \hat i + 7 \hat j - 8 \hat k )+\lambda \left((10-\frac {3}{\sqrt {10}})\hat i -\hat j +(6+ \frac {9}{\sqrt {10}})\hat k\right)=0$

8 0

3 years ago

the ratio of the length of an arc of the circle to the circumference of the circle is 3:7, if the diameter of the circle is 14cm

konstantin123 [22]

The perimeter and area of this minor circle are equal to 32.857 units and 88.345 units² respectively.

<h3>How to calculate the perimeter of the minor circle?</h3>

First of all, we would determine the circumference of this circle by using this formula:

C = πD

C = 22/7 × 14

C = 44 units.

For the length of an arc of the circle:

Arc length/Circumference = 3/7

Arc length = 3/7 × C

Arc length = 3/7 × 44

Arc length = 18.857 units.

Now, we can calculate the perimeter:

Perimeter = 2r + arc length

Perimeter = 2(14/2) + 18.857

Perimeter = 14 + 18.857

Perimeter = 32.857 units.

<h3>The area of the minor circle.</h3>

Mathematically, the area of a minor circle is given by this formula:

Area = θr²/2

But, we would have to find the angle in radians:

18.857/32.857 = θ/360

θ = 3.6059 radians.

Substituting the given parameters into the formula, we have;

Area = (3.6059 × 7²)/2

Area = 88.345 units².

Read more on circumference here: brainly.com/question/14478195

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4 0

1 year ago

9. Shawn wants to make the perimeter of the

Vesna [10]

Answer:

Scale factor of 2.5

Step-by-step explanation:

7 0

2 years ago

The red line segment on the number line below represents the segment from A to B, where A = -5 and B = 5. Find the value of the

Allushta [10]

If A=-5 and B=5 and you are looking for the distance it might be easiest to draw it out on a number line and count the numbers. Answer should be: distance from A to B is 10.

6 0

3 years ago