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STALIN [3.7K]
2 years ago
10

The value of the digit 8 in 1,857 is how many times the value of the digit 8 in 3,083?

Mathematics
1 answer:
KATRIN_1 [288]2 years ago
5 0

Answer:

10 times

Step-by-step explanation:

The 8 in 1,857 is in the hundreds place and the 8 in 3,083 is in the tens place.

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Multiple Choice
mihalych1998 [28]

Answer:

D is correct

Step-by-step explanation:

A graph of that sort will make a perfectly mirrored "V" shape, and with no offset, the bottom point will be on the y axis.  This means that the first equation will intercept the y axis at zero, the second will intercept the y-axis at -15.

"A" may seem correct also, as the second graph will intercept the x-axis at -15, but it is not complete, as it will intercept that axis at +15 as well.

7 0
3 years ago
Which of the following ordered pairs make this equation true? Y = 2x - 7 A. (1,9) B. (-3, 4) C. (4,1) D. (-1, 5)​
gregori [183]

Answer:

(4,1)

Step-by-step explanation:

Using substitution you can determine that 2*4-7 is in fact equal to 1

3 0
3 years ago
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Effectus [21]
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7 0
4 years ago
I need geometry help please.
irina1246 [14]

Lets remember the property "alternative-interior angles" when we have parallels lines:

Given two paralles lines, the following is true:

In our question, we have:

We know that the angles J + P + K=180

So, the triangles have all the angles the same, so they are similar by angle-angle-angle, and they have one side congruent, AC=CD, so the triangles are congruents.

6 0
2 years ago
What is the nth term?
kumpel [21]

Let a_k denote the <em>k</em>th term of the sequence. Then

a_k=a_1+d(k-1)

where <em>d</em> is the common difference between consecutive terms in the sequence and <em>a</em>₁ is the first term.

The sum of the first <em>n</em> terms is

S_n=\displaystyle\sum_{k=1}^na_k=a_1+a_2+\cdots+a_{n-1}+a_n

From the formula for a_k, we get

S_n=\displaystyle\sum_{k=1}^n(a_1+d(k-1))=a_1\sum_{k=1}^n1+d\sum_{k=1}^n(k-1)

S_n=\displaystyle na_1+d\sum_{k=0}^{n-1}k

S_n=na_1+\dfrac{d(n-1)n}2

S_n=\dfrac n2(2a_1-d+dn)

So we have d=-5, and 2a_1-d=16 so that a_1=\frac{11}2.

Then the <em>n</em>th term in the sequence is

a_n=\dfrac{11}2-5(n-1)=\boxed{\dfrac{21-10n}2}

7 0
4 years ago
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