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True [87]
3 years ago
8

If the Williams family has a monthly income of $3,000, how much do they spend on food each month?

Mathematics
2 answers:
vodomira [7]3 years ago
6 0
You take $3,000 and multiply it by .11 (11% in decimal form). it equals $330. they have room to spend $330 on food a month
KATRIN_1 [288]3 years ago
5 0
3000 x 11% = 330
They spent $330.00 on food each month.
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What is 1,000 to the power of 4?
quester [9]
The answer to what is 1,000 to the power of 4 is 1,000,000
8 0
3 years ago
a stone is thrown downward with an intitial velocity of 29.5 m/sec will travel a distance of a meters, where s(t) = 4.9t^2 + 29.
Alenkinab [10]

Answer:

t = 9 s

Step-by-step explanation:

A stone is thrown downward with an intitial velocity of 29.5 m/sec will travel a distance of a meters, where

s(t) = -4.9t^2 + 29.4+132.3

We need to find when the stone will hit the ground. It means when s(t) = 0. So,

-4.9t^2 + 29.4t+132.3=0\\\\t=-3\ and\ t=9

So, the stone will take 9 seconds to hit the ground. Hence, this is the required solution.

5 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
There were 7 field goals made in 14 attempts. how many goals should be made in 8 attempts
trapecia [35]
There would be 4

hope this helps you
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It would be shifted up
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