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1 year ago

Find the volumeof eachfigure, round to th nearest hundredth I'd necessary

Mathematics

1 answer:

ad-work [718]1 year ago

6 0

Determine the base of right triangle by using pythagoras theorem.

$\begin{gathered} b=\sqrt[]{(15.7)^2-(8.5)^2} \\ =\sqrt[]{174.24} \\ =13.2 \end{gathered}$

Determine the area of base of figure.

$\begin{gathered} A=\frac{1}{2}\cdot13.2\cdot8.5 \\ =56.1 \end{gathered}$

Determine the volume of the figure.

$\begin{gathered} V=A\cdot6.2 \\ =56.1\cdot6.2 \\ =347.82 \end{gathered}$

So volume of the figure is 347.82 yards square.

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What is the total surface area

Stels [109]

We have two right triangles and three different rectangles.

The formula of an area of a right triangle:

$A_T=\dfrac{1}{2}l_1l_2$

l₁, l₂ - legs

We have l₁ = 20cm and l₂ = 21cm. Substitute:

$A_T=\dfrac{1}{2}(20)(21)=(10)(21)=210\ cm^2$

The formula of an area of a rectangle:

$A_R=lw$

l - length

w - width

We have:

rectangle #1: l = 22cm, w = 29cm

$A_{R1}=(22)(29)=638\ cm^2$

rectangle #2: l = 22cm, w = 21cm

$A_{R2}=(22)(21)=462\ cm^2$

rectangle #3: l = 22cm, w = 20cm

$A_{R3}=(22)(20)=440\ cm^2$

The total Surface Area of the triangular prism:

$S.A.=2A_T+A_{R1}+A_{R2}+A_{R3}\\\\S.A.=2\cdot210+638+462+440=1960\ cm^2$

3 0

3 years ago

Match the parabolas represented by the equations with their vertices. y = x2 + 6x + 8 y = 2x2 + 16x + 28 y = -x2 + 5x + 14 y = -

GaryK [48]

Consider all parabolas:

$y = x^2 + 6x + 8,\\y=x^2+6x+9-9+8,\\y=(x^2+6x+9)-1,\\y=(x+3)^2-1.$

When x=-3, y=-1, then the point (-3,-1) is vertex of this first parabola.

$y = 2x^2 + 16x + 28=2(x^2+8x+14),\\y=2(x^2+8x+16-16+14),\\y=2((x^2+8x+16)-16+14),\\y=2((x+4)^2-2)=2(x+4)^2-4.$

When x=-4, y=-4, then the point (-4,-4) is vertex of this second parabola.

$y =-x^2 + 5x + 14=-(x^2-5x-14),\\y=-(x^2-5x+\dfrac{25}{4}-\dfrac{25}{4}-14),\\y=-((x^2-5x+\dfrac{25}{4})-\dfrac{25}{4}-14),\\y=-((x-\dfrac{5}{2})^2-\dfrac{81}{4})=-(x-\dfrac{5}{2})^2+\dfrac{81}{4}.$

When x=2.5, y=20.25, then the point (2.5,20.25) is vertex of this third parabola.

$y =-x^2 + 7x + 7=-(x^2-7x-7),\\y=-(x^2-7x+\dfrac{49}{4}-\dfrac{49}{4}-7),\\y=-((x^2-7x+\dfrac{49}{4})-\dfrac{49}{4}-7),\\y=-((x-\dfrac{7}{2})^2-\dfrac{77}{4})=-(x-\dfrac{7}{2})^2+\dfrac{77}{4}.$

When x=3.5, y=19.25, then the point (3.5,19.25) is vertex of this fourth parabola.

$y =2x^2 + 7x +5=2(x^2+\dfrac{7}{2}x+\dfrac{5}{2}),\\y=2(x^2+\dfrac{7}{2}x+\dfrac{49}{16}-\dfrac{49}{16}+\dfrac{5}{2}),\\y=2((x^2+\dfrac{7}{2}x+\dfrac{49}{16})-\dfrac{49}{16}+\dfrac{5}{2}),\\y=2((x+\dfrac{7}{4})^2-\dfrac{9}{16})=2(x+\dfrac{7}{4})^2-\dfrac{9}{8}.$

When x=-1.75, y=-1.125, then the point (-1.75,-1.125) is vertex of this fifth parabola.

$y =-2x^2 + 8x +5=-2(x^2-4x-\dfrac{5}{2}),\\y=-2(x^2-4x+4-4-\dfrac{5}{2}),\\y=-2((x^2-4x+4)-4-\dfrac{5}{2}),\\y=-2((x-2)^2-\dfrac{13}{2})=-2(x-2)^2+13.$