Ask question

Ask question

All categories

Murljashka [212]

3 years ago

11

3/5+1/3= ...............................................................................................................

1 answer:

Mrac [35]3 years ago

7 0

Answer:

Step-by-step explanation:

you add two fractions:

3/5+1/3=

find the common denominator = it is 15 (3*5)

9/15+5/15=14/15

You might be interested in

Ax+by=(a-b) , bx - ay =(a+b) simultaneous linear equations using cross multiplication

Molodets [167]

Answer:

$x=1\,,\,y=-1$

Step-by-step explanation:

Given: $ax+by=a-b\,,\,bx-ay=a+b$

To solve: the given linear equations

Solution:

Consider the equations:

$A_1x+B_1y+C_1=0\\A_2x+B_2y+C_2=0$

By method of cross multiplication:

$\frac{x}{B_1C_2-B_2C_1}=\frac{y}{C_1A_2-C_2A_1}=\frac{1}{A_1B_2-A_2B_1}$

For equations: $ax+by=a-b\,,\,bx-ay=a+b$

$ax+by-(a-b)=0\\bx-ay-(a+b)=0$

Take $A_1=a\,,\,B_1=b\,,\,C_1=-(a-b)\,,\,A_2=b\,,\,B_2=-a\,,\,C_2=-(a+b)$

So,

$\frac{x}{-b(a+b)-a(a-b)}=\frac{y}{-b(a-b)+a(a+b)}=\frac{1}{-a^2-b^2}\\\frac{x}{-ab-b^2-a^2+ab}=\frac{y}{-ab+b^2+a^2+ab}=\frac{1}{-(a^2+b^2)}\\\frac{x}{-(a^2+b^2)}=\frac{y}{a^2+b^2}=\frac{1}{-(a^2+b^2)}\\\frac{x}{-(a^2+b^2)}=\frac{1}{-(a^2+b^2)}\,,\,\frac{y}{a^2+b^2}=\frac{1}{-(a^2+b^2)}\\x=1\,,\,y=-1$

6 0

3 years ago

Jeff and jill are collecting coats for kids. They hope to collect 20 coats per day to reach their goal. They have already collec

weeeeeb [17]

None of the answers are correct in this question

6 0

3 years ago

SAFE<br> 42s + 32s + s - 42s

Marrrta [24]

75 S is the answer :)

5 0

3 years ago

Find the equation of the line which passes through the point (7,2) and is perpendicular to y=5x-2

MariettaO [177]

Answer:

$\displaystyle y=-\frac{1}{5}x+\frac{17}{5}$

Step-by-step explanation:

<u>Equation of a line</u>

A line can be represented by an equation of the form

$y=mx+b$

Where x is the independent variable, m is the slope of the line, b is the y-intercept and y is the dependent variable.

We need to find the equation of the line passing through the point (7,2) and is perpendicular to the line y=5x-2.

Two lines with slopes m1 and m2 are perpendicular if:

$m_1.m_2=-1$

The given line has a slope m1=5, thus the slope of our required line is:

$\displaystyle m_2=-\frac{1}{m_1}=-\frac{1}{5}$

The equation of the line now can be expressed as:

$\displaystyle y=-\frac{1}{5}x+b$

We need to find the value of b, which can be done by using the point (7,2):

$\displaystyle 2=-\frac{1}{5}*7+b$

Operating:

$\displaystyle 2=-\frac{7}{5}+b$

Multiplying by 5:

$10=-7+5b$

Operating:

$10+7=5b$

Solving for b:

$\displaystyle b=\frac{17}{5}$

The equation of the line is:

$\boxed{\displaystyle y=-\frac{1}{5}x+\frac{17}{5}}$

7 0

3 years ago

Joy is training for a triathlon. She runs a distance of 2x miles, cycles a distance of (x2 + 7) miles, and swims a distance of (

Tpy6a [65]

Add

2x+x²+7+x-2=
x²+2x+x+7-2=
x²+3x+5=total distance

6 0

3 years ago

Read 2 more answers