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2 years ago

What is 4/5 plus 2/3

Mathematics

2 answers:

omeli [17]2 years ago

6 0

Answer:

1 $\frac{7}{15}$ or

$\frac{22}{15}$

Step-by-step explanation:

Setler [38]2 years ago

4 0

Answer:

22/15 or 1 7/15

Step-by-step explanation:

4/5 + 2/3

So first you have to change the denominator to the same digit by finding a number they both go in to so:

12/15 and 10/15

then add them together and you'll get: 22/15

simplified it'll be 1 7/15

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Find the center and the radius

Paha777 [63]

Answer:

Center: $(0,9)$

Radius: $15$

Step-by-step explanation:

-According to the following equation, the center would be $(0,9)$ , because the center is known as $(h,k)$ from the equation of a circle. To find the radius, you would only need to square root the simplified radius from the following the equation and you would get the radius 15:

The equation of a circle:

$(x-h)^2+(y-k)^2 =r^2$

The following equation:

$x^2+(y-9)^2 =225$

Center: $(0,9)$ Radius: $\sqrt{225} = 15$

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2 years ago

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7 0

3 years ago

8. 6r + 5 = 4 6 - 7y -20

Rudiy27

Answer: r=-7/6y+7/2

Step-by-step explanation:

Step 1: Add -5 to both sides.

6r+5+−5=−7y+26+−5

6r=−7y+21

Step 2: Divide both sides by 6.

6r/6=-7y+21/6

3 0

2 years ago

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

2 years ago

WHAT VALUE IS EQUIVALENT TO (7+3)^2 + (8-1) X 5 EQUAL

sdas [7]

Answer:

135

Step-by-step explanation:

(7+3)^2 + (8-1) * 5

Parentheses first

(10)^2 + (7) * 5

Exponents next

100 + 7*5

Multiply

100 + 35

Add

135

7 0

2 years ago