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saul85 [17]
3 years ago
5

What is the distance between the points (14, 29) and (14, 58) in the coordinate plane?

Mathematics
2 answers:
Snowcat [4.5K]3 years ago
7 0
Answer: 29 units

∞∞∞∞∞∞∞∞∞
<em>Explanation:</em>
∞∞∞∞∞∞∞∞∞

\textnormal {Formula : Distance }= \sqrt{(Y_2 - Y_1) ^2+(X_2 - X_1)^2}

\textnormal {Plug }(X_1 ,Y_1) = (14, 29) \textnormal { and } (X_2 , Y_2) =  (14, 58) \textnormal { into the formula: }

\textnormal {Distance }= \sqrt{(58 -29) ^2+(14-14)^2} =  \sqrt{29^2 + 0^2} =  \sqrt{841} = 29 \textnormal { units}


ss7ja [257]3 years ago
6 0
Distance= \sqrt{(x_{2}-x_{1}) ^{2}+(y_{2}-y_{1})^2} = \sqrt{(58-29)^{2}+(14-14)^{2}} =29&#10;&#10;But because x_{1}=x_{2}, &#10;&#10; and it is a vertical segment, &#10;&#10;it is enough to find abs(y_{2} -y_{1})=|58-29|=29
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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
Evaluate the expression for x = -2, y = 3, z = -4. -4x + 5y + 6z *
babunello [35]

Answer:

-1

Step-by-step explanation:

Plug in given values

-4x + 5y + 6z

-4(-2) + 5(3) + 6(-4)

8 + 15 -24

23-24

-1

6 0
3 years ago
Read 2 more answers
A square has vertices at the points A(-4,-4), B(-5,-4), C(-5,-3), and D(-4,-3). What is the area of this square? A. 1 square uni
laiz [17]

Answer:

Step-by-step explanation:

The length of AB = 1 unit.

area of square = 1² unit² = 1 square unit.

8 0
3 years ago
20-{3+2[20÷4×5]}+16= -17 but I have to show how I got this.
aleksandr82 [10.1K]

20-{3+2(5*5)}+16

=20-{3+2*25}+16

=20-(3+50)+16

=20-53+16

=36-53

=-17

7 0
3 years ago
Please help a step by step would be nice answer is also ok
nasty-shy [4]
I would choose A I’m not to sure tho
6 0
3 years ago
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