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Sauron [17]
3 years ago
15

Comes with a picture multiple-choice answer please help I need immediate help Thank youuuu

Mathematics
1 answer:
Nimfa-mama [501]3 years ago
4 0

Answer:

The answer is A. 31,250.

Step-by-step explanation:

To solve for the 7th term in the geometric sequence, start by plugging 7 into the explicit formula, and the formula will look like a_{7}=2*5^{(7-1)}.

Next, subtract 1 from 7, and the formula will look like a_{7}=2*5^{6}.

Then, raise 5 to the power of 6, which will look like 2*15625.

Finally, multiply 2 by 15,625, and the answer will be 31,250.

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-3x(x - 4) + 2 - (5 - x)
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Answer:

- 3x(x - 4) + 2 - (5 - x) \\ =   - 3 {x}^{2}  + 12x + 2 - 5 + x \\  =  - 3 {x}^{2}  + 13x - 3

I hope I helped you^_^

3 0
3 years ago
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On a certain map, 1.25 inches represents 20 miles. Longwood and Milltown are 5 inches apart on the map. What is the actual dista
Elanso [62]
Divide 5 by 1.25. That equals 4. Multiply 4 by 20 and that equals 80. So its 80 miles between Longwood and Milltown.
3 0
4 years ago
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How do I complete this equation
Nesterboy [21]

Answer:

130°

Step-by-step explanation:

All the interior angles should add up to 540 degrees because it's a pentagon

∠T ≅∠S, so ∠T is also 115°

∠P and ∠R are both 90°

Add everything up and solve for ∠Q:

115 + 115 + 90 + 90 + m∠Q = 540

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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

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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.
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3 years ago
(help asap!) What is the value of the function at x = 2?<br><br> Enter your answer in the box.
cupoosta [38]

Answer:

Step-by-step explant

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