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Alisiya [41]
2 years ago
15

Help me please. dont make me post this question again or ima be throwing hands. uwu

Mathematics
2 answers:
Agata [3.3K]2 years ago
5 0

Answer:

a) In 5 weeks he would have $50

b)w | m

    1 : $10

2: $20

3:$30

Step-by-step explanation:

RUDIKE [14]2 years ago
4 0
5 weeks 50 dollars each week
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I need help please.
enot [183]

Answer:

The answer is D. 2

Step-by-step explanation:

exponential form

exponents equal

move constant to the right

subtract the number

divide both sides by 2

6 0
3 years ago
Translate the following statement into an equation and solve for y: "y divided by 27 is-18."​
Yanka [14]

Answer:

y= -486

Step-by-step explanation:

We first translate the statement into an equation:

\frac{y}{27}=-18

Then we simplify:

y= -486

8 0
2 years ago
Read 2 more answers
Find two consecutive odd integers such that 65 more than the lesser is four times the greate
lyudmila [28]
Let the lesser number be x  and y the greater. Then

x + 65 = 4y ,  Also

y - x =  2
Adding:-

y + 65 = 4y + 2
63 = 3y
y = 21
and
x = 21-2 = 19

Answer the 2 odd numbers are 19 and 21.


8 0
3 years ago
If the sum of the even integers between 1 and k, inclusive, is equal to 2k, what is the value of k?
Alisiya [41]
If k is odd, then

\displaystyle\sum_{n=1}^{\lfloor k/2\rfloor}2n=2\dfrac{\left\lfloor\frac k2\right\rfloor\left(\left\lfloor\frac k2\right\rfloor+1\right)}2=\left\lfloor\dfrac k2\right\rfloor^2+\left\lfloor\dfrac k2\right\rfloor

while if k is even, then the sum would be

\displaystyle\sum_{n=1}^{k/2}2n=2\dfrac{\frac k2\left(\frac k2+1\right)}2=\dfrac{k^2+2k}4

The latter case is easier to solve:

\dfrac{k^2+2k}4=2k\implies k^2-6k=k(k-6)=0

which means k=6.

In the odd case, instead of considering the above equation we can consider the partial sums. If k is odd, then the sum of the even integers between 1 and k would be

S=2+4+6+\cdots+(k-5)+(k-3)+(k-1)

Now consider the partial sum up to the second-to-last term,

S^*=2+4+6+\cdots+(k-5)+(k-3)

Subtracting this from the previous partial sum, we have

S-S^*=k-1

We're given that the sums must add to 2k, which means

S=2k
S^*=2(k-2)

But taking the differences now yields

S-S^*=2k-2(k-2)=4

and there is only one k for which k-1=4; namely, k=5. However, the sum of the even integers between 1 and 5 is 2+4=6, whereas 2k=10\neq6. So there are no solutions to this over the odd integers.
5 0
3 years ago
Can someone help me…
aleksandr82 [10.1K]

Answer:

378$

Step-by-step explanation:

since u got 700ft^2 you would multiply 700 with 0.54 and get 378$

7 0
1 year ago
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