5 + (-6) = ? Math:adding with rational integers

Is (9, 90) a solution to the equation y = 10x?<br>yes

Galina-37 [17]

Use the substitution method

(9,90) y= 10x

90= 10(9)

90= 90

Answer is yes

5 0

3 years ago

Read 2 more answers

Jose left the White House and drove toward the recycling plant at an average speed of 40 km/h. Rob left some time later driving

Aleks04 [339]

Let $x$ represent Jose's drived distance and $y$ represent Rob's.
So what you need to do is to solve the equation:
$40x=48*5$
$40x=240$
$x=240/40=6$

So Jose drove for 6 hours before Rob caught him.

4 0

3 years ago

On a coordinate plane, a line is drawn from point j to point k. point j is at (negative 15, negative 5) and point k is at (25, 1

Genrish500 [490]

The required x- and y- coordinates of point e, which partitions the directed line segment is (17, 11)

<h3>Midpoint of coordinates</h3>

The middle point of two coordinates is known as its midpoint. The formula for calculating the midpoint of a coordinate is expressed as:

$m(x,y)=(\frac{mx_1+nx_2}{m+n}, \frac{my_1+ny_2}{m+n})$

Given the coordinate points J(-15, -5) and k(25, 15) partitioned in the ratio 1:4, the x- and y- coordinates of point e, which partitions the directed line segment is given as:

$m(x,y)=(\frac{1(-15)+4(25)}{1+4}, \frac{1(-5)+4(15)}{1+4})\\m(x,y)=(\frac{85}{5}, \frac{55}{5} )\\m(x, y) = (17,11)$

Hence the required x- and y- coordinates of point e, which partitions the directed line segment is (17, 11)

Learn more on midpoint here: brainly.com/question/5566419

#SPJ4

7 0

2 years ago

The half-life of cesium-137 is 30 years. Suppose we have a 180-mg sample. (a) Find the mass that remains after t years. (b) How

DedPeter [7]

Answer:

a) $Q(t) = 180e^{-0.023t}$

b) 11.4mg of cesium-137 remains after 120 years.

c) 225.8 years.

Step-by-step explanation:

The following equation is used to calculate the amount of cesium-137:

$Q(t) = Q(0)e^{-rt}$

In which Q(t) is the amount after t years, Q(0) is the initial amount, and r is the rate at which the amount decreses.

(a) Find the mass that remains after t years.

The half-life of cesium-137 is 30 years.

This means that Q(30) = 0.5Q(0). We apply this information to the equation to find the value of r.

$Q(t) = Q(0)e^{-rt}$

$0.5Q(0) = Q(0)e^{-30r}$

$e^{-30r} = 0.5$

Applying ln to both sides of the equality.

$\ln{e^{-30r}} = \ln{0.5}$

$-30r = \ln{0.5}$

$r = \frac{\ln{0.5}}{-30}$

$r = 0.023$

So

$Q(t) = Q(0)e^{-0.023t}$

180-mg sample, so Q(0) = 180

$Q(t) = 180e^{-0.023t}$

(b) How much of the sample remains after 120 years?

This is Q(120).

$Q(t) = 180e^{-0.023t}$

$Q(120) = 180e^{-0.023*120}$

$Q(120) = 11.4$

11.4mg of cesium-137 remains after 120 years.

(c) After how long will only 1 mg remain?

This is t when Q(t) = 1. So

$Q(t) = 180e^{-0.023t}$

$1 = 180e^{-0.023t}$

$e^{-0.023t} = \frac{1}{180}$

$e^{-0.023t} = 0.00556$

Applying ln to both sides

$\ln{e^{-0.023t}} = \ln{0.00556}$

$-0.023t = \ln{0.00556}$

$t = \frac{\ln{0.00556}}{-0.023}$

$t = 225.8$

225.8 years.

8 0

3 years ago

Read 2 more answers

Matthew currently has enough money to buy 45 toy cars. If the cost of each car was 10 cents less, Matthew could buy 5 more cars.

Law Incorporation [45]

The answer is $45

Let's make the system of equations.
x - the cost of the car
y - the amount Matthew has

<span>Matthew currently has enough money to buy 45 toy cars:
y =45x
</span><span>If the cost of each car was 10 cents less, Matthew could buy 5 more cars:
y = (x - 10)*(45 + 5)

45x = (x - 0.10) * 50
45x = 50x - 5
5 = 50x - 45x
5 = 5x
x = 5 / 5
x = 1

y = 45x = 45 * 1 = $45</span>

5 0

3 years ago

5 + (-6) = ? Math:adding with rational integers​

5 + (-6) = ? Math:adding with rational integers