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3 years ago

7

what is the surface area of a rectangular prisim with a base length of 9 yd, base width of 7 yd,and a height of 4 yd?

2 answers:

12345 [234]3 years ago

7 0

Answer:254 yds

Step-by-step explanation:

krek1111 [17]3 years ago

3 0

Answer:

254 yds²

Step-by-step explanation:

There are 6 faces of the prism we need to calculate the area of for a rectangular prism.

The base and top of the prism measure 9x7 yards each, so there are two faces with an area of 63 yds² (9 x 7 = 63)

The sides of the prism measure 4x7 yards each, so there are two faces with an area of 28 yds² (4 x 7 = 28)

The front and back face of the prism measure 9x4 yards each, so there are two faces with an area of 36 yds² (9 x 4 = 36)

The total surface area is

2(63) + 2(28) + 2(36) = 254 yds²

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Step-by-step explanation:

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23+4y=63 what does y equal?

Sladkaya [172]

Answer:

y=10

Step-by-step explanation:

1. Make sure that the variable is left alone on the other side of the equal sign

2. remove all the no. on the variable side

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3 years ago

Find an equation of the tangent to the curve at the given point by both eliminating the parameter and without eliminating the pa

konstantin123 [22]

Answer:

$y=2x-8$

Step-by-step explanation:

The given parametric equation is;

$x=6+ln(t),y=t^2+3$

<h3><u>BY ELIMINATING THE PARAMETER</u></h3>

To eliminate the parameter we make $t$ the subject in one equation and put it inside the other.

We make $t$ the subject in $x=6+ln(t)$ because it is easier.

$\Rightarrow x-6=ln(t)$

$\Rightarrow {e}^{x-6}=e^{ln(t)}$

$\Rightarrow {e}^{x-6}=t$

Or

$t={e}^{x-6}$

We now substitute this into $y=t^2+3$ .

This gives us;

$y=(e^{x-6})^2+3$ .

$\Rightarrow y=e^{2(x-6)}+3$ .

We have now eliminated the parameter.

The equation of the tangent at (6,4) is given by;

$y-y_1=m(x-x_1)$

where the gradient function is given by;

$\frac{dy}{dx}=2e^{2(x-6)}$

We substitute $x=6$ into the gradient function to obtain the gradient.

$\Rightarrow m=2e^{2(6-6)}$

$\Rightarrow m=2e^0$

$\Rightarrow m=2$

The equation of the tangent becomes

$y-4=2(x-6)$

We simplify to obtain

$y=2x-12+4$

$y=2x-8$

<h3><u>WITHOUT ELIMINATING THE PARAMETER</u></h3>

The given parametric equation is;

$x=6+ln(t),y=t^2+3$

For $x=6+ln(t)$

$\frac{dx}{dt}=\frac{1}{t}$

For $y=t^2+3$

$\frac{dy}{dt}=2t$

The slope is given by;

$\frac{dy}{dx}=\frac{\frac{dy}{dt} }{\frac{dx}{dt} }$

$\frac{dy}{dx}=\frac{2t }{\frac{1}{t} }$

$\frac{dy}{dx}=2t^2$

At the point, (6,4), we plug in any of the values into the parametric equation and find the corresponding value for $t$ .

Notice that

When $x=6$ , $6=6+\ln(t)$

$6-6=\ln(t)$

$0=\ln(t)$

$e^0=e^\ln(t)$

$1=t$

when $y=4$ , $4=t^2+3$

$4-3=t^2$

$1=t^2$

$t=\pm1$

But the slope is the same when we plug in any of these values for t.

$\frac{dy}{dx}=2(\pm1)^2=2$

The equation of the tangent becomes

$y-4=2(x-6)$

We simplify to obtain

$y=2x-12+4$

$y=2x-8$

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Answer:

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