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777dan777 [17]
2 years ago
5

703 to the nearest hundred

Mathematics
2 answers:
Andrej [43]2 years ago
4 0

Answer:

700

Step-by-step explanation:

Yuki888 [10]2 years ago
3 0

Your answer is 700.

Since it is 703 and not 750 or above then it is 700.

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PLEASE HELP ME!!! I'LL MARK YOU BRAINLIEST IF YOU ANSWER THESE CORRECTLY!!
alisha [4.7K]

Answer:

1) Volume = ⅓×pi×r²×h

= ⅓×3.14×7²×14

= 718.01 in³

2) ⅓×3.14×5²×10

= 261.67 cm³

3) r = 12÷2 = 6

Volume = (4/3)×3.14×6³

= 904.32 mi³

4) 3.142×4²×5

= 251.2 ft³

5) radius = 2

(4/3)×3.14×2³ = 33.49 ft³

6) radius = 2

3.142×2²×3 = 37.68 m³

4 0
3 years ago
B. How many hundreds are in 4362? How many<br> whole hundreds?
zhannawk [14.2K]

Answer:

3

Step-by-step explanation:

4362

3= hundreds place value

So that means there are 3 whole hundreds

6 0
2 years ago
Read 2 more answers
Find the sum of the positive integers less than 200 which are not multiples of 4 and 7​
taurus [48]

Answer:

12942 is the sum of positive integers between 1 (inclusive) and 199 (inclusive) that are not multiples of 4 and not multiples 7.

Step-by-step explanation:

For an arithmetic series with:

  • a_1 as the first term,
  • a_n as the last term, and
  • d as the common difference,

there would be \displaystyle \left(\frac{a_n - a_1}{d} + 1\right) terms, where as the sum would be \displaystyle \frac{1}{2}\, \displaystyle \underbrace{\left(\frac{a_n - a_1}{d} + 1\right)}_\text{number of terms}\, (a_1 + a_n).

Positive integers between 1 (inclusive) and 199 (inclusive) include:

1,\, 2,\, \dots,\, 199.

The common difference of this arithmetic series is 1. There would be (199 - 1) + 1 = 199 terms. The sum of these integers would thus be:

\begin{aligned}\frac{1}{2}\times ((199 - 1) + 1) \times (1 + 199) = 19900 \end{aligned}.

Similarly, positive integers between 1 (inclusive) and 199 (inclusive) that are multiples of 4 include:

4,\, 8,\, \dots,\, 196.

The common difference of this arithmetic series is 4. There would be (196 - 4) / 4 + 1 = 49 terms. The sum of these integers would thus be:

\begin{aligned}\frac{1}{2}\times 49 \times (4 + 196) = 4900 \end{aligned}

Positive integers between 1 (inclusive) and 199 (inclusive) that are multiples of 7 include:

7,\, 14,\, \dots,\, 196.

The common difference of this arithmetic series is 7. There would be (196 - 7) / 7 + 1 = 28 terms. The sum of these integers would thus be:

\begin{aligned}\frac{1}{2}\times 28 \times (7 + 196) = 2842 \end{aligned}

Positive integers between 1 (inclusive) and 199 (inclusive) that are multiples of 28 (integers that are both multiples of 4 and multiples of 7) include:

28,\, 56,\, \dots,\, 196.

The common difference of this arithmetic series is 28. There would be (196 - 28) / 28 + 1 = 7 terms. The sum of these integers would thus be:

\begin{aligned}\frac{1}{2}\times 7 \times (28 + 196) = 784 \end{aligned}.

The requested sum will be equal to:

  • the sum of all integers from 1 to 199,
  • minus the sum of all integer multiples of 4 between 1\! and 199\!, and the sum integer multiples of 7 between 1 and 199,
  • plus the sum of all integer multiples of 28 between 1 and 199- these numbers were subtracted twice in the previous step and should be added back to the sum once.

That is:

19900 - 4900 - 2842 + 784 = 12942.

8 0
3 years ago
What percentage of 440 is 325
GuDViN [60]

The answer would be 73.86

3 0
2 years ago
Read 2 more answers
F(x) = 4x^2 - 3x + 2kx + 1 
dybincka [34]
f(x)=4x^2-3x+2kx+1=4x^2+(2k-3)x+1\\\\a=4;\ b=2k-3;\ c=1\\\\function\ has\ two\ zeros\ when\ \Delta=b^2-4ac > 0\\\\\Delta=(2k-3)^2-4\cdot4\cdot1=4k^2-12k+9-16=4k^2-12k-7 > 0\\\\a_k=4;\ b_k=-12;\ c_k=-7\\\\\Delta_k=(-12)^2-4\cdot4\cdot(-7)=144+112=256\\\\k_1=\frac{-b_k-\sqrt{\Delta_k}}{2a_k};\ k_2=\frac{-b_k+\sqrt{\Delta_k}}{2a_k}

\sqrt{\Delta_k}=\sqrt{256}=16\\\\k_1=\frac{12-16}{2\cdot4}=\frac{-4}{8}=-\frac{1}{2};\ k_2=\frac{12+16}{2\cdot4}=\frac{28}{8}=\frac{7}{2}\\\\a_k=4 > 0\ (up\ parabola\ arms-see\ the\ picture)\\\\Answer:k\in(-\infty;-\frac{1}{2})\ \cup\ (\frac{7}{2};\ \infty)

6 0
2 years ago
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