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3 years ago

3. Five less than twice a number is 7.

Mathematics

2 answers:

Ainat [17]3 years ago

8 0

Answer: 2a -5 = 7 and a = 6

Step-by-step explanation:

If 5 is less than twice a, the the expression will be 2a -5, since 2 a is the same as 2 times a, and 5 less than 2a is the same as subtracting 5 from 2a. "is" is the same as equal to, so the whole expression has to equal 7.

2a - 5 = 7 You can solve for a by adding 5 to both sides.

+5 +5

2a = 12 Now divide both sides by 2.

a = 6.

Vera_Pavlovna [14]3 years ago

7 0

Answer:

2x-5=7 As an expression

x=6

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3 years ago

The two lines y= 6x+15 and y= mx+4 intersect at x= -2. What is the y-coordinate of their intersection point

taurus [48]

Answer:

The y-coordinate of their intersection point is 3

That is y=3

Step-by-step explanation:

Given two lines are y=6x+15 and y=mx+4

Given that the two lines intersect at x=-2

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Equating the two lines

6x+15=mx+4

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3 years ago

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3 years ago

Alex wants to fence in an area for a dog park. He has plotted three sides of the fenced area at the points E (1, 5), F (3, 5), a

hram777 [196]

Answer:

$(1,1)$

Step-by-step explanation:

Given: E, F, G, H denote the three coordinates of the area fenced

To find: coordinates of point H

Solution:

According to distance formula,

length of side joining points $(x_1,y_1)\,,\,(x_2,y_2)$ is equal to $\sqrt{(x_2-x_1)^2+(y_2-y_1)^2}$

So,

$EF=\sqrt{(3-1)^2+(5-5)^2}=2\,\,units\\FG=\sqrt{(6-3)^2+(1-5)^2}=\sqrt{9+16}=5\,\,units\\GH=\sqrt{(x-6)^2+(y-1)^2}\\EH=\sqrt{(x-1)^2+(y-5)^2}$

Perimeter of a figure is the length of its outline.

$EF+FG+GH+EH=16\\2+5+\sqrt{(x-6)^2+(y-1)^2}+\sqrt{(x-1)^2+(y-5)^2}=16\\\sqrt{(x-6)^2+(y-1)^2}+\sqrt{(x-1)^2+(y-5)^2}=16-2-5\\\sqrt{(x-6)^2+(y-1)^2}+\sqrt{(x-1)^2+(y-5)^2}=9$

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

$\sqrt{(1-6)^2+(1-1)^2}+\sqrt{(1-1)^2+(1-5)^2}=9\\\sqrt{25}+\sqrt{16}=9\\5+4=9\\9=9$

This is true.

So, the point $(1,1)$ satisfies the equation $\sqrt{(x-6)^2+(y-1)^2}+\sqrt{(x-1)^2+(y-5)^2}=9$

So, point H is $(1,1)$ .

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3 years ago