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3 years ago

Solve the equation.8(4-x) = 7x + 2

Mathematics

2 answers:

oksian1 [2.3K]3 years ago

8 0

Answer:

x = 2.

Step-by-step explanation:

8(4-x) = 7x + 2

32 - 8x = 7x + 2

32 - 2 = 7x + 8x

15x = 30

x =2.

GalinKa [24]3 years ago

7 0

32-8x = 7x + 2

32= 15x + 2

30= 15x

X= 2

Hope this helps!

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Find the unknown length y in the following pair of the similar triangles.

yulyashka [42]

Answer:

y = 9.75

Step-by-step explanation:

Since both triangles are said to be similar, the proportion or ratio of the sides of both triangles would be equal.

Therefore, 39/y = 23/5.75 = 35/8.75

Let's find y

39/y = 23/5.75

==>Cross multiply

39*5.75 = 23*y

224.25 = 23y

==>Divide both sides by 23

224.25/23 = y

9.75 = y

y = 9.75

Check:

Thus,

39/9.75 = 23/5.75 = 35/8.75 = 4

Our answer y, which is 9.75, is correct.

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3 years ago

a model of an apartment is shown 1\4 in. represents 3 feet in the actual apartment. Find the actual area of the master bedroom

Hatshy [7]

Answer:12 feet by 9 feet

Step-by-step explanation:

1/4 = 3 feet, so every 1/4 of an inch there is three feet, e.g. 3/4 inches of that scale is 9 feet, 1 1/2 inches would be 18 feet, make sense?

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3 years ago

What is y+1=2(x-4) written in standard form

IrinaVladis [17]

Answer:

2x - y = 9

Step-by-step explanation:

The equation of a line in standard form is

Ax + By = C ( A is a positive integer and B, C are integers )

Rearrange y + 1 = 2(x - 4) into this form

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3 years ago

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The hypotenuse of a right triangle has endpoints A(4, 1) and B(–1, –2). On a coordinate plane, line A B has points (4, 1) and (n

GarryVolchara [31]

Answer:

$(-1,1),(4,-2)$

Step-by-step explanation:

Given: The hypotenuse of a right triangle has endpoints A(4, 1) and B(–1, –2).

To find: coordinates of vertex of the right angle

Solution:

Let C be point $(x,y)$

Distance between points $(x_1,y_1),(x_2,y_2)$ is given by $\sqrt{(x_2-x_1)^2+(y_2-y_1)^2}$

$AC=\sqrt{(x-4)^2+(y-1)^2}\\BC=\sqrt{(x+1)^2+(y+2)^2}\\AB=\sqrt{(4+1)^2+(1+2)^2}=\sqrt{25+9}=\sqrt{34}$

ΔABC is a right angled triangle, suing Pythagoras theorem (square of hypotenuse is equal to sum of squares of base and perpendicular)

$34=\left [ (x-4)^2+(y-1)^2 \right ]+\left [ (x+1)^2+(y+2)^2 \right ]$

Put $(x,y)=(-1,1)$

$34=\left [ (-1-4)^2+(1-1)^2 \right ]+\left [ (-1+1)^2+(1+2)^2 \right ]\\34=25+9\\34=34$

which is true. So, $(-1,1)$ can be a vertex

Put $(x,y)=(4,-2)$

$34=\left [ (4-4)^2+(-2-1)^2 \right ]+\left [ (4+1)^2+(-2+2)^2 \right ]\\34=9+25\\34=34$

which is true. So, $(4,-2)$ can be a vertex

Put $(x,y)=(1,1)$

$34=\left [ (1-4)^2+(1-1)^2 \right ]+\left [ (1+1)^2+(1+2)^2 \right ]\\34=9+4+9\\34=22$

which is not true. So, $(1,1)$ cannot be a vertex

Put $(x,y)=(2,-2)$

$34=\left [ (2-4)^2+(-2-1)^2 \right ]+\left [ (2+1)^2+(-2+2)^2 \right ]\\34=4+9+9\\34=22$

which is not true. So, $(2,-2)$ cannot be a vertex

Put $(x,y)=(4,-1)$

$34=\left [ (4-4)^2+(-1-1)^2 \right ]+\left [ (4+1)^2+(-1+2)^2 \right ]\\34=4+25+1\\34=30$

which is not true. So, $(4,-1)$ cannot be a vertex

Put $(x,y)=(-1,4)$

$34=\left [ (-1-4)^2+(4-1)^2 \right ]+\left [ (-1+1)^2+(4+2)^2 \right ]\\34=25+9+36\\34=70$

which is not true. So, $(-1,4)$ cannot be a vertex

So, possible points for the vertex are $(-1,1),(4,-2)$

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