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3 years ago

If the replacement set is the set of integers, find the solution set for the inequality:

Mathematics

2 answers:

kakasveta [241]3 years ago

8 0

It would be D as the correct answer

Andreyy893 years ago

7 0

Answer:

c. {....., -8, -7, -6}

Step-by-step explanation:

x - 9 < -15

x < -6

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How do I find the perimeter and area?

jeyben [28]

Perimeter:
add the length of each side
area:
if it really is a rhombus, then you do
(top side+bottom side) times the height divided by 2

4 0

3 years ago

Sue answer 43 out of 60 questions correctly. What percentage of her answers are incorrect round the percentage to one Decimal Pl

Irina18 [472]

Answer: 28.3 %

Step-by-step explanation:

60-43 = 17

17/60 = 0.283

7 0

3 years ago

Drag each figure to show if it is similar to the figure shown or why it is not similar

Margarita [4]

Answer:

Step-by-step explanation:

The given figure is a rectangle having length = 9 yd and width = 3 yd

Now we will check each figure

Figure one- rectangle with length = 18 yd and width = 9 yd

For similarity of both rectangles ratio of length should be equal to ratio of widths.

$\frac{3}{9}\neq\frac{9}{18}$

So rectangles are not similar.

Figure 2- rectangle with length = 3 yd and width = 6 yd

$\frac{9}{3}\neq \frac{3}{6}$

So figures are not similar.

Figure 3 - rectangle with length = 3 yd and width = 1 yd

$\frac{9}{3}=\frac{3}{1}$

Therefore, figures are similar.

Figure 4 -rectangle with length = 8 yd and width = 2 yd

$\frac{3}{2}\neq \frac{9}{8}$

So given rectangles are not similar.

Figure 5 - rectangle with length = 18 yd and width = 6 yd

$\frac{9}{18}=\frac{3}{6}$

So rectangles are similar.

Figure 6 - trapezoid with larger base = 18 yd and height = 6 yd

Since shapes of the given figures are not same so they are not similar.

8 0

3 years ago

Find the measure of each angle. ∠EFG and ∠LMN are supplementary angles, m∠EFG=(3x+17)º, and m∠LMN=(12x−5)º.

krek1111 [17]

Answer:

∠EFG=(3× 11.2+17)º = 50.6°

∠LMN= (12×11.2−5)º = 129.4°

Step-by-step explanation:

We know that if two angles are supplementary two each other sum of these two angles is equal to 180°.

It is given that ∠EFG and ∠LMN are supplementary angles.

Therefore, ∠EFG +∠LMN = 180°

Given that: ∠EFG=(3x+17)º and ∠LMN=(12x−5)º

⇒ (3x+17)º + (12x−5)º =180°

⇒ (15x+12)° = 180°

⇒15x = 168°

⇒ x = 168°/15

x = 11.2°

Then, ∠EFG=(3× 11.2+17)º = 50.6°

Also, ∠LMN= (12×11.2−5)º = 129.4°

5 0

3 years ago

Help please 50 points!

uranmaximum [27]

Answer:

A and B

Step-by-step explanation:

we would like to solve the following equation:

$\rm\displaystyle 5 - 2x = \sqrt{ {2x}^{2} + x - 1 } - x$

to do so isolate -x to the left hand side and change its sign:

$\rm\displaystyle 5 - 2x + x = \sqrt{ {2x}^{2} + x - 1 }$

simplify addition:

$\rm\displaystyle 5 - x = \sqrt{ {2x}^{2} + x - 1 }$

square both sides:

$\rm\displaystyle (5 - x {)}^{2} = (\sqrt{ {2x}^{2} + x - 1 } {)}^{2}$

simplify square of the right hand side:

$\rm\displaystyle (5 - x {)}^{2} = {2x}^{2} + x - 1$

use (a-b)²=a²-2ab+b² to expand the left hand side:

$\rm\displaystyle {x}^{2} - 10x + 25= {2x}^{2} + x - 1$

swap the equation:

$\rm\displaystyle {2x}^{2} + x - 1 = {x}^{2} - 10x + 25$

isolate the right hand side expression to the left hand side and change every sign:

$\rm\displaystyle {2x}^{2} + x - 1 - {x}^{2} + 10x - 25 = 0$

simplify:

$\rm\displaystyle {x}^{2} + 11x - 26= 0$

rewrite the middle term as 13x-2x:

$\rm\displaystyle {x}^{2} + 13x - 2x - 26= 0$

factor out x:

$\rm\displaystyle x({x}^{} + 13)- 2x - 26= 0$

factor out -2:

$\rm\displaystyle x({x}^{} + 13)- 2(x + 13)= 0$

group:

$\rm\displaystyle (x- 2)(x + 13)= 0$

by Zero product property we obtain:

$\displaystyle \begin{cases}x- 2 = 0\\ x + 13= 0 \end{cases}$

solve for x:

$\displaystyle \begin{cases}x = 2\\ x = - 13 \end{cases}$

to check for extraneous solutions we can define the domain of equation recall that a square root of a function is always greater than or equal to 0 therefore

$\rm\displaystyle 5 - x \geq0$

solve the inequality for x:

$\rm\displaystyle x \leqslant 5$

since 2 and -13 is less than 5 both solutions are valid for x hence,

$\displaystyle \begin{cases}x _{1} = 2\\ x_{2} = - 13 \end{cases}$

and we're done!

4 0

3 years ago

Read 2 more answers