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2 years ago

If IK=JK, find mlJ A. 72° B. 82° C. 122° D. 134°

Mathematics

1 answer:

yanalaym [24]2 years ago

3 0

Answer:

D.134

Step-by-step explanation:

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Point A is located at (?5, 2) on a coordinate grid. Point A is translated 1 unit to the left and 5 units down to create point A'

svp [43]

ANSWER

C) 5.1 units

EXPLANATION

The coordinates of A are (5,2).

If this point is translated 1 unit to the left and 5 units down to create point A'.

Then A' has coordinates

$A(5,2)\rightarrow \: A'(5-1,2-5)$

$A(5,2)\rightarrow \: A'(4, - 3)$

The distance between A and A' is calculated using the distance formula,

$|AA'| = \sqrt{ {(5 - 4)}^{2} + {(2 - - 3}^{2} }$

$|AA'| = \sqrt{ {(1)}^{2} + {(5)}^{2} }$

$|AA'| = \sqrt{ 26}$

$|AA'| = 5.099$

We round to the nearest tenth to obtain,

$|AA'| =5.1 \: units$

The measurement that is closed to the distance is 5.1 units

4 0

4 years ago

Ax + c = R (solve for x)

sp2606 [1]

Ax + c = R (subtract c from each side)

Ax + c - c = R - c

Ax = R - c (divide A from each side)

Ax/A = (R-c)/A

x = $\frac{R-c}{A}$ <-answer

4 0

3 years ago

Read 2 more answers

Find the length of AB. A 0140 AB = [?] 140° 8 m B Round your answer to the nearest hundredth.

lana [24]

$\huge{ \mathfrak{ \underline{ Answer} \: \: ✓ }}$

Radius (r) = 8m

Angle made at centre $(\theta)$ = 140°

$\boxed{ \mathrm{length \: \: of \: \: arc = \dfrac{ \theta}{360 \degree} \times 2\pi r}}$

$\dfrac{140}{360} \times 2 \times 3.14 \times 8$

$\dfrac{7}{18} \times 50.24$

$\dfrac{351.64}{18}$

$19.54 \:m$

_____________________________

$\mathrm{ ☠ \: TeeNForeveR \:☠ }$

6 0

3 years ago

Hello, can you help me with a probability question, please:)

sladkih [1.3K]

Ok what is it that u need

6 0

3 years ago

The diagram shows a 5 cm x 5 cm x 5 cm cube.

mylen [45]

Answer:

~8.66cm

Step-by-step explanation:

The length of a diagonal of a rectangular of sides a and b is

$\sqrt{a^2+b^2}$

in a cube, we can start by computing the diagonal of a rectangular side/wall containing A and then the diagonal of the rectangle formed by that diagonal and the edge leading to A. If the cube has sides a, b and c, we infer that the length is:

$\sqrt{\sqrt{a^2+b^2}^2 + c^2} = \sqrt{a^2+b^2+c^2}$

Using this reasoning, we can prove that in a n-dimensional space, the length of the longest diagonal of a hypercube of edge lengths $a_1, a_2, a_3, \ldots, a_n$ is

$\sqrt{a_1^2 + a_2^2 + a_3^2 + \ldots + a_n^2}$

So the solution here is

$\sqrt{(5cm)^2 + (5cm)^2 + (5cm)^2} = \sqrt{75cm^2} = 5\sqrt{3cm^2} \approx 5\cdot 1.732cm = 8.66cm$

5 0

3 years ago

Read 2 more answers