Ask question

Ask question

All categories

andreyandreev [35.5K]

3 years ago

11

Luc carved an ice sculpture of a dragon with wings that weighed 42 pounds when he moved it one of the wings broke of and now it

weighs 31 pounds 2 ounces how much did the broken wing weigh?

1 answer:

prisoha [69]3 years ago

5 0

The broken wing weighed 10 pounds and 14 ounces. There are 16 ounces in a pound. I hope this helps!

You might be interested in

Explain how you can compare the size of a product to the size of a factor when multiplying fractions without actually doing mult

MAXImum [283]

Answer: If you multiply a number by a fraction less than 1, the resulting product will be smaller than the original fraction.

Imagine, if you had half a pizza and you ate one-half of that. You would multiply 1/2 by 1/2 and get an even smaller number. The result would be 1/4.

On the other hand, if you multiply a number by a mixed number that is larger than one, the result would be larger than when you started.

5 0

3 years ago

Following the formula (-b+-√b^2-4ac) / 2a, working out -√b^2 - 4ac) / 2a

lina2011 [118]

Answer:

Jsjsjsnsnsksjssshjsiosjs

8 0

3 years ago

I NEED HELP <br><br> THERE ARE TWO OTHER QUESTIONS AS WELL

Temka [501]

Answer:

Please find attached the drawing of ABCD, A'B'C'D', and A''B''C''D'', created with MS Excel

Step-by-step explanation:

The given composition of transformation is $r_{y-axis} \ \circ \ r_{x = - 4}$

The transformation on the right, $r_{x = - 4}$ , is carried out first as follows;

The point <em>A</em> with coordinate (-7, 7) reflected across the line x = -4, gives;

A(-7, 7) $r_{x = - 4}$ <em> </em>→ A'(-1, 7)

Similarly, we have;

B(-5, 7) $r_{x = - 4}$ → B'(-3, 7)

C(-5, 1) $r_{x = - 4}$ → C'(-3, 1)

D(-7, 1) $r_{x = - 4}$ → D'(-1, 1)

Therefore, the image of the given rectangle after reflection across the lie, x = -4, is A'(-1, 7), B'(-3, 7), C'(-3, 1), and D'(-1, 1)

The second transformation, $r_{y-axis}$ , is found as follows;

(x, y) $r_{y-axis}$ → (x, -y)

Therefore, we have;

A'(-1, 7) $r_{y-axis}$ → A''(-1, -7)

B'(-3, 7) $r_{y-axis}$ → B''(-3, -7)

C'(-3, 1) $r_{y-axis}$ → C''(-3, -1)

D'(-1, 1) $r_{y-axis}$ → D''(-1, -1)

The points ABCD, A'B'C'D', and A''B''C''D'' are plotted on MS Excel and the points joined with straight lines to form the drawing of the triangles as shown in the attached figure.

5 0

3 years ago

Math will give medals

castortr0y [4]

20 and 10
This is because when two negative integers are multiplied, the product will be positive. Multiplication with another positive integer will produce a positive product.

3 0

3 years ago

Find the absolute maximum and minimum values of f on the set

natta225 [31]

$f(x,y)=9+xy-x-2y$
$\implies\begin{cases}f_x=y-1=0\implies y=1\\f_y=x-2=0\implies x=2\end{cases}$

which tells you that the only critical point of $f(x,y)$ occurs at (2, 1), which does lie within the region $D$ . At this point, we get $f(2,1)=7$ .

Next we check along the boundaries of $D$ . They are the lines $x=1$ with $0\le y\le4$ , $y=0$ with $1\le x\le5$ , and $y=5-x$ with $1\le x\le5$ .

If $x=1$ , then $f(1,y)=8-y$ , which is monotonically decreasing and must therefore attain its maximum at $y=0$ and minimum at $y=4$ . We get $f(1,0)=8$ and $f(1,4)=4$ .

If $y=0$ , then $f(x,0)=9-x$ , which is also monotonically decreasing and attains its maximum at $x=1$ and minimum at $x=5$ . We get $f(1,0)=8$ and $f(5,0)=4$ .

If $y=5-x$ , then $f(x,5-x)=g(x)=-x^2+6x-1$ . We have $g'(x)=-2x+6=0\implies x=3$ , which suggests an extremum occurs at (3, 2). We get $f(3,2)=8$ .

So $f(x,y)$ has a minimum value of 4 at (1, 4) and (5, 0), and a maximum value of 8 at (1, 0) and (3, 2).

5 0

3 years ago