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kvasek [131]
4 years ago
9

Someone help me please thank you

Mathematics
2 answers:
Oksi-84 [34.3K]4 years ago
7 0
6/7 is the answer I hope it helps
vfiekz [6]4 years ago
4 0
1/1x6/7 =6/7yd is the answer
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PLEASE HELP QUICK
nydimaria [60]

we are given with a triangle with the sides 6 ft, 21 ft, 23 ft. although it follows the rule of triangles that state that no side of the triangle exceeds the sum of the lengths of two other sides, this does not follow Pythagorean theorem: 23^2-21^2 = 88 which is not equal to 6^2 or 36. This is not a right triangle, hence. Hope this helps!

6 0
3 years ago
Here is a test to how good your math is,
Marina CMI [18]

Answer:

Zero

Step-by-step explanation:

The distance AB is ...

... √((5-1)²+(5-2)²) = √(16+9) = 5

The largest right triangle that can be constructed with AB as the hypotenuse is one with an altitude of 5/2 = 2.5 units. Its area will be ...

... (1/2)·5·2.5 = 6.25 . . . . square units

It is not possible to construct the triangle ABC described.

_____

In order to achieve the given area, ∠C would need to be 87.75° or smaller. It could not be 90°.

7 0
3 years ago
Read 2 more answers
Bill currently has $45 in savings. He had been saving $9 each week. yesterday he spent $126 of the savings. for how many weeks h
serious [3.7K]

Answer: 9 weeks

Step-by-step explanation: Subtract 126(final amount) by 45(starting amount) and you will get 81(amount of money made of # of weeks). Divide 81 by 9(money made per week) and you will get 9. (# of weeks)

8 0
3 years ago
Figure a’b’c’d’f is a dilation or figure abcdf by a scale factor of 1/2. The dilation is centered at (-4,-1)
AfilCa [17]
B is true.

A dilation creates similar figures.  By definition, the proportion of the sides of similar figures is the same.  AB/A'B' would be the same as BC/B'C', since the numerator of both comes from the first figure and the denominator of both comes from the second figure.
8 0
4 years ago
Part 2: NO LINKS!! NOT MULTIPLE CHOICE! Please help me​
Dmitry_Shevchenko [17]

Answer:

(see attachment for tree diagram)

\sf Probability\:of\:an\:event\:occurring = \dfrac{Number\:of\:ways\:it\:can\:occur}{Total\:number\:of\:possible\:outcomes}

<h3><u>Part (a)</u></h3>

\textsf{P(Head) and P(3)}=\sf \dfrac{1}{2} \times \dfrac{1}{6}=\dfrac{1}{12}

<h3><u>Part (b)</u></h3>

\textsf{P(Tail) and P(even)}=\sf \dfrac{1}{2} \times \dfrac{3}{6}=\dfrac{3}{12}=\dfrac{1}{4}

<h3><u>Part (c)</u></h3>

\textsf{P(not 6)}=\sf 1-\textsf{P(6)}=1-\dfrac{1}{6}=\dfrac{5}{6}

\implies \textsf{P(Head) and P(not 6)}=\sf \dfrac{1}{2} \times \dfrac{5}{6}=\dfrac{5}{12}

<h3><u>Part (d)</u></h3>

As the six-sided die does not have a side labelled "7", the probability of rolling a 7 is zero.

\implies \textsf{P(Head) and P(7)}=\sf \dfrac{1}{2} \times 0=0

7 0
2 years ago
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