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3 years ago

Find the distance between these points. C(0, 4), T(-6, -3) √(37) √(85) √(109)

Mathematics

2 answers:

Leona [35]3 years ago

5 0

<span>√85</span>≈<span>9.21954445

Hope this helps. c:</span>

lorasvet [3.4K]3 years ago

5 0

Answer:

(B) $CT=\sqrt{85}$

Step-by-step explanation:

Given: Two points $C(0,4)$ and $T(-6,-3)$ .

To find: The distance between the given points $C(0,4)$ and $T(-6,-3)$ .

Solution: It is given taht the two points are $C(0,4)$ and $T(-6,-3)$ , we have to find the distance between these two points, thus using teh distance formula, we have:

$CT=\sqrt{(-6-0)^2+(-3-4)^2}$

⇒ $CT=\sqrt{36+49}$

⇒ $CT=\sqrt{85}$

which is the required distance between the two given points.

Therefore, option B is correct.

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There is a triangle with a perimeter of 63 cm, one side of which is 21 cm. Also, one of the medians is perpendicular to one of t

NemiM [27]

Answer:

21cm; 28cm; 14cm

Step-by-step explanation:

There is no info in the problem/s text which one of the triangle's side is 21 cm. That is why we have to try all possible variants.

Let the triangle is ABC . Let the AK is the angle A bisector and BM is median.

Let O is AK and BM cross point.

Have a look to triangle ABM. AO is the bisector and AOB=AOM=90 degrees (means that AO is as bisector as altitude)

=> triangle ABM is isosceles => AB=AM (1)

1. Let AC=21 So AM=21/2=10.5 cm

So AB=10.5 cm as well. So BC= P-AB-AC=63-21-10.5=31.5 cm

Such triangle doesn' t exist ( is impossible), because the triangle's inequality doesn't fulfill. AB+AC>BC ( We have 21+10.5=31.5 => AB+AC=BC)

2. Let AB=21 So AM=21 and AC=42 .So BC= P-AB-AC=63-21-42=0 cm- such triangle doesn't exist.

3. Finally let BC=21 cm. So AB+AC= 63-21=42 cm

We know (1) that AB=AM so AC=2*AB. So AB+AC=AB+2*AB=3*AB

=>3*AB=42=> AB=14 cm => AC=2*14=28 cm.

Let check if this triangle exists ( if the triangle's inequality fulfills).

BC+AB>AC 21+14>28 - correct=> the triangle with the sides' length 21cm,14 cm, 28cm exists.

This variant is the only possible solution of the given problem.

6 0

3 years ago

Write an appropriate direct variation equation if y = 45 when x = 5.

Firdavs [7]

45×5 or 45÷5 i hope this helps if it dont im so so very very sorry for you

4 0

3 years ago

Evaluate the expression. ( – 5+ – 9)× – 3÷ – 2

Ronch [10]

Answer:

$-21$

Step-by-step explanation:

$( - 5+ - 9) \times - 3 \div - 2$

Solve the brackets first.

$( - 14) \times - 3 \div - 2$

Division comes next.

$( - 14) \times - 3/-2$

Cancel the negative signs.

$- 14 \times 3/2$

Multiply the terms.

$-42/2$

$=-21$

3 0

3 years ago

Read 2 more answers

The integral of (5x+8)/(x^2+3x+2) from 0 to 1

Gnom [1K]

Compute the definite integral:
integral_0^1 (5 x + 8)/(x^2 + 3 x + 2) dx

Rewrite the integrand (5 x + 8)/(x^2 + 3 x + 2) as (5 (2 x + 3))/(2 (x^2 + 3 x + 2)) + 1/(2 (x^2 + 3 x + 2)):
= integral_0^1 ((5 (2 x + 3))/(2 (x^2 + 3 x + 2)) + 1/(2 (x^2 + 3 x + 2))) dx

Integrate the sum term by term and factor out constants:
= 5/2 integral_0^1 (2 x + 3)/(x^2 + 3 x + 2) dx + 1/2 integral_0^1 1/(x^2 + 3 x + 2) dx

For the integrand (2 x + 3)/(x^2 + 3 x + 2), substitute u = x^2 + 3 x + 2 and du = (2 x + 3) dx.
This gives a new lower bound u = 2 + 3 0 + 0^2 = 2 and upper bound u = 2 + 3 1 + 1^2 = 6: = 5/2 integral_2^6 1/u du + 1/2 integral_0^1 1/(x^2 + 3 x + 2) dx

Apply the fundamental theorem of calculus.
The antiderivative of 1/u is log(u): = (5 log(u))/2 right bracketing bar _2^6 + 1/2 integral_0^1 1/(x^2 + 3 x + 2) dx

Evaluate the antiderivative at the limits and subtract.
(5 log(u))/2 right bracketing bar _2^6 = (5 log(6))/2 - (5 log(2))/2 = (5 log(3))/2: = (5 log(3))/2 + 1/2 integral_0^1 1/(x^2 + 3 x + 2) dx

For the integrand 1/(x^2 + 3 x + 2), complete the square:
= (5 log(3))/2 + 1/2 integral_0^1 1/((x + 3/2)^2 - 1/4) dx

For the integrand 1/((x + 3/2)^2 - 1/4), substitute s = x + 3/2 and ds = dx.
This gives a new lower bound s = 3/2 + 0 = 3/2 and upper bound s = 3/2 + 1 = 5/2: = (5 log(3))/2 + 1/2 integral_(3/2)^(5/2) 1/(s^2 - 1/4) ds

Factor -1/4 from the denominator:
= (5 log(3))/2 + 1/2 integral_(3/2)^(5/2) 4/(4 s^2 - 1) ds

Factor out constants:
= (5 log(3))/2 + 2 integral_(3/2)^(5/2) 1/(4 s^2 - 1) ds

Factor -1 from the denominator:
= (5 log(3))/2 - 2 integral_(3/2)^(5/2) 1/(1 - 4 s^2) ds

For the integrand 1/(1 - 4 s^2), substitute p = 2 s and dp = 2 ds.
This gives a new lower bound p = (2 3)/2 = 3 and upper bound p = (2 5)/2 = 5:
= (5 log(3))/2 - integral_3^5 1/(1 - p^2) dp

Apply the fundamental theorem of calculus.
The antiderivative of 1/(1 - p^2) is tanh^(-1)(p):
= (5 log(3))/2 + (-tanh^(-1)(p)) right bracketing bar _3^5

Evaluate the antiderivative at the limits and subtract. (-tanh^(-1)(p)) right bracketing bar _3^5 = (-tanh^(-1)(5)) - (-tanh^(-1)(3)) = tanh^(-1)(3) - tanh^(-1)(5):
= (5 log(3))/2 + tanh^(-1)(3) - tanh^(-1)(5)

Which is equal to:

Answer: = log(18)

5 0

3 years ago

How do u do this in advanced math

UNO [17]

Answer is d = 50.

Step by step. These are vertical angles so the (d + 70) will equal the opposite angle of 120.

Equation is d + 70 = 120.

Solve for d. Subtract 70 from both sides of the equal sign to solve for d.

d + 70 - 70 = 120 - 70
d = 50

4 0

2 years ago