What is 1.9 inches to the nearest fourth inch?

If ΔRED ≅ ΔTAN, then which of the following statements is not true? RE ≅ TA ∠E ≅ ∠A RD≅ TA ∠D ≅ ∠N

Assoli18 [71]

Answer:

RD≅ TA

Step-by-step explanation:

RD≅ TA is not true statement as they are not corresponding sides.

4 0

3 years ago

Ashlynd's backpack has a mass of 30,000 grams. What is the mass of ashlynd's backpack on kilograms?

borishaifa [10]

1,000 grams = 1 kilogram
2,000 grams = 2 kilograms
3,000 grams = 3 kilograms
4,000 grams = 4 kilograms
5,000 grams = 5 kilograms
.
.
10,000 grams = 10 kilograms
.
.
20,000 grams = 20 kilograms
.
.
30,000 grams = 30 kilograms

8 0

3 years ago

Evaluate the following integral using trigonometric substitution

serg [7]

Answer:

The result of the integral is:

$\arcsin{(\frac{x}{3})} + C$

Step-by-step explanation:

We are given the following integral:

$\int \frac{dx}{\sqrt{9-x^2}}$

Trigonometric substitution:

We have the term in the following format: $a^2 - x^2$ , in which a = 3.

In this case, the substitution is given by:

$x = a\sin{\theta}$

So

$dx = a\cos{\theta}d\theta$

In this question:

$a = 3$

$x = 3\sin{\theta}$

$dx = 3\cos{\theta}d\theta$

So

$\int \frac{3\cos{\theta}d\theta}{\sqrt{9-(3\sin{\theta})^2}} = \int \frac{3\cos{\theta}d\theta}{\sqrt{9 - 9\sin^{2}{\theta}}} = \int \frac{3\cos{\theta}d\theta}{\sqrt{9(1 - \sin^{\theta})}}$

We have the following trigonometric identity:

$\sin^{2}{\theta} + \cos^{2}{\theta} = 1$

So

$1 - \sin^{2}{\theta} = \cos^{2}{\theta}$

Replacing into the integral:

$\int \frac{3\cos{\theta}d\theta}{\sqrt{9(1 - \sin^{2}{\theta})}} = \int{\frac{3\cos{\theta}d\theta}{\sqrt{9\cos^{2}{\theta}}} = \int \frac{3\cos{\theta}d\theta}{3\cos{\theta}} = \int d\theta = \theta + C$

Coming back to x:

We have that:

$x = 3\sin{\theta}$

So

$\sin{\theta} = \frac{x}{3}$

Applying the arcsine(inverse sine) function to both sides, we get that:

$\theta = \arcsin{(\frac{x}{3})}$

The result of the integral is:

$\arcsin{(\frac{x}{3})} + C$

8 0

2 years ago

Points J, K and L are collinear with J between L and K. If KJ = 2x-3, LJ= 4x-8, solve for x:

notsponge [240]

Your question is incomplete, here is the complete form.

Points J, K and L are collinear with J between L and K. If KJ = 2x - 3, LK = 9x + 7 and LJ = 4x - 8, solve for x:

Answer:

The value of x is -6 ⇒ B

Step-by-step explanation:

∵ J, K, and L are collinear

→ That means they form a straight segment

∵ J is between K and L

→ That means J divides LK into two segments KJ and LJ

∴ LK = KJ + LJ

∵ LK = 9x + 7

∵ KJ = 2x - 3

∵ LJ = 4x - 8

→ Substitute them in the equation above

∴ 9x + 7 = (2x - 3) + (4x - 8)

→ Add the like terms in the right side

∵ 9x + 7 = (2x + 4x) + (-3 - 8)

∴ 9x + 7 = 6x + -11

∴ 9x + 7 = 6x - 11

→ Subtract 7 from both sides

∵ 9x + 7 - 7 = 6x - 11 - 7

∴ 9x = 6x - 18

→ Subtract 6x from both sides

∵ 9x - 6x = 6x - 6x - 18

∴ 3x = -18

→ Divide both sides by 3

∵ $\frac{3x}{3}=\frac{-18}{3}$

∴ x = -6

∴ The value of x is -6

8 0

2 years ago

If you had a right triangle with legs measuring 40 yards and 100 yards long,

tia_tia [17]

9514 1404 393

Answer:

(c) 108 yd

Step-by-step explanation:

For hypotenuse c and legs a and b, the Pythagorean theorem tells you the relation is ...

c² = a² +b²

Solving for c, we have ...

c = √(a² +b²)

c = √(40² +100²) = √(1600 +10000) = √11600

c ≈ 107.70 . . . . yards

The length of the hypotenuse is about 108 yards.

6 0

2 years ago