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3 years ago

5

Hey I need this answered i have no clue what to put pls help

1 answer:

Molodets [167]3 years ago

4 0

Answer:

b is correct for Maria but c is correct for the boy

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On Saturday, many tickets fir a hockey game were sold. Children and adult were in the ratio of 4:7. There are 88 tickets in tota

ollegr [7]

Answer:

<em>56 adult tickets were sold</em>

Step-by-step explanation:

Ratio of ticket sold to Children and adult is in the ratio 4 : 7

Total ratio = 4 +7 = 11

Ratio of adult = 7

Total ticket sold = 88

Number of adult tickets sold = Ratio of adult/Total ratio * Total ticket sold

Number of adult tickets sold = 7/11 * 88

Number of adult tickets sold = 7 * 8

Number of adult tickets sold = 56 tickets

<em>Hence 56 adult tickets were sold</em>

3 0

2 years ago

X-3y=-6 what is the solution for this question ??

seraphim [82]

Answer: y = 1/3x + 2

8 0

3 years ago

What is the sum of the given polynomials in standard form ? (X^2-3x)+(-2x^2+5x-3)

Greeley [361]

$(x^2-3x)+(-2x^2+5x-3)=x^2-3x-2x^2+5x-3\\\\=(x^2-2x^2)+(-3x+5x)-3=-x^2+2x-3$

3 0

3 years ago

I need to drag one into the box to match them

ioda

Answer:

54÷9 = 6 so for the second box is 6 hot dogs per person

54÷27= 2 third box is 2 hot dogs per person

54÷54= 1 fourth box is 1 hot dog per person

5 0

3 years ago

Write a solution in Interval Notation - (you don't have to help me on all, 1 or 2 is fine c: )

Gemiola [76]

QUESTION 1

The given inequality is

$|m|-2>0$

We group like terms to get,

$|m|>2$

This implies that,

$-m>2$ or $m>2$ .

We simplify the inequality to get,

$m$ or $m>2$ .

We can write this interval notation to get,

$(-\infty,-2)\cup (2,+\infty)$ .

QUESTION 2

$|x-4|-3\:>\:5$ .

We group like terms to get,

$|x-4|\:>\:5+3$ .

$|x-4|\:>\:8$

We split the absolute value sign to get,

$-(x-4)\:>\:8$ or $x-4\:>\:8$

This implies that,

$x-4\:$ or $x-4\:>\:8$

$x\:$ or $x\:>\:8+4$

$x\:$ or $x\:>\:12$

We can write this interval notation to get,

$(-\infty,-4)\cup (12,+\infty)$ .

QUESTION 3

The given inequality is

$|6+9x|\leq 24$

We split the absolute value sign to obtain,

$-(6+9x)\leq 24$ or $(6+9x)\leq 24$

This simplifies to

$6+9x\ge -24$ and $6+9x\leq 24$

$9x\ge -24-6$ and $9x\leq 24-6$

$9x\ge -30$ and $9x\leq 18$

$x\ge -\frac{10}{3}$ and $x\leq 2$

$-\frac{10}{3}\leq x\leq2$

We write this in interval form to get,

$[-\frac{10}{3},2]$

QUESTION 4

The given inequality is

$|1-5a|>29$

We split the absolute value sign to get,

$-(1-5a)>29$ or $1-5a>29$

This simplifies to,

$1-5a\:$ or $1-5a\:>\:29$

This implies that,

$-5a\:$ or $-5a\:>\:29-1$

$-5a\:$ or $-5a\:>\:28$

$a\:>\:6$ or $a\:$

We write this in interval notation to get,

$(-\infty,-\frac{28}{5})\cup (6,+\infty)$

7 0

3 years ago