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Iteru [2.4K]
3 years ago
14

Whats the answer 6+1

Mathematics
2 answers:
RSB [31]3 years ago
7 0

7
My. Homie is 7
I know this because I learned to add when Dr.Seuss taught me
Sergeeva-Olga [200]3 years ago
6 0

Answer:

7

Step-by-step explanation:

Add 1 to six and u get seven.

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I need help on this plz
Fantom [35]
I’m timed on this I need help as well
4 0
3 years ago
A circle passes through points A(7,4), B(10,6), C(12,3). Show that AC must be the diameter of the circle.
Artist 52 [7]

so we have three points, A, B and C, if indeed AC is the diameter of the circle, then half the distance of AC is its radius, and the midpoint of AC is the center of the circle, morever, since B is also on the circle, the distance from B to the center must be the same radius distance.

in short, half the distance of AC must be equals to the distance of B to the midpoint of AC, if indeed AC is the diameter.

\bf ~~~~~~~~~~~~\textit{middle point of 2 points } \\\\ A(\stackrel{x_1}{7}~,~\stackrel{y_1}{4})\qquad C(\stackrel{x_2}{12}~,~\stackrel{y_2}{3}) \qquad \left(\cfrac{ x_2 + x_1}{2}~~~ ,~~~ \cfrac{ y_2 + y_1}{2} \right) \\\\\\ \left( \cfrac{12+7}{2}~~,~~\cfrac{3+4}{2} \right)\implies \left( \cfrac{19}{2}~~,~~\cfrac{7}{2} \right)=M\impliedby \textit{center of the circle}

now, let's check the distance from say A to the center, and check the distance of B to the center, if it's indeed the center, they'll be the same and thus AC its diameter.

\bf ~~~~~~~~~~~~\textit{distance between 2 points} \\\\ A(\stackrel{x_1}{7}~,~\stackrel{y_1}{4})\qquad M(\stackrel{x_2}{\frac{19}{2}}~,~\stackrel{y_2}{\frac{7}{2}})\qquad \qquad d = \sqrt{( x_2- x_1)^2 + ( y_2- y_1)^2} \\\\\\ AM=\sqrt{\left( \frac{19}{2}-7 \right)^2+\left( \frac{7}{2}-4 \right)^2} \\\\\\ AM=\sqrt{\left( \frac{5}{2}\right)^2+\left( -\frac{1}{2} \right)^2}\implies \boxed{AM\approx 2.549509756796392} \\\\[-0.35em] ~\dotfill

\bf ~~~~~~~~~~~~\textit{distance between 2 points} \\\\ B(\stackrel{x_1}{10}~,~\stackrel{y_1}{6})\qquad M(\stackrel{x_2}{\frac{19}{2}}~,~\stackrel{y_2}{\frac{7}{2}}) \\\\\\ BM=\sqrt{\left( \frac{19}{2}-10 \right)^2+\left( \frac{7}{2}-6 \right)^2} \\\\\\ BM=\sqrt{\left( -\frac{1}{2}\right)^2+\left( -\frac{5}{2} \right)^2}\implies \boxed{BM\approx 2.549509756796392}

6 0
3 years ago
Need help not sure if my question is true or false
Alika [10]
The answer is false. The radius is half the diameter, not 2 times it.
7 0
3 years ago
A class of 33 students elected a class treasurer. There were 11 votes for Candidate A and 15 votes for Candidate B. The remainin
vovangra [49]
To start out, notice that you want the percent of voters that chose candidate A, not the percent of the class that chose candidate A.

Your fraction should be "number that chose candidate A" out of "number of voters," which is the same thing as saying:  
"number that chose candidate A" divided by "number of voters"

1) The numerator of the fraction should be the number of votes for candidate A, which is 11.

2) The denominator of the fraction should be the number of voters. You're told that "t<span>here were 11 votes for Candidate A and 15 votes for Candidate B," so there are:
</span>11 + 15 \: total \: votes

3) Finally put parts 1 and 2 together into a fraction and multiply by 100 to get your percent. That is your final answer:
\frac{11}{11+15}  \times 100

----

Answer: Top right choice, \frac{11}{11+15} \times 100
<span>

</span>
5 0
3 years ago
What is the value of 2x^-2y^-2 for x = 3 and y = 2? a) 1/18 b) 72 c) 2(-6)^-4 d) 1/648
sladkih [1.3K]
2x^-2y^-2 = 2/x^2y^2 = 2/(2)^2 x (3)2 = 2/4 x 9 = 2/36 = 1/18
8 0
3 years ago
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