An equation of the line through the points (-2,5) and (3,-4) is??

A father is 60 years old and his son is half his age. How old was the boy when his father was four times his age?

Archy [21]

He would be 120 yrs old because 60 times 4 is 240 and the son is half of his age so 240 divided by 2 is 120 hope this helped

5 0

2 years ago

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A self-tanning lotion advertises that a 4 -oz bottle will provide seven applications. Jen found a great deal on a 13 -oz bott

Rainbow [258]

Answer:

22.75 applications

Step-by-step explanation:

From the above question, we understand that:

4 -oz bottle = 7 applications

13 -oz bottle = x applications

Cross Multiply

4 -oz bottle × x applications = 13 -oz bottle × 7 applications

x applications = 13 -oz bottle × 7 applications/4 -oz bottle

x applications = 22.75 applications

Therefore, the number of applications of the 13 -oz bottle self-tanner that should Jen expect is 22.75 applications.

5 0

2 years ago

A patient needs to receive 1 teaspoon of a medication per day for one week. How many milliliters of the medication must be dispe

BabaBlast [244]

Answer:

B. 35 milliliters,

Step-by-step explanation:

1 teaspoonful = 5 mls so the total volume of medicine

= 7 * 5

= 35 milliliters.

6 0

3 years ago

Please solve this equation

tangare [24]

Answer:

∠2 = 41°

Step-by-step explanation:

This is because, all three interior angles in a triangle will have a sum of 180 degrees.

So given that we have two angles, 90 degrees(right angle) and 49 degrees, we can replace angle 2 as our unknown value(a variable) and make it equal to 180 degrees then solve for x;

49 + 90 + x = 180

139 + x = 180

-139 -139

x = 41°, ∠2 = 41°

3 0

2 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 $(1,1)$ .

7 0

3 years ago