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3 years ago

A grocery store sells a bag of 6 oranges for $4.20. If Ayden spent $2.80 on oranges, how many did he buy?

Mathematics

1 answer:

aleksley [76]3 years ago

5 0

Answer:

4 oranges.

Step-by-step explanation:

First, you find the unit rate of cost per orange, which is $0.70, then divide that by $2.80.

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PLEASE HELP WILL MEDAL BRAINLIEST ANSWER!

babunello [35]

The amount after 7 years is given by:
$A=15000(1+\frac{0.048}{4})^{4\times7}=20,948.15$
The answer is: $20,948.15

3 0

3 years ago

How many kilometers are in 35 MI

Aleksandr-060686 [28]

Answer:

56.327km

Step-by-step explanation:

To find miles from kilometers times the number of kilometers(35) by 1.609 which approximately equals 56.327km.

8 0

3 years ago

Ben and Tom each take a driving test.

GaryK [48]

Answer:

(b)0.56

Step-by-step explanation:

(a)The tree diagram is attached.

P(Ben Fails)=1-0.8=0.2

P(Tom Fails)=1-0.7=0.3

(b)Probability that both will pass their driving test.

P(both will pass)=0.8 X 0.7 =0.56

(c)Probability that only one of them will pass their driving test.

P(Tom Bass, Ben Fails OR Ben Pass, Tom Fails)

=(0.7 X 0.2)+(0.8 X 0.3)

=0.14+0.24

=0.38

3 0

3 years ago

Tamika has two bags of trail mix. Each has a combination of 60 pieces of fruit and nuts. Ratio of fruit to nuts in the first bag

madam [21]

Answer:

the second bag because the ratio is greater

Step-by-step explanation:

Bag1

fruits 3 : 5 nuts so 3+5=8 and 60/8=7.5 but 7.5*3=22.5

ratio is 3/5=0.428

Bag2

fruits 7 : 10 nuts so 10+7=17 and (60/17)*7 ≈ 24.7

ratio is 7/10=0.7

4 0

3 years ago

Read 2 more answers

Write down the explicit solution for each of the following: a) x’=t–sin(t); x(0)=1

Kay [80]

Answer:

a) x=(t^2)/2+cos(t), b) x=2+3e^(-2t), c) x=(1/2)sin(2t)

Step-by-step explanation:

Let's solve by separating variables:

$x'=\frac{dx}{dt}$

a) x’=t–sin(t), x(0)=1

$dx=(t-sint)dt$

Apply integral both sides:

$\int {} \, dx=\int {(t-sint)} \, dt\\\\x=\frac{t^2}{2}+cost +k$

where k is a constant due to integration. With x(0)=1, substitute:

$1=0+cos0+k\\\\1=1+k\\k=0$

Finally:

$x=\frac{t^2}{2} +cos(t)$

b) x’+2x=4; x(0)=5

$dx=(4-2x)dt\\\\\frac{dx}{4-2x}=dt \\\\\int {\frac{dx}{4-2x}}= \int {dt}\\$

Completing the integral:

$-\frac{1}{2} \int{\frac{(-2)dx}{4-2x}}= \int {dt}$

Solving the operator:

$-\frac{1}{2}ln(4-2x)=t+k$

Using algebra, it becomes explicit:

$x=2+ke^{-2t}$

With x(0)=5, substitute:

$5=2+ke^{-2(0)}=2+k(1)\\\\k=3$

Finally:

$x=2+3e^{-2t}$

c) x’’+4x=0; x(0)=0; x’(0)=1

Let $x=e^{mt}$ be the solution for the equation, then:

$x'=me^{mt}\\x''=m^{2}e^{mt}$

Substituting these equations in c)

$m^{2}e^{mt}+4(e^{mt})=0\\\\m^{2}+4=0\\\\m^{2}=-4\\\\m=2i$

This becomes the solution m=α±βi where α=0 and β=2

$x=e^{\alpha t}[Asin\beta t+Bcos\beta t]\\\\x=e^{0}[Asin((2)t)+Bcos((2)t)]\\\\x=Asin((2)t)+Bcos((2)t)$

Where A and B are constants. With x(0)=0; x’(0)=1:

$x=Asin(2t)+Bcos(2t)\\\\x'=2Acos(2t)-2Bsin(2t)\\\\0=Asin(2(0))+Bcos(2(0))\\\\0=0+B(1)\\\\B=0\\\\1=2Acos(2(0))\\\\1=2A\\\\A=\frac{1}{2}$

Finally:

$x=\frac{1}{2} sin(2t)$

7 0

3 years ago