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3 years ago

9

Write a number sentence that does not involve regrouping using this sum and two 2- digit addends

1 answer:

melisa1 [442]3 years ago

5 0

Using the sum (addition) and two 2-digit addends, 12 + 34 can be answered without regrouping.

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Which number is halfway between

Anarel [89]

You are looking for the average. To find the average, one adds all terms and divides by the number of terms.
((1/2)+(5/6))/2
((3/6)+(5/6))/2
(8/6)/2
(4/6)
(2/3)
The answer is B) 2/3.

6 0

3 years ago

Find the Area please help me

Serggg [28]

Step-by-step explanation:

Area for a right angle triangle = 1/2 bh

b = Base

h = Height

A = 1/2×bh

= 1/2 × 2c^3d^2 × 4c

= (2c^3d^2 × 4c) / 2

= c^3d^2 × 2c

= 2c^4 × d^2

7 0

2 years ago

In a right triangle, the acute angles measure 2x+50 and 3x degrees

xenn [34]

2x+50 + 3x= 90 degrees
5x +50 = 90
5x =40
x=8

Angle1= 2(8) +50= 66
Angle2= 3(8)=24

66+24+90= 180

6 0

3 years ago

Read 2 more answers

Pls help me and can you give tips on exponents??!

Marina CMI [18]

Answer:

Basically, exponents represent the number of times you multiply that number by itself. For example, 3² = 9, because you multiply 3 times 3.

Step-by-step explanation:

7 0

2 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