Ask question

Ask question

All categories

3 years ago

6

Write a real world problem that could represent the bar diagram shown

2 answers:

aalyn [17]3 years ago

6 0

Answer:

matt hgad 12 cookies his 2 other friends came over and divided them equally with them now each of them had four

Step-by-step explanation:

Sphinxa [80]3 years ago

5 0

Answer:

Matt had 12 cookies his 2 other friends came over and divided them equally with them now each of them had four.

Step-by-step explanation:

You might be interested in

Simplify : 5y-2y+4=10

Mnenie [13.5K]

Answer:

See below.

Step-by-step explanation:

5y - 2y + 4 = 10

5y - 2y = 6

3y = 6

y = 2

Check:

5y - 2y + 4 = 10

5(2) - 2(2) + 4 = 10

10 - 4 + 4 = 10

6 + 4 = 10

10 = 10

The value of y = 2

7 0

3 years ago

Read 2 more answers

Please help me out with this guys. No Links Please!

Sergio [31]

I think it is 29%? I am not 100% sure though...

5 0

2 years ago

Ghella [55]

Answer: 8

Step-by-step explanation:

4 0

3 years ago

Expanded -8(-2x-5)<br> please give answer

nika2105 [10]

Answer:

-80

Step-by-step explanation:

We have to do the parentheses first because it's part of PEDMAS.

-2x-5= 10

Next, we take 10 times -8

-8 x 10= -80

There you go! Hope this helps!

8 0

3 years ago

Read 2 more answers

Let Q(x, y) be the predicate "If x < y then x2 < y2," with domain for both x and y being R, the set of all real numbers.

Levart [38]

Answer:

a) Q(-2,1) is false

b) Q(-5,2) is false

c)Q(3,8) is true

d)Q(9,10) is true

Step-by-step explanation:

Given data is $Q(x,y)$ is predicate that $x$ then $x^{2}$ . where $x,y$ are rational numbers.

a)

when $x=-2, y=1$

Here $-2$ that is $x$ satisfied. Then

$(-2)^{2}$

$4$ this is wrong. since $4>1$

That is $x^{2}$ $>y^{2}$ Thus $Q(x,y)$ $=Q(-2,1)$ is false.

b)

Assume $Q(x,y)=Q(-5,2)$ .

That is $x=-5, y=2$

Here $-5$ that is $x$ this condition is satisfied.

Then

$(-5)^{2}$

$25$ this is not true. since $25>4$ .

This is similar to the truth value of part (a).

Since in both $x$ satisfied and $x^{2} >y^{2}$ for both the points.

c)

if $Q(x,y)=Q(3,8)$ that is $x=3$ and $y=8$

Here $3$ this satisfies the condition $x$ .

Then $3^{2}$

$9$ This also satisfies the condition $x^{2}$ .

Hence $Q(3,8)$ exists and it is true.

d)

Assume $Q(x,y)=Q(9,10)$

Here $9$ satisfies the condition $x$

Then $9^{2}$

$81$ satisfies the condition $x^{2}$ .

Thus, $Q(9,10)$ point exists and it is true. This satisfies the same values as in part (c)

6 0

3 years ago