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2 years ago

What is 4/5 plus 2/3 and in simplest terms

Mathematics

2 answers:

Butoxors [25]2 years ago

8 0

4/5 plus 2/3 and in simplest terms is 22 / 15

<h3>Step by Step calculation :</h3>

Given that

= 4 / 5 + 2 / 3

We will multiply 4 by 3 to get 12 and then multiply 2 by 5 to get 10

= 3 * 4 / 5 + 5 * 2 /3

= 12 / 5 + 10 / 3

Next we give both terms new denominators 5 × 3 = 15

= 12 / 5 + 10 / 3

= (12 + 10) / 15

Then, the fraction would be :

= (12 + 10) / 15

= 22 / 15

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nevsk [136]2 years ago

5 0

Answer:

$\huge\boxed{\fcolorbox{red}{orange}{Answer : }}$

Given~

$\frac{4}{5} + \frac{2}{3}$

Step-by-step explanation:

So we multiply 4 by 3, and get 12.

Then we multiply 2 by 5, and get 10.

Next we give both terms new denominators 5 × 3 = 15.

So now our fractions look like this~

$\frac{12 + 10}{15}$

$\frac{22}{15}$

<h3>Hope It's Helps~</h3>

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Easy math problems!!!!!

Alexandra [31]

Answer: 36

Step-by-step explanation:

7 0

2 years ago

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Rhonda took $20 with her to spend on popcorn and candy for herself and her friends at the movie theater. The price for each bag

MrRa [10]

The linear equation that represents the scenario is 5x + 2.5y = 20

<h3>How to determine the equation?</h3>

The given parameters are:

Popcorn = $5 per bag
Candy = $2.5 per bag i.e. half the price of pop corn

Let x represents popcorn and y represents candy.

So, we have:

5x + 2.5y = 20

The intercepts in this case are the maximum number of candy and popcorn Rhonda can buy

See attachment for the graph

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5 0

2 years ago

A triangle is graphed in the coordinate plane. The vertices of the triangle have coordinates (–3, 1), (1, 1), and (1, –2). What

vladimir2022 [97]

Answer:

The perimeter of the triangle is $12\ units$

Step-by-step explanation:

Let

$A(-3,1),B(1,1),C(1,-2)$

we know that

The perimeter of triangle is equal to

$P=AB+BC+AC$

the formula to calculate the distance between two points is equal to

$d=\sqrt{(y2-y1)^{2}+(x2-x1)^{2}}$

step 1

Find the distance AB

$A(-3,1),B(1,1)$

substitute in the formula

$AB=\sqrt{(1-1)^{2}+(1+3)^{2}}$

$AB=\sqrt{(0)^{2}+(4)^{2}}$

$AB=4\ units$

step 2

Find the distance BC

$B(1,1),C(1,-2)$

substitute in the formula

$BC=\sqrt{(-2-1)^{2}+(1-1)^{2}}$

$BC=\sqrt{(-3)^{2}+(0)^{2}}$

$BC=3\ units$

step 3

Find the distance AC

$A(-3,1),C(1,-2)$

substitute in the formula

$AC=\sqrt{(-2-1)^{2}+(1+3)^{2}}$

$AC=\sqrt{(-3)^{2}+(4)^{2}}$

$AC=5\ units$

step 4

Find the perimeter

$P=AB+BC+AC$

substitute the values

$P=4+3+5=12\ units$

6 0

3 years ago

Complete the inequality 17/20 ___ 88%

Margarita [4]

Answer:

17/20 < 88%

Step-by-step explanation:

17/20 = .85

.85 < .88

17/20 < 88%

5 0

3 years ago

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How do you prove each of the following theorems using either a two-column, paragraph, or flow chart proof?

lilavasa [31]

All the theorems are proved as follows.

<h3>What is a Triangle ?</h3>

A triangle is a polygon with three sides , three vertices and three angles.

1. The Triangle sum Theorem

According to the Triangle Sum Theorem, the sum of a triangle's angles equals 180 degrees.

To create a triangle ABC, starting at point A, move 180 degrees away from A to arrive at point B.

We turn 180 degrees from B to C and 180 degrees from C to return to A, giving a total turn of 360 degrees to arrive to A.

180° - ∠A + 180° - ∠B + 180° - ∠C = 360°

- ∠A - ∠B - ∠C = 360° - (180°+ 180°+ 180°) = -180°

∠A + ∠B + ∠C = 180°

(Hence Proved)

2. Isosceles Triangle Theorem

Considering an isosceles triangle ΔABC

with AB = AC, we have by sine rule;

$\rm \dfrac{sinA}{BC} = \dfrac{sinB}{AC} = \dfrac{sinC}{AB}\\$

as AB = AC

sin B = sin C

angle B = angle C

3.Converse of the Isosceles theorem

Consider an isosceles triangle ΔABC with ∠B= ∠C, we have by sine rule;

$\rm \dfrac{sinA}{BC} = \dfrac{sinB}{AC} = \dfrac{sinC}{AB}\\$

as ∠B= ∠C ,

AB = AC

4. Midsegment of a triangle theorem

It states that the midsegment of two sides of a triangle is equal to (1/2)of the third side parallel to it.

Given triangle ABC with midsegment at D and F of AB and AC respectively, DF is parallel to BC

In ΔABC and ΔADF

∠A ≅ ∠A

BA = 2 × DA, BC = 2 × FA

Hence;

ΔABC ~ ΔADF (SAS similarity)

BA/DA = BC/FA = DF/AC = 2

Hence AC = 2×DF

5.Concurrency of Medians Theorem

A median of a triangle is a segment whose end points are on vertex of the triangle and the middle point of the side ,the medians of a triangle are concurrent and the point of intersection is inside the triangle known as Centroid .

Consider a triangle ABC , X,Y and Z are the midpoints of the sides

Since the medians bisect the segment AB into AZ + ZB

BC into BX + XB

AC into AY + YC

Where:

AZ = ZB

BX = XB

AY = YC

AZ/ZB = BX/XB = AY/YC = 1

AZ/ZB × BX/XB × AY/YC = 1 and

the median segments AX, BY, and CZ are concurrent (meet at point within the triangle).

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8 0

1 year ago