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3 years ago

Help find the distance between two points?

Mathematics

1 answer:

miv72 [106K]3 years ago

7 0

Answer:

Step-by-step explanation:

Calculate the distance using the distance formula

d = √ (x₂ - x₁ )² + (y₂ - y₁ )²

with (x₁, y₁ ) = B(0, 0) and (x₂, y₂ ) = ( $\frac{\sqrt{6} }{2}$ , $\frac{\sqrt{58} }{2}$ )

AB = $\sqrt{((\frac{\sqrt{6} }{2})^2 }$ + $\{(\frac{\sqrt{58} }{2})^2 }$

= $\sqrt{\frac{6}{4} }$ + $\frac{58}{4}$

= $\sqrt{\frac{64}{4} }$

= $\sqrt{16}$

= 4

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Square A has a side length (2x-7) and Square B has side length (-4x+18). How much bigger is the perimeter of Square B than Squar

mihalych1998 [28]

How much bigger is B than A , so Side length B- A = (-4x+18) - (2x-7) = -6x+25
So perimeter = 4x length = 4x(-6x+25) = -24x+100 = 100-24x
How much bigger? 100-24X

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3 years ago

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29 mint plants and 29 other herb plants at the garden center. Considering this data, how many of the next 62 herb plants stocked

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Answer: 31 mint plants

Step-by-step explanation:

We will have mint plants over total plants:

29/29+29 = 29/58 = 1/2

Now, we will multiply 62 by 1/2 to guess how many mint plants will be stocked.

62* 1/2 = 31

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2 years ago

Point X is the incenter of ΔABC.

Gnoma [55]

The measures of GX = 13

The measures of $\angle ABX=22^0$

EX = 4z + 1 and XF = 2z + 7

Take EX and XF equal time each other and then solve for z and then you will get GX because EX = XF = GX are congruent.

EX = XF

<h3>What is congruent?</h3>

Two figures or objects are congruent if they have the same shape and size, or if one has the same shape and size as the mirror image of the other.

4z + 1 = 2z + 7

2z = 6

z = 3

As, EX = XF = GX

EX = 4(3) + 1

EX = 13

Therefore, GX = 13 ------ ( EX = XF = GX )

To find the angle ABX, divide the angle ABC by 2 you will get

$\angle ABX=22^0$

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2 years ago

Find the general solution to each of the following ODEs. Then, decide whether or not the set of solutions form a vector space. E

Ipatiy [6.2K]

Answer:

(A) $y=ke^{2t}$ with $k\in\mathbb{R}$ .

(B) $y=ke^{2t}/2-1/2$ with $k\in\mathbb{R}$

(D) $y=k_1e^{2t}+k_2e^{-2t}+e^{3t}/5$ with $k_1,k_2\in\mathbb{R}$ ,

Step-by-step explanation

(A) We can see this as separation of variables or just a linear ODE of first grade, then $0=y'-2y=\frac{dy}{dt}-2y\Rightarrow \frac{dy}{dt}=2y \Rightarrow \frac{1}{2y}dy=dt \ \Rightarrow \int \frac{1}{2y}dy=\int dt \Rightarrow \ln |y|^{1/2}=t+C \Rightarrow |y|^{1/2}=e^{\ln |y|^{1/2}}=e^{t+C}=e^{C}e^t} \Rightarrow y=ke^{2t}$ . With this answer we see that the set of solutions of the ODE form a vector space over, where vectors are of the form $e^{2t}$ with $t$ real.

(B) Proceeding and the previous item, we obtain $1=y'-2y=\frac{dy}{dt}-2y\Rightarrow \frac{dy}{dt}=2y+1 \Rightarrow \frac{1}{2y+1}dy=dt \ \Rightarrow \int \frac{1}{2y+1}dy=\int dt \Rightarrow 1/2\ln |2y+1|=t+C \Rightarrow |2y+1|^{1/2}=e^{\ln |2y+1|^{1/2}}=e^{t+C}=e^{C}e^t \Rightarrow y=ke^{2t}/2-1/2$ . Which is not a vector space with the usual operations (this is because $-1/2$ ), in other words, if you sum two solutions you don't obtain a solution.

(C) This is a linear ODE of second grade, then if we set $y=e^{mt} \Rightarrow y''=m^2e^{mt}$ and we obtain the characteristic equation $0=y''-4y=m^2e^{mt}-4e^{mt}=(m^2-4)e^{mt}\Rightarrow m^{2}-4=0\Rightarrow m=\pm 2$ and then the general solution is $y=k_1e^{2t}+k_2e^{-2t}$ with $k_1,k_2\in\mathbb{R}$ , and as in the first items the set of solutions form a vector space.

(D) Using C, let be $y=me^{3t}$ we obtain that it must satisfies $3^2m-4m=1\Rightarrow m=1/5$ and then the general solution is $y=k_1e^{2t}+k_2e^{-2t}+e^{3t}/5$ with $k_1,k_2\in\mathbb{R}$ , and as in (B) the set of solutions does not form a vector space (same reason! as in (B)).

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3 years ago

Please help me. I really need help

Flauer [41]

A. 38

Let X = seventh graders

6th graders are twice the number of 7th graders; therefore, 6th graders will be 2X.
The total number of students is 57.

2X + X = 57
3X = 57
X = 57 / 3
X = 19

IF X represents the 7th graders; there are 19 seventh graders present.
Since 6th graders are represented by 2X; Then, 2 * 19 = 38 is the number of 6th graders present.

Add both numbers to check: 19 + 38 = 57.

6 0

4 years ago

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