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natta225 [31]
2 years ago
6

How do I solve number 15

Mathematics
1 answer:
Rudik [331]2 years ago
6 0
If the rectangle's area is X
then :
x+20=44
  -20   -20
  x=24
the rectangle's area is 24 square inches.
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Estimate the area of Alberta in square miles. Show your reasoning
ella [17]

Answer:  About 278,250\ mi^2

Step-by-step explanation:

The missing figure is attached.

Notice in the first picture that Alberta has a complex shape.

You can calculate the area of a complex shape by decomposing it into polygons whose areas can be calculated easily.

Observe the second picture. Notice that it can be descompose into two polygons: A trapezoid and a rectangle.

The area of the trapezoid  can be calcualted with the formula:

A_t=\frac{h}{2}(B+b)

Where "h" is the height, "B" is the long base and "b" is short base.

And the area of the rectangle can be found with the formula:

 A_r=lw

Wkere "l" is the lenght and "w" is the width.

Then, the apprximate area of Alberta is:

A=\frac{h}{2}(B+b)+lw

Substituting vallues, you get:

A=(\frac{(410\ mi-180\ mi)}{2})(760\ mi+470\ mi)+(180\ mi)(760\ im)\\\\A=141,450\ mi^2+136,800\ mi^2\\\\A=278,250\ mi^2

Therefore, the area of of Alberta is about 278,250\ mi^2.

5 0
3 years ago
Please help it’s geometry! I’ll make brainliest
expeople1 [14]
GX is the angle bisector. Answer is C
4 0
2 years ago
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Which line has a slope of 1 and a y-intercept of 2?
Vedmedyk [2.9K]

Answer:

bottom left :)

Step-by-step explanation:

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3 years ago
A coordinate plane with a line passing through (negative 3, 2) and (0, 3)
Arturiano [62]

Let's check

\\ \sf\longmapsto Slope=m=\dfrac{3-2}{0+3}=\dfrac{1}{3}

  • Y intercept is 3✓

Now

let f(x) be y

\\ \sf\longmapsto y=mx+b

\\ \sf\longmapsto y=\dfrac{1}{3}x+3=f(x)\checkmark

5 0
2 years ago
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Divide. (18x^3 + 12x^2 - 3x) ÷ 6x^2
nlexa [21]

\bold{[ \ Answer \ ]}

\boxed{\bold{\frac{x^3\left(6x^2+4x-1\right)}{2}}}

\bold{[ \ Explanation \ ]}

  • \bold{Divide: \ \left(18x^3\:+\:12x^2\:-\:3x\right)\:\div \:6x^2}

\bold{-------------------}

  • \bold{Rewrite}

\bold{18x^3+12x^2-3x \ = \ x^2\frac{x\left(6x^2+4x-1\right)}{2}}

  • \bold{Rewrite}

\bold{x^2\frac{x\left(6x^2+4x-1\right)}{2}}

  • \bold{Multiply \ Fractions \ (a\cdot \frac{b}{c}=\frac{a\:\cdot \:b}{c})}

\bold{\frac{x\left(6x^2+4x-1\right)x^2}{2}}

  • \bold{Rewrite}

\bold{x\left(6x^2+4x-1\right)x^2 \ = \ x^3\left(6x^2+4x-1\right)}

  • \bold{Simplify}

\bold{\frac{x^3\left(6x^2+4x-1\right)}{2}}

\boxed{\bold{[] \ Eclipsed \ []}}

3 0
3 years ago
Read 2 more answers
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