All categories

2 years ago

Need Help ASAP How many weeks will it take to save $150 if I save $2.50 per week?

Mathematics

2 answers:

Doss [256]2 years ago

8 0

150 divided by 2.50 equals 60 so the answer is 60 weeks :)

Sergeu [11.5K]2 years ago

5 0

Answer: 60 weeks.

Step-by-step explanation: 150 divided by 2.50 = 60

You might be interested in

-3(4x+5) please help<br>

Natasha_Volkova [10]

Answer:
-12x-15
explanation:
you multiply -3 by 4x, gives u -12x and then you multiply -3 by 15 which gives you -15
then you put it in an equation
-12x-15

7 0

3 years ago

-17-3(2+13)+24 What is the answer?

yawa3891 [41]

The answer will be - 38

6 0

3 years ago

Read 2 more answers

Explaining your answer can either include a counterexample, a diagram or a written explanation to support your answer.

anzhelika [568]

Answer:

sorry I didn't understanding this

6 0

2 years ago

Write the equation of a line in slope-intercept form that passes through the point (0,-7) and has a slope of 3/4.

kondor19780726 [428]

Answer:

y=3/4x-7

Step-by-step explanation:

(0,-7) is the y-intercept

3/4 is the slope

y=mx+b

m is the slope

b is the y-intercept

plug it in:

y=3/4x-7

please mark brainliest!

4 0

2 years ago

Consider the following initial value problem, in which an input of large amplitude and short duration has been idealized as a de

Ganezh [65]

Answer:

a. $\mathbf{Y(s) = L \{y(t)\} = \dfrac{7}{s(s+1)}+ \dfrac{e^{-3s}}{s+1}}$

b. $\mathbf{y(t) = \{7e^t + e^3 u (t-3)-7\}e^{-t}}$

Step-by-step explanation:

The initial value problem is given as:

$y' +y = 7+\delta (t-3) \\ \\ y(0)=0$

Applying laplace transformation on the expression $y' +y = 7+\delta (t-3)$

to get $L[{y+y'} ]= L[{7 + \delta (t-3)}]$

$l\{y' \} + L \{y\} = L \{7\} + L \{ \delta (t-3\} \\ \\ sY(s) -y(0) +Y(s) = \dfrac{7}{s}+ e ^{-3s} \\ \\ (s+1) Y(s) -0 = \dfrac{7}{s}+ e^{-3s} \\ \\ \mathbf{Y(s) = L \{y(t)\} = \dfrac{7}{s(s+1)}+ \dfrac{e^{-3s}}{s+1}}$

Taking inverse of Laplace transformation

$y(t) = 7 L^{-1} [ \dfrac{1}{(s+1)}] + L^{-1} [\dfrac{e^{-3s}}{s+1}] \\ \\ y(t) = 7L^{-1} [\dfrac{(s+1)-s}{s(s+1)}] +L^{-1} [\dfrac{e^{-3s}}{s+1}] \\ \\ y(t) = 7L^{-1} [\dfrac{1}{s}-\dfrac{1}{s+1}] + L^{-1}[\dfrac{e^{-3s}}{s+1}] \\ \\ y(t) = 7 [1-e^{-t} ] + L^{-1} [\dfrac{e^{-3s}}{s+1}]$