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3 years ago

How do I evaluate x+y=10

1 answer:

8 0

Hi there! Evaluate means to solve so we are going to solve x+y=10. Add both sides to -y, x+y+-y=10+-y=X=-y+10. Therefore, the answer is x=-y+10

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Which of the following correctly uses absolute value to show the distance between -80 and 15? (5 points)

netineya [11]

Answer:

it's answer is b

|-80+15| = |-65| = 65 units

Hope this helps :)

5 0

3 years ago

Read 2 more answers

Which angles below are complementary?

tigry1 [53]

9514 1404 393

Answer:

∠KXN and ∠QXT

Step-by-step explanation:

The measure of each angle is the difference of the scale values that the rays intercept. (The same scale needs to be used for each ray.) Of course, complementary angles have a sum of 90°.

Here, we'll refer to angle aXb as "ab".

NP = 95 -35 = 60

PQ = 35 -20 = 15 . . . not complementary to NP

KN = 165 -95 = 70

QT = 20 -0 = 20 . . . complementary to KN ⇒ your answer

JK = 180 -165 = 15

PQ = 35 -20 = 15 . . . not complementary to JK

JK = 15

NK = KN = 70 . . . not complementary to JK

6 0

3 years ago

Bethany is 62 in tall Bethany's brother scott is 3 inches taller than her she is write Scott's height in feet and inches

MrRissso [65]

Scott is 62+3=65 inches tall, and there are 12 inches in a foot, so you get 65/12=about 5 whole feet. 5 whole feet is 60 inches in total, so we have a leftover 5 inches, meaning that Scott is 5 feet and 5 inches tall.

8 0

3 years ago

In figure AB and CD bisect each other at O. State the 3 pairs of equal parts in ∆AOC and ∆BOD. Is ∆AOC ≅ ∆BOD? Give reasons

GuDViN [60]

Answer:

SEE EXPLANATION

Step-by-step explanation:

$In\:\triangle AOC \:\&\:\triangle BOD$

$AO \cong OB.... (given)$

$\angle AOC \cong\angle BOD$

$(vertical \: \angle s)$

$CO \cong OD.... (given)$

$\therefore \triangle AOC \:\cong\:\triangle BOD$

$(SAS \: postulate)$

So, the 3 pairs of equal parts in ∆AOC and ∆BOD are:

$AC = BD$

$m\angle OAC = m\angle OBD$

$m\angle OCA = m\angle ODB$

7 0

3 years ago

Expand (2x+2)^6<br> How would you find the answer using the binomial theorem?

Yanka [14]

Answer:

Step-by-step explanation:

$\displaystyle\\\sum\limits _{k=0}^n\frac{n!}{k!*(n-k)!}a^{n-k}b^k .\\\\k=0\\\frac{n!}{0!*(n-0)!}a^{n-0}b^0=C_n^0a^n*1=C_n^0a^n.\\\\ k=1\\\frac{n!}{1!*(n-1)!} a^{n-1}b^1=C_n^1a^{n-1}b^1.\\\\k=2\\\frac{n!}{2!*(n-2)!} a^{n-2}b^2=C_n^2a^{n-2}b^2.\\\\k=n\\\frac{n!}{n!*(n-n)!} a^{n-n}b^n=C_n^na^0b^n=C_n^nb^n.\\\\C_n^0a^n+C_n^1a^{n-1}b^1+C_n^2a^{n-2}b^2+...+C_n^nb^n=(a+b)^n.$

$\displaystyle\\(2x+2)^6=\frac{6!}{(6-0)!*0!} (2x)^62^0+\frac{6!}{(6-1)!*1!} (2x)^{6-1}2^1+\frac{6!}{(6-2)!*2!}(2x)^{6-2}2^2+\\\\ +\frac{6!}{(6-3)!*3!} (2a)^{6-3}2^3+\frac{6!}{(6-4)*4!} (2x)^{6-4}b^4+\frac{6!}{(6-5)!*5!}(2x)^{6-5} b^5+\frac{6!}{(6-6)!*6!}(2x)^{6-6}b^6. \\\\$

$(2x+2)^6=\frac{6!}{6!*1} 2^6*x^6*1+\frac{5!*6}{5!*1}2^5*x^5*2+\\\\+\frac{4!*5*6}{4!*1*2}2^4*x^4*2^2+ \frac{3!*4*5*6}{3!*1*2*3} 2^3*x^3*2^3+\frac{4!*5*6}{2!*4!}2^2*x^2*2^4+\\\\+\frac{5!*6}{1!*5!} 2^1*x^1*2^5+\frac{6!}{0!*6!} x^02^6\\\\(2x+2)^6=64x^6+384x^5+960x^4+1280x^3+960x^2+384x+64.$

8 0

2 years ago