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3 years ago

What equation relates the number of quarters x and the number of quarters x and the number of dime y to the goal of $10

1 answer:

8 0

.25x+.10y=$10

.25 because that is the value of the quarter, times x the number or quarters you have and .10 because that is the value of the dime times y, the number of dimes you have.

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Find the 53rd term of the arithmetic sequence<br> 27,11, -5

vivado [14]

Answer:

The 53rd term of this arithmetic sequence is -805.

Step-by-step explanation:

The general rule of an arithmetic sequence is the following:

$a_{n+1} = a_{n} + d$

In which d is the common diference between each term, that is, $d = a_{3} - a_{2} = a_{2} - a_{1}$ .

To find the nth term of the sequence, this equation can be written as:

$a_{n} = a_{1} + (n-1)d$

27,11, -5

So $a_{1} = 27, a_{2} - a_{1} = 11 - 27 = -16[/tex[tex]a_{n} = a_{1} + (n-1)d$

$a_{53} = a_{1} + (52)d = 27 + 52*(-16) = -805$

The 53rd term of this arithmetic sequence is -805.

3 0

2 years ago

[888-18(240:6+216:6)]:4<br><br> Rezultatul nu este nr negativ

Goshia [24]

-120 Thats the answer

3 0

3 years ago

Please answer correctly! I will mark you as Brainliest!

jasenka [17]

Answer:

I have actually done this before, I got it right and I chose C!

hope this helps. love u guys!

6 0

3 years ago

Read 2 more answers

What is the slope and y-intercept of the equation on the graph?

Fynjy0 [20]

The answer is B :)
hope this helps you out

3 0

2 years ago

Read 2 more answers

A small metal bar, whose initial temperature was 10° C, is dropped into a large container of boiling water. How long will it tak

IgorC [24]

Answer

According to newton law of cooling

$\dfrac{d T}{dt} = k (T - T_a)$

and

$T = Ce^{kt} + T_a$

now At Ta = 100

T₀ = 10 °C

T₁ = 12° C

$T(0) = Ce^{0} + T_a$

10 = C + 100

C = -90

now,

$12 = -90e^k + 100$

$e^k = \dfrac{88}{90}$

$k = ln{\dfrac{88}{90}}$

at time equal to t

T(t) = 70

$T(t) = -90 (\dfrac{88}{90})^t + 100$

$70 = -90(\dfrac{88}{90})^t + 100$

$(\dfrac{88}{90})^t = \dfrac{1}{3}$

$t\times ln(\dfrac{88}{90}) = ln(\dfrac{1}{3})$

t = 48.88 s

hence, after 48.88 s the temperature of the body will be 70°C

b) time taken to reach 98°C

$T(t) = -90 (\dfrac{88}{90})^t + 100$

$98 = -90(\dfrac{88}{90})^t + 100$

$(\dfrac{88}{90})^t = \dfrac{1}{45}$

$t\times ln(\dfrac{88}{90}) = ln(\dfrac{1}{45})$

t = 390 s

hence, after 390 s the temperature of the body will be 98°C

5 0

2 years ago