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miv72 [106K]
3 years ago
5

What is 10 times as much as 700?

Mathematics
2 answers:
Stels [109]3 years ago
8 0
7000 because 10 groups of 700 equals 7000
Hunter-Best [27]3 years ago
3 0
7,000 is your answer
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It is not a function because both (2, 2) and (2,−2) have the same x-coordinate. It is a function because both (2, 2) and (2,−2)
olya-2409 [2.1K]

Answer:

It is not a function because both (2,2) and (2,-2) have the same x-coordinate.

Step-by-step explanation:

To be a function for the same x value there should NOT be two different y values.

So, in (2, 2) and (2,-2)

For the same x value 2, there are two different y values 2, -2.

So, this is not a function.

3 0
3 years ago
Can someone please help me
salantis [7]

Using Pythagorean theorem

\\ \sf\longmapsto P^2=H^2-B^2

\\ \sf\longmapsto P^2=6^2-5^2

\\ \sf\longmapsto P^2=36-25

\\ \sf\longmapsto P^2=11

\\ \sf\longmapsto P=3.2

4 0
2 years ago
Read 2 more answers
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Triss [41]

Answer:

Area = 152ft²

Step-by-step explanation:

Simpifly:

A = 13(13-6) + 9(13-6)

A = 13(7) + 9(7)

A = 22(7)

A = 22(7)

A = 20(7)+2(7)

A = 140 + 14

A = 152

5 0
2 years ago
There are three power plants [X, Y, Z] that at any given time each one either generates electricity or idles. Event A is that pl
insens350 [35]

We're told that

P(A\cap B)=0.15

P(A\cup B)^C=0.06\implies P(A\cup B)=0.94

P(B\mid A)=P(B^C\mid A)=0.5

where the last fact is due to the law of total probability:

P(A)=P(A\cap B)+P(A\cap B^C)

\implies P(A)=P(B\mid A)P(A)+P(B^C\mid A)P(A)

\implies 1=P(B\mid A)+P(B^C\mid A)

so that B\mid A and B^C\mid A are complementary.

By definition of conditional probability, we have

P(B\mid A)=P(B^C\mid A)

\implies\dfrac{P(A\cap B)}{P(A)}=\dfrac{P(A\cap B^C)}{P(A)}

\implies P(A\cap B)=P(A\cap B^C)

We make use of the addition rule and complementary probabilities to rewrite this as

P(A\cap B)=P(A\cap B^C)

\implies P(A)+P(B)-P(A\cup B)=P(A)+P(B^C)-P(A\cup B^C)

\implies P(B)-[1-P(A\cup B)^C]=[1-P(B)]-P(A\cup B^C)

\implies2P(B)=2-[P(A\cup B)^C+P(A\cup B^C)]

\implies2P(B)=[1-P(A\cup B)^C]+[1-P(A\cup B^C)]

\implies2P(B)=P(A\cup B)+P(A\cup B^C)^C

\implies2P(B)=P(A\cup B)+P(A^C\cap B)\quad(*)

By the law of total probability,

P(B)=P(A\cap B)+P(A^C\cap B)

\implies P(A^C\cap B)=P(B)-P(A\cap B)

and substituting this into (*) gives

2P(B)=P(A\cup B)+[P(B)-P(A\cap B)]

\implies P(B)=P(A\cup B)-P(A\cap B)

\implies P(B)=0.94-0.15=\boxed{0.79}

8 0
3 years ago
How can I find the exact distance between the points(√7,-√2)and (4√7,5√2)
chubhunter [2.5K]
\bf \textit{distance between 2 points}\\ \quad \\
\begin{array}{lllll}
&x_1&y_1&x_2&y_2\\
%  (a,b)
&({{ \sqrt{7}}}\quad ,&{{ -\sqrt{2}}})\quad 
%  (c,d)
&({{4\sqrt{7}}}\quad ,&{{ 5\sqrt{2}}})
\end{array}~~~
%  distance value
d = \sqrt{({{ x_2}}-{{ x_1}})^2 + ({{ y_2}}-{{ y_1}})^2}
\\\\\\


\bf d=\sqrt{[4\sqrt{7}-\sqrt{7}]^2~+~[5\sqrt{2}-(-\sqrt{2})]^2}
\\\\\\
d=\sqrt{(4\sqrt{7}-\sqrt{7})^2~+~(5\sqrt{2}+\sqrt{2})^2}\implies d=\sqrt{(3\sqrt{7})^2~+~(6\sqrt{2})^2}
\\\\\\
d=\sqrt{3^2\cdot 7~~+~~6^2\cdot 2}\implies d=\sqrt{63+72}\implies d=\sqrt{135}
\\\\\\
\begin{cases}
135=5\cdot 3\cdot 3\cdot 3\\
\qquad 5\cdot 3^2\cdot 3\\
\qquad 15\cdot 3^2
\end{cases}\implies d=\sqrt{15\cdot 3^2}\implies d=3\sqrt{15}
8 0
3 years ago
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