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2 years ago

My question is down there

Mathematics

1 answer:

slavikrds [6]2 years ago

7 0

I don’t really see the question ?

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Suppose Set B contains 69 elements and the total number elements in either Set A or Set B is 107. If the Sets A and B have 13 el

erastovalidia [21]

Set A or B refers to the union of A and B, in other words how many element there are in TOTAL. We are told Set B has 69 elements, but that 13 of them overlap with Set A. This means Set B has 69-13=56 unique elements, and all others are therefore part of set A. If there are 107 elements in total, Set A has 107-56=51 elements.

4 0

2 years ago

A local coffee shop serves 2/9 as many customers as a shop run by a national coffee chain. The shop run by the chain serves 502

Inessa [10]

Answer:

about 112 customers

Step-by-step explanation:

we know that

To find out how many customers does the local shop serve in a day, multiply the fraction 2/9 by the number of customers that the shop run serve in a day

Let

x ----> the number of customers that the shop run serve in a day

y ----> the number of customers that the local shop serve in a day

$y=\frac{2}{9}x$

we have

x=502

substitute the value of x in the equation

$y=\frac{2}{9}(502)$

$y=111.6$

Round up

about 112 customers

8 0

2 years ago

Read 2 more answers

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

2 years ago

The Chord CE is a diameter of O

miv72 [106K]

Answer:

False.

Step-by-step explanation:

8 0

2 years ago

The force (F) needed to cause the acceleration (a) of an object of mass, m, is given by the equation, F=ma where m is the object

KATRIN_1 [288]

From Newton’s Second Law of Motion, the force needed to accelerate a moving object of mass m at acceleration a is given by ;

F = ma

In this case, m = 5kg, a = 6m/s^2 and we want to find F

F = 5 x 6

F = 30N

Therefore, a force of 30N is needed to accelerate a 5kg object at 6m/s^2.

8 0

2 years ago