Use the equation s=5q+8 to find the value of s when q=1.

Meaghan and Lily are dividing 43.61 by 7. Meaghan says that the first digit of the quotient is in the ones place. Lily says it i

Ivahew [28]

Answer: Meaghan is right.

Step-by-step explanation:

For a number like:

123.45

the tens place would be the "2" (second number at the left of the decimal point)

The ones place would be the "3" (first number at the left of the decimal point)

Let's suppose that Lily is correct.

Then the quotient of:

43.61/7

Will be a number with two digits in the left of the decimal point.

The smallest number that meets this condition, is the number 10.

Then let's see:

7*10 = 70

then:

70/7 = 10

and:

43.61 < 70

(70 is larger than 43.61)

then:

43.61/7 < 70/7 = 10

Then:

(43.61/7) < 10

This means that our quotient is smaller than 10, then the first digit of the quotient can not be on the tens place.

Then Meaghan is the correct one.

We also could perform the quotient to find:

43.61/7 = 6.23

So yes, Meaghan is correct.

5 0

3 years ago

una tienda de ropa realiza un descuento de 5% por toda compra,si un pantalon sin descuento cuesta 420 pesos cuanto pagare por el

lukranit [14]

Inglés por favoro siento, pero no puedo ayudarle si no soy capaz de entender la pregunta. Haga clic en el enlace y luego escriba su pregunta y luego tratar de publicar

www.spanishdict.com/translation

7 0

3 years ago

If ashley receives 150 dimes from her grandfather how many nickels did she save?

olga2289 [7]

Ashely has 30 nickles.

8 0

3 years ago

Read 2 more answers

PLEASE SOLVE THIS FOR ME. HELP ASAP!!!

hjlf

Answer:

m = -3

b = 4

Equation: y = -3x + 4

Step-by-step explanation:

The equation for a straight line is y = mx + b.

The line crosses the y-axis at (0,4). So, b (the y-intercept) will be 4.

Now, to find m (the slope). The next point on the line is (1,1). To get there, you go down 3 units (-3) and to the right 1 unit.

Slope is known as rise (change in x-axis) over run (change in y-axis). So, -3 ÷ 1 = -3. m will be -3.

For the equation of the line, just plug the numbers into the equation for a straight line.

y = mx + b

y = -3x + 4

Hope that helps.

4 0

3 years ago

You are traveling down a country road at a rate of 95 feet/sec when you see a large cow 300 feet in front of you and directly in

emmainna [20.7K]

Answer:

1) You can rely solely on your brakes because when doing so the car will just travel 250ft from the point you hit your brakes till the point the car stopped completely, leaving you 50ft away from the cow.

2) See attached picture.

j(t) represents the distance from the point you hit the brake t seconds after you hit it in feet

j'(t) represents the velocity of the car t seconds after the brakes have been hit in ft/s.

j"(t) represents the acceleration of the car t seconds after the brakes have been hit in $ft/s^{2}$

3) yes, any time after t=5.28 will not accurately model the path of the car since at that exact time the car will reach a velocity of 0ft/s and unless another force is applied to the car, then the car will not move after that time.

4) $j(t)=\left \{ {{95t-9t^{2}; 0\le t$

$j'(t)=\left \{ {{95-18t; 0\leq t$

(see attached picture for graph)

Step-by-step explanation:

1) In this part of the problem we need to find the time when the speed of the car is 0. Gets to a complete stop. For this we will need to take the derivative of the position function so we get:

$j(t)=95t-9t^2$

$j'(t)=95-18t$

and we set the first derivative equal to zero so we get:

95-18t=0

and solve for t

-18t=-95

$t=\frac{95}{18}$

t=5.28s

so now we calculate the position of the car after 5.28 seconds, so we get:

$j(5.28)=95(5.28)-9(5.28)^{2}$

$j(5.28)=250.69ft$

so we have that the car will stop 250.69ft after he hit the brakes, so there will be about 50ft between the car and the cow when the car stops completely, so he can rely just on the breaks.

2) For answer 2 I take the second derivative of the function so I get:

$j(t)=95t-9t^{2}$

$j'(t)=95-18t$

j"(t)=-18

and then we graph them. (See attached picture)

j(t) represents the distance from the point you hit the brake t seconds after you hit it in feet

j'(t) represents the velocity of the car t seconds after the brakes have been hit in ft/s.

j"(t) represents the acceleration of the car t seconds after the brakes have been hit in $ft/s^{2}$

3) yes, any time after t=5.28 will not accurately model the path of the car since at that exact time the car will reach a velocity of 0ft/s and unless another force is applied to the car, then the car will not move after that time.

4) $j(t)=\left \{ {{95t-9t^{2}; 0\le t$

$j'(t)=\left \{ {{95-18t; 0\leq t$

(see attached picture for graph)

5 0

3 years ago