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4 years ago

(3)2y+3×4 written as an expression of two factors

Mathematics

1 answer:

goldfiish [28.3K]4 years ago

4 0

Answer:

6/12

Step-by-step explanation:

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If 2/3 and -3 are the roots of the equation px^2+7x+q=0, find the values of p and q.

weqwewe [10]

Answer:

p = 3

q = -6

Step-by-step explanation:

Work backwards from the roots:

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

3x = 2

3x - 2 = 0

Other root:

x = -3

x + 3 = 0

Multiply them together:

(3x - 2)(x + 3) = 0

3x² - 9x + 2x - 6 = 0

3x² - 7x - 6 = 0

p = 3

q = -6

4 0

3 years ago

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I’ll give you Brainliest

IRINA_888 [86]

Answer:

line k im pretty sure and 34

6 0

3 years ago

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A model of a building is made using a scale of 1 inch = 25 feet. what is the height of the actual building of the height of the

Solnce55 [7]

1 in. = 25 ft.

12.5 in. = ___ft

25/1=25

12.5*25=312.5

So the answer to your question is 312.5 feet.

Hope this helps! :)

7 0

3 years ago

for triangle ABC find the measure of line a b given measurement of angle a is 55 measurement of angle B is 44 and b equals 68

S_A_V [24]

Use the Law of Sines:

sin B   sin 81 sin 44
---------- = ----------- = --------------
side AC 68 side AB

Then (side AB)(sin 81) = 68(sin 44), or

   68 sin 44
side AB =   ----------------------- = 47.83 units long
sin 81

3 0

4 years ago

The line segment AB with endpoints A (-3, 6) and B (9, 12) is dilated with a scale

Gwar [14]

Answer:

C) (-2, 4), (6,8) is the correct answer.

Step-by-step explanation:

Given that line segment AB:

A (-3, 6) and B (9, 12) is dilated with a scale factor 2/3 about the origin.

First of all, let us calculate the distance AB using the distance formula:

$D = \sqrt{(x_2-x_1)^2+(y_2-y_1)^2}$

Here,

$x_2=9\\x_1=-3\\y_2=12\\y_1=6$

Putting all the values and finding AB:

$AB = \sqrt{(9-(-3))^2+(12-6)^2}\\\Rightarrow AB = \sqrt{(12)^2+(6)^2}\\\Rightarrow AB = \sqrt{144+36}\\\Rightarrow AB = \sqrt{180}\\\Rightarrow AB = 6\sqrt{5}\ units$

It is given that AB is dilated with a scale factor of $\frac{2}{3}$ .

$x_2'=\dfrac{2}{3}\times x_2=\dfrac{2}{3}\times9=6\\x_1'=\dfrac{2}{3}\times x_1=\dfrac{2}{3}\times-3=-2\\y_2'=\dfrac{2}{3}\times y_2=\dfrac{2}{3}\times 12=8\\y_1'=\dfrac{2}{3}\times y_1=\dfrac{2}{3}\times 6=4$

So, the new coordinates are A'(-2,4) and B'(6,8).

Verifying this by calculating the distance A'B':

$A'B' = \sqrt{(6-(-2))^2+(8-4)^2}\\\Rightarrow A'B' = \sqrt{(8)^2+(4)^2}\\\Rightarrow A'B' = \sqrt{64+16}\\\Rightarrow A'B' = \sqrt{80}\\\Rightarrow A'B' = 4\sqrt{5}\ units = \dfrac{2}{3}\times AB$

So, option C) (-2, 4), (6,8) is the correct answer.

5 0

3 years ago