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3 years ago

TRULY IN NEED OF HELP PLEASEE

Mathematics

2 answers:

Alexeev081 [22]3 years ago

7 0

Answer:
x=10
y=7.5
CE = 6
BD=4.5

Explanation:
Use similar triangles. Since you know one side length of the larger triangle, divide it by the hypotenuse of the smaller triangle to get the factor you are multiplying you side lengths by.

12.5 / 5
= 2.5

Multiply every other side length by 2.5

4 x 2.5 = 10
x = 10

3 x 2.5 = 7.5
y = 7.5

To get CE and BD subtract your values:

CE = 10 - 4
CE = 6

BD = 7.5 - 3
BD = 4.5

maria [59]3 years ago

5 0

Answer:

no lo se, sol necesitaba puntos

Step-by-step explanation:

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Solve for p in the literal equation 3p + 41 =q.

denis23 [38]

Answer:

The answer is 44 because 3+41=44

7 0

3 years ago

The equations -6x + 3y = 6 and x² + y = 10 intersect at the

eimsori [14]

The values of the letters in coordinates (a, b) and (c,d) are;

a = -4

b = -6

c = 2

d = 6

We are given two equations;

-6x + 3y = 6 ---(eq 1)

x² + y = 10 ---(eq 2)

We are told that they intersect at coordinates; (a, b) and (c, d).

Let us make y the subject in eq 2 to get;

y = 10 - x² --(eq 3)

Let us put 10 - x² into eq 1 to get;

-6x + 3(10 - x²) = 6

expanding further gives;

-6x + 30 - 3x² = 6

rearranging gives;

3x² + 6x - 24 = 0

Using online quadratic equation solver, we have;

x = -4 and x = 2

Putting x = -4 into eq 3 gives;

y = 10 - (-4)²

y = 10 - 16

y = -6

Putting x = 2 into eq 3 gives;

y = 10 - (2)²

y = 10 - 4

y = 6

Thus, the coordinates are; (-4, -6) and (2, 6)

Comparing with (a, b) and (c,d), we have;

a = -4

b = -6

c = 2

d = 6

Read more at; brainly.com/question/15165519

5 0

3 years ago

Use the method of reduction of order to find a second solution to t^2y' + 3ty' – 3y = 0, t> 0 Given yı(t) = t y2(t) = Preview

Anestetic [448]

Let $y_2(t)=tv(t)$ . Then

${y_2}'=tv'+v$

${y_2}''=tv''+2v'$

and substituting these into the ODE gives

$t^2(tv''+2v')+3t(tv'+v)-3tv=0$

$t^3v''+5t^2v'=0$

$tv''+5v'=0$

Let $u(t)=v'(t)$ , so that $u'(t)=v''(t)$ . Then the ODE is linear in $u$ , with

$tu'+5u=0$

Multiply both sides by $t^4$ , so that the left side can be condensed as the derivative of a product:

$t^5u'+5t^4u=(t^5u)'=0$

Integrating both sides and solving for $u(t)$ gives

$t^5u=C\implies u=Ct^{-5}$

Integrate again to solve for $v(t)$ :

$v=C_1t^{-6}+C_2$

and finally, solve for $y_2(t)$ by multiplying both sides by $t$ :

$tv=y_2=C_1t^{-5}+C_2t$

$y_1(t)=t$ already accounts for the $t$ term in this solution, so the other independent solution is $y_2(t)=t^{-5}$ .

6 0

3 years ago

Can anyone tell me what the answer please

Tasya [4]

What you're looking at in this image are two alternate exterior angles.

Alternate exterior angles, are, essentially, exterior angles of a transversal that runs through two parallel lines. These angles are on alternating sides. These angles, assuming the lines are parallel, must be equal.

There's a proof behind why these angles are equal, but I won't bore you with the specifics as the question doesn't require it.

So - we know that these two angles must be equal to prove that the lines A and B are parallel. Knowing this, we can write an equation:

5x + 20 = 3x + 60

Subtract 20 from both sides:

5x + 20 - 20 = 3x + 60 - 20

5x = 3x + 40

Subtract 3x from both sides:

5x - 3x = 3x - 3x + 40

2x = 40

x = 20

If you have any questions on how I got to the answer, just ask!

- breezyツ

3 0

3 years ago

the sum of two numbers is 133. four times the smaller of the two numbers equals three times the greater number. find the numbera

Evgesh-ka [11]

67

2x/2=132/2
x =66

x+1=
66+1=67
Therefore the smaller number is 66 and the greater 67.

8 0

2 years ago