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ELEN [110]
2 years ago
12

Decide whether the statement below is true or false.

Mathematics
2 answers:
GREYUIT [131]2 years ago
8 0

Answer:

true  rectangle have two sets of parallel sides and has four right angles.

Step-by-step explanation:

goldenfox [79]2 years ago
6 0
True
because it just.... is ig
You might be interested in
pablo made a shirt using 3/8 yards of red fabric and 1/4 yards of green fabric. how many more yards of red fabric did pablo use?
In-s [12.5K]
The answer is 1/8 more yards of red fabric
4 0
4 years ago
Read 2 more answers
Need help solving this problem:
enot [183]
$24,000 ---------> City or State Bank
$3,000 < City Bank < $11000
State Bank< $18,000
City Bank = .06 of interest
State Bank = .11 of interest

<u>Bold</u>= Answer

<u>State Bank - (18, 000)(.11)-----> $1980</u>
<u>                                                          } Total Interest = $2,430</u>
<u>City Bank - (6000)(.06)---------->$ 360 </u>
____________________________________________________________________
State Bank - (13000)(.11) ----------> $1,430
                                                             } Total Interest = $2,090
City Bank - (11000)(.06) -------------> $660
____________________________________________________________________
State Bank- (17000)(.11) ---------->$ 1870
                                                               } Total Interest = $2,290
City Bank- (7000)(.06) ---------------->$ 420

5 0
3 years ago
When Sabine set off to climb Mt. Marcy, she had 21 gummi bears in her
Dima020 [189]

You know how many she started and finished with.

To find the amount she ate, subtract the end amount from the start amount.

Subtract 6 from 21

She ate 15 gummy bears.

8 0
3 years ago
Please help me with this question thank you
gulaghasi [49]

Answer:

\mathrm{The\:solution\:is} :

x=\frac{-y^2-1}{2-y^2}\space\left\{y\ge \:0\right\}

Step-by-step explanation:

Given

y=\sqrt{\frac{2x+1}{x-1}}

Taking square of both sides

y^2=\left(\sqrt{\frac{2x+1}{x-1}}\right)\:^2

\mathrm{Subtract\:}\left(\sqrt{\frac{2x+1}{x-1}}\right)^2\mathrm{\:from\:both\:sides}

y^2-\left(\sqrt{\frac{2x+1}{x-1}}\right)^2=\left(\sqrt{\frac{2x+1}{x-1}}\right)^2-\left(\sqrt{\frac{2x+1}{x-1}}\right)^2

\mathrm{Simplify}

y^2-\left(\sqrt{\frac{2x+1}{x-1}}\right)^2=0

As we know that \mathrm{Apply\:Difference\:of\:Two\:Squares\:Formula:\:}x^2-y^2=\left(x+y\right)\left(x-y\right)

\mathrm{Factor\:}y^2-\left(\sqrt{\frac{2x+1}{x-1}}\right)^2:\quad \left(y+\sqrt{\frac{2x+1}{x-1}}\right)\left(y-\sqrt{\frac{2x+1}{x-1}}\right)

so

\left(y+\sqrt{\frac{2x+1}{x-1}}\right)\left(y-\sqrt{\frac{2x+1}{x-1}}\right)=0        

\mathrm{Using\:the\:Zero\:Factor\:Principle:\quad \:If}\:ab=0\:\mathrm{then}\:a=0\:\mathrm{or}\:b=0\:\left(\mathrm{or\:both}\:a=0\:\mathrm{and}\:b=0\right)

\mathrm{Solve\:}\:y+\sqrt{\frac{2x+1}{x-1}}=0

\mathrm{Subtract\:}y\mathrm{\:from\:both\:sides}

y+\sqrt{\frac{2x+1}{x-1}}-y=0-y

\sqrt{\frac{2x+1}{x-1}}=-y

\mathrm{Square\:both\:sides}

\left(\sqrt{\frac{2x+1}{x-1}}\right)^2=\left(-y\right)^2

\mathrm{Expand\:}\left(\sqrt{\frac{2x+1}{x-1}}\right)^2

\left(\sqrt{\frac{2x+1}{x-1}}\right)^2

\mathrm{Apply\:radical\:rule}:\quad \sqrt{a}=a^{\frac{1}{2}}

=\left(\left(\frac{2x+1}{x-1}\right)^{\frac{1}{2}}\right)^2

=\frac{2x+1}{x-1}

so equation  \left(\sqrt{\frac{2x+1}{x-1}}\right)^2=\left(-y\right)^2 becomes

\frac{2x+1}{x-1}=y^2

now

\mathrm{Solve\:}\:\frac{2x+1}{x-1}=y^2

\frac{2x+1}{x-1}=y^2

\mathrm{Multiply\:both\:sides\:by\:}x-1

\frac{2x+1}{x-1}\left(x-1\right)=y^2\left(x-1\right)

2x+1=y^2\left(x-1\right)

2x+1=xy^2-y^2         ∵  y^2\left(x-1\right):\quad xy^2-y^2

2x=xy^2-y^2-1

2x-xy^2=-y^2-1

x\left(2-y^2\right)=-y^2-1         ∵ \mathrm{Factor}\:2x-xy^2:\quad x\left(2-y^2\right)

\mathrm{Divide\:both\:sides\:by\:}2-y^2

\frac{x\left(2-y^2\right)}{2-y^2}=-\frac{y^2}{2-y^2}-\frac{1}{2-y^2}

x=\frac{-y^2-1}{2-y^2}

so

y+\sqrt{\frac{2x+1}{x-1}}=0:\quad x=\frac{-y^2-1}{2-y^2}\space\left\{y\le \:0\right\}

similarly

y-\sqrt{\frac{2x+1}{x-1}}=0:\quad x=\frac{-y^2-1}{2-y^2}\space\left\{y\ge \:0\right\}

\mathrm{Verify\:Solutions}:\quad x=\frac{-y^2-1}{2-y^2}

\mathrm{Check\:the\:solutions\:by\:plugging\:them\:into\:}y^2=\left(\sqrt{\frac{2x+1}{x-1}}\right)^2

\mathrm{Remove\:the\:ones\:that\:don't\:agree\:with\:the\:equation.}

\mathrm{Plug}\quad x=\frac{-y^2-1}{2-y^2}

y^2=\left(\sqrt{\frac{2\left(\frac{-y^2-1}{2-y^2}\right)+1}{\left(\frac{-y^2-1}{2-y^2}\right)-1}}\right)^2

\mathrm{Subtract\:}\left(\sqrt{\frac{2\frac{-y^2-1}{2-y^2}+1}{\frac{-y^2-1}{2-y^2}-1}}\right)^2\mathrm{\:from\:both\:sides}

y^2-\left(\sqrt{\frac{2\cdot \frac{-y^2-1}{2-y^2}+1}{\frac{-y^2-1}{2-y^2}-1}}\right)^2=\left(\sqrt{\frac{2\cdot \frac{-y^2-1}{2-y^2}+1}{\frac{-y^2-1}{2-y^2}-1}}\right)^2-\left(\sqrt{\frac{2\cdot \frac{-y^2-1}{2-y^2}+1}{\frac{-y^2-1}{2-y^2}-1}}\right)^2

y^2-\left(\sqrt{\frac{2\cdot \frac{-y^2-1}{2-y^2}+1}{\frac{-y^2-1}{2-y^2}-1}}\right)^2=0

\mathrm{Factor\:}y^2-\left(\sqrt{\frac{2\frac{-y^2-1}{2-y^2}+1}{\frac{-y^2-1}{2-y^2}-1}}\right)^2:\quad \left(y+\sqrt{\frac{2\cdot \frac{-y^2-1}{2-y^2}+1}{\frac{-y^2-1}{2-y^2}-1}}\right)\left(y-\sqrt{\frac{2\cdot \frac{-y^2-1}{2-y^2}+1}{\frac{-y^2-1}{2-y^2}-1}}\right)

so

\left(y+\sqrt{\frac{2\cdot \frac{-y^2-1}{2-y^2}+1}{\frac{-y^2-1}{2-y^2}-1}}\right)\left(y-\sqrt{\frac{2\cdot \frac{-y^2-1}{2-y^2}+1}{\frac{-y^2-1}{2-y^2}-1}}\right)=0

\mathrm{Using\:the\:Zero\:Factor\:Principle:\quad \:If}\:ab=0\:\mathrm{then}\:a=0\:\mathrm{or}\:b=0\:\left(\mathrm{or\:both}\:a=0\:\mathrm{and}\:b=0\right)

\mathrm{Solve\:}\:y+\sqrt{\frac{2\cdot \frac{-y^2-1}{2-y^2}+1}{\frac{-y^2-1}{2-y^2}-1}}=0:\quad y\le \:0

\mathrm{Solve\:}\:y-\sqrt{\frac{2\cdot \frac{-y^2-1}{2-y^2}+1}{\frac{-y^2-1}{2-y^2}-1}}=0:\quad y\ge \:0

\mathrm{True\:for\:all}\:y

Therefore,  \mathrm{The\:solution\:is} :

x=\frac{-y^2-1}{2-y^2}\space\left\{y\ge \:0\right\}

6 0
4 years ago
Solve.<br><br> -15 = -0.2k<br><br> k =
Olin [163]
-5 = 0.2k
divide both sides by 0.2 to isolate k
-25 = k

might be wrong tho lol
6 0
3 years ago
Read 2 more answers
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