Jordan is starting to save $150 to buy a new cell phone. In December, he saved $5. In January, he plans twice as much as he did in December, for a total savings of $15 ($5 + $10). If Jordan continues to save twice as much he saved from the previous month, in which month will his total savings will be enough to buy the phone.

At the Month May Jordan as able to get enough money to buy the phone.

Work:

December:$5

January:$10

February:$20

March:$40

April:$80

May:$160 <--Answer

In the problem Jordan saves $5 at December, then at January Jordan saved $10 which is twice as December's saved money amount. February, Jordan saved $20, which is twice as January's saved money amount. March is $40, which is twice as February's saved money amount. For April it is $80, which is twice as March's money amount, May is $160 which is twice as April's money amount. Or another way to say this is by starting at $5. Then, keep multiplying by 2 for each month until you get $150 or more. When you do, stop at the month that gets you $150 or more. When i did this exact method, I got $160 on the Month of May. $160 is more than enough for Jordan to buy his phone. So the month May is the correct answer where Jordan can get enough money for his phone.

The fifth month; By April Jordan would have saved more than enough to buy a new phone.

Step-by-step explanation:

Dec $5

Jan $10

Feb $20

Mar $40

Apr $80

Dec + Jan = $15($5+$10)

Dec+ Jan + feb = $35($5+$10+$20=$35)

Dec+ Jan + feb + mar = $75 ($5+$10+$20+$40=$75)

Dec+ Jan + feb + mar + apr = $155

($5+$10+$20+$40+$80=$155)

a ≈ 172.05

step-by-step explanation:

the pictures.

answer: (-8,4) are the coordinates of a

step-by-step explanation:

let the side of square abcd is x .

thus the area of square abcd is .

now, by the diagram area of triangle ead= 1/2 × ea × ad = 1/2 × ea × x

( because ad is the side of square)

and, according to the question,

the areas of square abcd and right ead are equal.

⇒

⇒ea=2x

thus, ea: ab=2: 1

therefore, point a divides line segment eb into the ratio 2: 1.

⇒ by sectional formula coordinates of a are,

=

=

=(-8, 4)

thus the coordinates of a are (-8,4)