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mrs_skeptik [129]
3 years ago
14

Plz help will be thanked and will pick brainly

Mathematics
1 answer:
arsen [322]3 years ago
6 0

Answer:

275 units^2

Step-by-step explanation:

The formula for the area of trapezoid is:

Area=((b1+b2)/2)*h

In the given trapezoid, as it can be seen

b1=16

h=11

The lower base will be calculated using all the lengths in the lower base

b2=9+16+9

b=34

Putting the values in formula

Area=((16+34)/2)*11

Area=(50/2)*11

Area=25*11

=275 units^2

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Solve the equation for y <br> 1/2y - 21/4 = -35/4
mina [271]

Answer:

1/2y - 21/4 = -35/4

1/2y  = 21/4  -35/4

1/2y  =   - 7/4

y =   - \frac{7}{2}

5 0
3 years ago
The sentence 3 + (5 + 2) = (5 + 2) + 3 illustrates
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The associative property
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What does ‘b” represent in y=mx+b
tigry1 [53]

Answer:

In the equation y = mx + b for a straight line, the. number b is called the y-intercept of the line.

Step-by-step explanation:

5 0
3 years ago
10 2/6 - 7 5/6 i know its prob easy but i need help
miv72 [106K]

Answer: 2 1/3

Explanation:

10 2/6 - 7 5/6

You can make both a fraction by multiplying the denominator by the whole number, and then adding the numerator to that number, and keeping the denominator the same. So, 10*6 = 60 and 60 + 2 = 62 and you keep the denominator as 6, which would make 62/6

7*6 = 42 and 42 + 5 = 47 so 7 5/6 becomes 47/6

10 2/6 is equivalent to 62/6

7 5/6 is equivalent to 47/6

This just makes it easier to look at.

Now you just work through the equation.

62/6 - 47/6 = 15/6

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3 0
3 years ago
Write the equation -4x^2+9y^2+32x+36y-64=0 in standard form. Please show me each step of the process!
IgorC [24]
Hey there, hope I can help!

-4x^2+9y^2+32x+36y-64=0

\mathrm{Add\:}64\mathrm{\:to\:both\:sides} \ \textgreater \  9y^2+32x+36y-4x^2=64

\mathrm{Factor\:out\:coefficient\:of\:square\:terms} \ \textgreater \  -4\left(x^2-8x\right)+9\left(y^2+4y\right)=64

\mathrm{Divide\:by\:coefficient\:of\:square\:terms:\:}4
-\left(x^2-8x\right)+\frac{9}{4}\left(y^2+4y\right)=16

\mathrm{Divide\:by\:coefficient\:of\:square\:terms:\:}9
-\frac{1}{9}\left(x^2-8x\right)+\frac{1}{4}\left(y^2+4y\right)=\frac{16}{9}

\mathrm{Convert}\:x\:\mathrm{to\:square\:form}
-\frac{1}{9}\left(x^2-8x+16\right)+\frac{1}{4}\left(y^2+4y\right)=\frac{16}{9}-\frac{1}{9}\left(16\right)

\mathrm{Convert\:to\:square\:form}
-\frac{1}{9}\left(x-4\right)^2+\frac{1}{4}\left(y^2+4y\right)=\frac{16}{9}-\frac{1}{9}\left(16\right)

\mathrm{Convert}\:y\:\mathrm{to\:square\:form}
-\frac{1}{9}\left(x-4\right)^2+\frac{1}{4}\left(y^2+4y+4\right)=\frac{16}{9}-\frac{1}{9}\left(16\right)+\frac{1}{4}\left(4\right)

\mathrm{Convert\:to\:square\:form}
-\frac{1}{9}\left(x-4\right)^2+\frac{1}{4}\left(y+2\right)^2=\frac{16}{9}-\frac{1}{9}\left(16\right)+\frac{1}{4}\left(4\right)

\mathrm{Refine\:}\frac{16}{9}-\frac{1}{9}\left(16\right)+\frac{1}{4}\left(4\right) \ \textgreater \  -\frac{1}{9}\left(x-4\right)^2+\frac{1}{4}\left(y+2\right)^2=1

Refine\;once\;more\;-\frac{\left(x-4\right)^2}{9}+\frac{\left(y+2\right)^2}{4}=1

For me I used
\frac{\left(y-k\right)^2}{a^2}-\frac{\left(x-h\right)^2}{b^2}= 1
As\;\mathrm{it\;\:is\:the\:standard\:equation\:for\:an\:up-down\:facing\:hyperbola}

I know yours is an equation which is why I did not go any further because this is the standard form you are looking for. I would rewrite mine to get my hyperbola standard form. However the one I have provided is the form you need where mine would be.
\frac{\left(y-\left(-2\right)\right)^2}{2^2}-\frac{\left(x-4\right)^2}{3^2}=1

Hope this helps!
4 0
3 years ago
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