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3 years ago

12

What is the greatest common factor of 60, 84 and 72?

2 answers:

frez [133]3 years ago

8 0

The answer in 12, plz thanks if I was helpful, have a great day ;)

Arada [10]3 years ago

7 0

The answer is B. 12 .

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Jackson is starting an exercise program. The first day he will spend 30 minutes on the treadmill. He will increase his time on t

yulyashka [42]

Answer:

40 minutes.

Step-by-step explanation:

time spent on the thread mill as a function of d(day) = t(d)

= 30 +2(d-1) minutes

[where d represents day number]

(when d is 1 the time spent should be 30

so substitute d=1 and check it will be 30 only.

on first day he will spend 30 minutes so, I added 30 .

and on every additional day( these additional days are excluding first day), he will increase the time by 2 minutes.so, I added 2(d-1) to initial 30 minutes)

so, T(6), the minutes he will spend on the treadmill on day 6=

=30+2(6-1)

=30+2(5)

=30+10

=40 minutes

3 0

3 years ago

HELP NOW PLZ AND I WILL LOVE YOU FOREVER

mr_godi [17]

Answer:

12/4+13=16 2+22/4=7.5

Step-by-step explanation:

you answer is the last option im 100 percent sure.

8 0

3 years ago

A car rental company offers two plans for renting a car. Plan A: $35 per day and $O.25 per mile. Plan B: $55 per day with free u

salantis [7]

The answer for this question is 90.25$

4 0

3 years ago

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

7 0

3 years ago

Read 2 more answers

5 + 4x - 2x = 8x + 11 + 12<br><br> Explain step by step PLEASE

Yuliya22 [10]

Answer:

The solution is x = -3

Step-by-step explanation: Please brainliest!

4 0

2 years ago

Read 2 more answers