All categories

3 years ago

Question 7 (5 points)

Mathematics

2 answers:

Reika [66]3 years ago

6 0

Answer:

d) 9b + 3c

Step-by-step explanation:

Sladkaya [172]3 years ago

3 0

C is the correct answer 9b+30

You might be interested in

Which statements are true when using algebra tiles to solve the equation 8x + (–4) = 11x + 5?

kogti [31]

Answer:

See explanation

Step-by-step explanation:

You are given the equation $8x+(-4)=11x+5$

1. Add 8 negative x-tiles to both sides of the equation to create a zero pair on the left side. The equation then will have form

$8x+(-4)+(-8x)=11x+5+(-8x)\\ \\-4=3x+5$

2. Add 5 negative unit tiles to both sides of the equation to create a zero pair on the right side. The equation now is

$-4+(-5)=3x+5+(-5)\\ \\-9=3x$

3. Divide both sides by 3:

$\dfrac{3x}{3}=\dfrac{-9}{3}\\ \\x=-3$

7 0

3 years ago

Read 2 more answers

The square of a number is 15 more than 2 times the number

UkoKoshka [18]

x^2 = 2x + 15

x^2 -2x - 15 = 0

Factor:

(x-5)(x+3) = 0

x = 5 or x = -3

3 0

3 years ago

PLEASE HELP!!!

devlian [24]

Addition or Subtraction
Addition or Subtraction
Multiplication or Division
Multiplication or Division

all of order. I'll have to confirm this though

8 0

3 years ago

Read 2 more answers

How many 3/4s are in 2? Draw a Diagram

Iteru [2.4K]

Answer:

2 2/3

Step-by-step explanation:

2/1 x 4/3 = 8/3 = 2 2/3

8 0

2 years ago

The set of points (-3, 4), (-1, 1), (-3,-2), and (-5,1) identifies the vertices of a quadrilateral. Which is the most specific d

KIM [24]

Answer:

Option (4). Rhombus

Step-by-step explanation:

From the figure attached,

Distance AB = $\sqrt{(x_{2}-x_{1})^2+(y_{2}-y_{1})^2}$

= $\sqrt{(1-4)^2+(-5+3)^2}$

= $\sqrt{(-3)^2+(-2)^2}$

= $\sqrt{13}$

Distance BC = $\sqrt{(4-1)^2+(-3+1)^2}$

= $\sqrt{9+4}$

= $\sqrt{13}$

Distance CD = $\sqrt{(-2-1)^2+(-3+1)^2}$

= $\sqrt{9+4}$

= $\sqrt{13}$

Distance AD = $\sqrt{(1+2)^2+(-5+3)^2}$

= $\sqrt{9+4}$

= $\sqrt{13}$

Slope of AB ( $m_{1}$ ) = $\frac{y_{2}-y_{1}}{x_{2}-x_{1}}$

= $\frac{4-1}{-3+5}$

= $\frac{3}{2}$

Slope of BC ( $m_{2}$ ) = $\frac{4-1}{-3+1}$

= $-\frac{3}{2}$

If AB and BC are perpendicular then,

$m_{1}\times m_{2}=-1$

But it's not true.

[ $m_{1}\times m_{2}=(\frac{3}{2})(-\frac{3}{2})$ = - $\frac{9}{4}$ ]

It shows that the consecutive sides of the quadrilateral are not perpendicular.

Therefore, ABCD is neither square nor a rectangle.

Slope of diagonal BD = $\frac{4+2}{-3+3}$

= Not defined (parallel to y-axis)

Slope of diagonal AC = $\frac{1-1}{-1+5}$

= 0 [parallel to x-axis]

Therefore, both the diagonals AC and BD will be perpendicular.

And the quadrilateral formed by the given points will be a rhombus.

5 0

3 years ago