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3 years ago

Help me solve this problem plsssss

Mathematics

1 answer:

Arisa [49]3 years ago

6 0

Answer:

It is a Right angle triangle ,

So, Apply Pythagoras Theorem ,

$y = \sqrt{ {2x}^{2} + {(x + 3)}^{2} }$

$y = \sqrt{4 {x}^{2} + {x}^{2} + 3 + 6x }$

$y = \sqrt{5 {x}^{2} + 6x + 9}$

$y = \sqrt{5 \times 4 + 6 \times 2 + 9}$

$y = \sqrt{41}$

<h2>Hope it helps you...</h2>

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It's geometry, not really sure what to do

Grace [21]

Answer:

K'= (-1,-1)

J'= (-1,-5)

L'= (0,-3)

Step-by-step explanation:

What you do here is, input the (x,y) coordinates into the translation.

For example, the original point K is (-3,5). Insert this into the translation.

(-3,5) → (-3+2, 5-8) = (-1,-3)

Repeat this for the next coordinates of L and J.

J= (-3,3)

(-3,3) → (-3+2, 3-8) = (-1,-5)

L= (-2, 5)

(-2, 5) → (-2+2, 5-8) = (0,-3)

5 0

3 years ago

over the summer steven"s tutor told him he needed to complete 15 total workbook pages. he finished 5 workbook pages the first we

Eduardwww [97]

He has 2/3 left to complete.
He did 5 out of 15 which is 5/15 or 1/3. Therefore he has 1-1/3 or 2/3 left.

7 0

3 years ago

Graph the image of the given triangle after the transformation with the rule (x, y)→(x, −y) .

Zarrin [17]

Firstly, we will select three corner points

A=(2,3)

B=(6,8)

C=(7,4)

we are given

the transformation with the rule (x, y)→(x, −y)

y--->-y

so, it is reflected about x-axis

so, we will multiply y-value by -1

we get new points as

A=(2,-3)

B=(6,-8)

C=(7,-4)

now, we can locate these points and draw graph

we get

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2 years ago

Read 2 more answers

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Fudgin [204]

(-3,2) (0,0) i think. it’s been a whole

4 0

3 years ago

Suppose x=c1e−t+c2e3tx=c1e−t+c2e3t. Verify that x=c1e−t+c2e3tx=c1e−t+c2e3t is a solution to x′′−2x′−3x=0x′′−2x′−3x=0 by substitu

Harrizon [31]

The correct question is:

Suppose x = c1e^(-t) + c2e^(3t) a solution to x''- 2x - 3x = 0 by substituting it into the differential equation. (Enter the terms in the order given. Enter c1 as c1 and c2 as c2.)

Answer:

x = c1e^(-t) + c2e^(3t)

is a solution to the differential equation

x''- 2x' - 3x = 0

Step-by-step explanation:

We need to verify that

x = c1e^(-t) + c2e^(3t)

is a solution to the differential equation

x''- 2x' - 3x = 0

We differentiate

x = c1e^(-t) + c2e^(3t)

twice in succession, and substitute the values of x, x', and x'' into the differential equation

x''- 2x' - 3x = 0

and see if it is satisfied.

Let us do that.

x = c1e^(-t) + c2e^(3t)

x' = -c1e^(-t) + 3c2e^(3t)

x'' = c1e^(-t) + 9c2e^(3t)

Now,

x''- 2x' - 3x = [c1e^(-t) + 9c2e^(3t)] - 2[-c1e^(-t) + 3c2e^(3t)] - 3[c1e^(-t) + c2e^(3t)]

= (1 + 2 - 3)c1e^(-t) + (9 - 6 - 3)c2e^(3t)

= 0

Therefore, the differential equation is satisfied, and hence, x is a solution.

4 0

2 years ago

Help me solve this problem plsssss ​

Help me solve this problem plsssss