Help me asap!!! 3(m+7) = 6 (m+2)

Ainat [17]

So you It’s like this: you owe somebody three dollars (-3) but then that same person owes you four dollars 4 instead of exchanging a bunch of money around 4+(-3) how much is owed to you? $1...

4 0

3 years ago

Please help me guys like I really need it

Lesechka [4]

Inequality 1: y ≤ 5x + 1

Inequality 2: y > x - 2

#1. The slope is 5, the y-intercept is 1, the shaded area is below the line and the line is solid, which means that y is less than or equal to the equation. (you add the "equal to" because the line is a solid line)

#2. The slope is 1, the y-intercept is -2, the shaded area is above the line and the line is dotted, which means that y is greater than the equation

6 0

2 years ago

A rectangle has a length of 12 mm and a width of 15 mm. A new rectangle was created by multiplying all of the dimensions by a sc

solmaris [256]

Answer:

the new perimeter will be 1/3 time the perimeter of the original rectangle

Step-by-step explanation:

8 0

3 years ago

Paula and Kawai shared $312 in the ratio of 6:7. How much money did each person get?

Maslowich

Answer:

Paula get $\$144$ and Kawai get $\$168$ .

Step-by-step explanation:

Given: Paula and Kawai shared $\$312$ in the ratio of $6:7$ .

To find: How much money did each person get?

Solution:

We have,

Paula and Kawai shared $\$312$ in the ratio of $6:7$ .

So, let Paula get $\$6x$ and Kawai get $\$7x$ .

As per the question,

$6x+7x=312$

$\implies13x=312$

$\implies x=\frac{312}{13} =24$

Therefore, Paula get $6\times 24=\$144$ , and Kawai get $7\times 24=\$168$ .

Hence, Paula get $\$144$ and Kawai get $\$168$ .

6 0

2 years ago

Which equation has the solutions x = -3 ± √3i/2 ?

Maurinko [17]

Answer:Answer is option C : [ $x^{2}$ + 3x + 3 ] =0

Note: None of options matches with given question.

instead of "-3" , there should be "- $\frac{3}{2}$ ".

Step-by-step explanation:

Note: None of options matches with given question.

instead of "-3" , there should be " $\frac{3}{2}$ ".

Here, First thing you have to observe the nature of roots.

∴ x = - $\frac{3}{2}$ + $\frac{\sqrt{3}}{2}$ i and x = - $\frac{3}{2}$ - $\frac{\sqrt{3}}{2}$

∴ [ x+( $\frac{3}{2}$ - $\frac{\sqrt{3}}{2}$ i) ][ x+( $\frac{3}{2}$ + $\frac{\sqrt{3}}{2}$ i) ]=0

∴ [ $x^{2}$ + x( $\frac{3}{2}$ + $\frac{\sqrt{3}}{2}$ i)+ x( $\frac{3}{2}$ - $\frac{\sqrt{3}}{2}$ i) + ( $\frac{3}{2}$ - $\frac{\sqrt{3}}{2}$ i)( $\frac{3}{2}$ + $\frac{\sqrt{3}}{2}$ i) ]=0

∴ [ $x^{2}$ + $\frac{3}{2}$ x + $\frac{\sqrt{3}}{2}$ ix + $\frac{3}{2}$ x - $\frac{\sqrt{3}}{2}$ ix + (3- $\frac{\sqrt{3}}{2}$ i)(3+ $\frac{\sqrt{3}}{2}$ i) ] =0

∴ [ $x^{2}$ + 3x + ( $\frac{3}{2}$ - $\frac{\sqrt{3}}{2}$ i)( $\frac{3}{2}$ + $\frac{\sqrt{3}}{2}$ i) ] =0

∴ [ $x^{2}$ + 3x + $\frac{9}{4}$ - ( $\frac{\sqrt{3}}{2}$ i)( $\frac{\sqrt{3}}{2}$ i) ] =0

∴ [ $x^{2}$ + 3x + $\frac{9}{4}$ - ( $\frac{3}{4}$ ) $i^{2}$ ] =0

∴ [ $x^{2}$ + 3x + $\frac{9}{4}$ + ( $\frac{3}{4}$ ) ] =0

∴ [ $x^{2}$ + 3x + $\frac{12}{4}$ ] =0

∴ [ $x^{2}$ + 3x + 3 ] =0

Thus, Answer is option C : [ $x^{2}$ + 3x + 3 ] =0

6 0

3 years ago