All categories

3 years ago

Can someone show me how to solve? please take a picture of results.

Mathematics

2 answers:

marshall27 [118]3 years ago

6 0

2x = 2√17 and 3x = 3√17

Step-by-step explanation:

Step 1: Given length of the rectangle is 3x and breadth is 2x. Area is 102 m². Area of the rectangle = length × breadth

102 = 2x × 3x = 6x²

x² = 102/6 = 17

x = √17

Length 2x = 2√17 and Breadth = 3x = 3√17

BaLLatris [955]3 years ago

6 0

Answer:

The answer to your question is 2x = 2 $\sqrt{17}$ and 3x = 3 $\sqrt{17}$

Step-by-step explanation:

Data

A = 102 m²

2x = ?

3x = ?

Formula

Area of a rectangle = base x height

Process

1.- Substitute values in the formula

102 = (2x)(3x)

2.- Simplify

102 = 6x²

102/6 = x²

17 = x²

x = $\sqrt{17}$

3.- Find 2x and 3x

2x = 2 $\sqrt{17}$

3x = 3 $\sqrt{17}$

You might be interested in

Distribute -2(c - 8)

Ulleksa [173]

Answer:

Step-by-step explanation:

<h2>give me brainliest please</h2>

given:

$-2(c-8)$

$-2(c) + -2(-8)$

$-2c + (--16)$

$-2c + (+16)$

$-2c+16$

5 0

3 years ago

Read 2 more answers

ABCD- parallelogram, If the perimeter of Triangle CPQ is 15cm, Find the perimeter of triangle BAQ. Find the perimeter of triangl

melamori03 [73]

Answer:

The answer is below

Step-by-step explanation:

A parallelogram is a quadrilateral (has 4 sides and 4 angle) with two pair of parallel and opposite sides. Opposite sides of a parallelogram are parallel and equal.

Given parallelogram ABCD:

AB = CD = 18 cm; BC = AD = 8 cm

∠P = ∠P, ∠PDA = ∠PCQ (corresponding angles are equal).

Hence ΔPCQ and ΔPDA are similar by angle-angle similarity theorem. For similar triangles, the ratio of their corresponding sides equal. Therefore:

$\frac{CD}{PC}= \frac{AD}{CQ}\\\\\frac{18}{6}=\frac{8}{x} \\\\x=\frac{6*8}{18}=\frac{8}{3}\ cm$

Perimeter of CPQ = CP + CQ + PQ

15 = 6 + 8/3 + PQ

PQ = 15 - (6 + 8/3)

PQ = 6.33

∠CQP = ∠AQB (vertical angles), ∠QCP = ∠QBA (alternate angles are equal).

Hence ΔCPQ and ΔABQ are similar by angle-angle similarity theorem

$\frac{AQ}{QP}=\frac{AB}{CP} \\\\\frac{AQ}{6.33} =\frac{18}{6} \\\\AQ=\frac{18}{6}*6.33\\\\AQ = 19$

$\frac{BQ}{CQ}=\frac{AB}{CP} \\\\\frac{BQ}{8/3} =\frac{18}{6} \\\\BQ=\frac{18}{6}*\frac{8}{3} \\\\BQ =8$

Perimeter of BAQ = AB + BQ + AQ = 18 + 8 + 19 = 45cm

PA = AQ + PQ = 19 + 6.33 = 25.33

PD = CD + DP = 18 + 6 = 24

Perimeter of PDA = PA + PD + AD = 24 + 25.33 + 8 = 57.33 cm

7 0

3 years ago

Given that the radius of circle O has a slope of 2.5. What is the slope of the line tangent to circle O at point A?

GuDViN [60]

The problem ask to calculate the slope of the line tangent to a circle O at point A if the radius of the circle is 2.5. So based on the problem the tangent line formed a right angle so it means that the slope is also 2.5 and the answer is letter B. 2.5

3 0

3 years ago

What figure can be made with exactly 420 squares?

Natali [406]

5. The 5th figure will have 30 tiles.
6. The 6th figure will have 42 tiles.
7. The pattern is non-linear (quadratic).
8. The 20th figure can be made with 420 tiles.

_____
The Nth figure is made with n*(n+1) tiles. To answer question 8, you must solve
.. n(n +1) = 420
.. n^2 +n -420 = 0
.. (n +21)(n -20) = 0
.. n = 20 or -21 . . . . the negative solution is extraneous

4 0

3 years ago

Find the distance between the two points in simplest radical form. (3,8) and (5, -1)

ruslelena [56]

Answer: irk

Step-by-step explanation:

7 0

3 years ago

Can someone show me how to solve? please take a picture of results.​

Can someone show me how to solve? please take a picture of results.