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3 years ago

6

Write and solve an equation. On Monday Anna picked some apples. On Tuesday she picked 15 more apples than on Monday. If Anna pic

ked 32 apples on Tuesday, how many apples, a, did she pick on Monday? Solve for a.

1 answer:

Ket [755]3 years ago

6 0

32 minus 15 equals 17. Anna picked 17 apples on Monday.

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Ramona deposited $4,190.51 into a savings account with an interest rate of 5.2% compounded twice a year. About how long will it

kap26 [50]

<span>A = P(1+r/2)^2t A/P = (1+r/2)^2t ln(A/P) = 2t ln((1+r/2)) ln(A/P)/ln(1+r/2) =2t ln(9000/4190.51)/ln(1+0.052/2)=2t t=15 yrs therfore ans is C.14 years ,11 months</span>

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3 years ago

PLZZ HELP ILL GIVE BRAINLIEST AND DO NOT TYPE NO LINK WITH YALL LAZY AHH

Reil [10]

Answer:

D

Step-by-step explanation:

8 is LESS THAN 1/2x

so youd take those numbers and multiply by 1/2

C, B, and A are all under 8, so that would be wrong

when you multiply 1/2 with 17, you get 8.5 which is bigger than 8. then everything else above that is more than 8, therefore D is the right answer.

3 0

3 years ago

Read 2 more answers

(01.03 MC) Traveling at 65 miles per hour, how many minutes, rounded to the nearest whole number, does it take to drive 125 mile

mariarad [96]

It should take 115 minutes

4 0

2 years ago

djverab [1.8K]

Answer:

x=7

Step-by-step explanation:

Find slope first

$m-\frac{y2-y1}{x2-x1}$

$m=\frac{6-5}{7-7}$

$m=\frac{1}{0}\\$

m is undefined- so it is a vertical line

x=7

6 0

3 years ago

13. The least common multiple of two non-zero integers a and b is the unique positive integer m such that (i) m is a common mult

Vlad [161]

Answer:

[a,b] divides n

Step-by-step explanation:

Let us denote the least common multiple of a and b [a,b]=m.

We want to prove that m divides n, where n is a multiple of a and b.

We suppose m does not divide n, then by the Division Theorem, there exists q and r integers such that:

(1) ... n=mq+r, where 0<r<m

As n is a multiple of a and b, there exists s and t integers such that:

sa=n and tb=n

Same thing happens to m as it is the least common multiple, there exists u and v such that:

ua=m and vb=m

So (1) has the following form:

n=mq+r ⇒ sa=uaq+r ⇒sa-uaq=r⇒(s-uq)a=r and

n=mq+r ⇒ tb=vbq+r ⇒ tb-vbq=r⇒ (t-vq)b=r

So r is a multiple of a and b, but r<m which is a contradiction as, m is the least common multiple of a and b. So this concludes the proof.

So this means that $\frac{ab}{m}$ is and integer.

As m= vb, then $\frac{m}{b}$ is an integer, lets say $\frac{m}{b}$ =v; and as m=ua, then $\frac{m}{a}$ =u.

So $\frac{ab}{m}$ v= $\frac{ab}{m}$ $\frac{m}{b}$ =a, so $\frac{ab}{m}$ divides a; on the other hand, $\frac{ab}{m}$ u= $\frac{ab}{m}$ $\frac{m}{a}$ =b, so $\frac{ab}{m}$ divides b. From this we can conclude that $\frac{ab}{m}$ is a common divisor of a and b.

4 0

3 years ago