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3 years ago

Jack is 4 times as old as Lacy. 9 years from now the sum of their ages will be 63. How old are they now?

Mathematics

2 answers:

11Alexandr11 [23.1K]3 years ago

7 0

Let Jack be J and Lacy be L.

First equation: J=4L

Second equation: (J+9)+(L+9)=63

Solving using substitution:

4L+9+L+9=63

5L+18=63

5L= 45

L = 9

L=9 and J=4L = 4*9 = 36

So Jack is 36 years old and Lacy is 9 years old.

Contact [7]3 years ago

5 0

Answer:

jack is 54 and lacey is 13.5. I think

Step-by-step explanation:

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At Janine party the orchestra has 10 guitar players and 8 flute players. At brown party the orchestra has 12 guitar players and

jenyasd209 [6]

The brown Party has the higher ratio of guitar to flute players which is 12:7 compared to 10:8( Jaine)

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3 years ago

6x=12 with explanation plz

lord [1]

Answer:

X = 2

Step-by-step explanation:

6(2) = 12

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2 years ago

Bethany paid $408.75 for a pair of diamond stud earrings. The sale price was $375. What is the Percent of Sales Tax she paid for

9966 [12]

Answer:

She paid 9% of sales tax

Step-by-step explanation:

The sale price of diamond = $375

Let x be percent of sales tax

So, amount of tax = $x\% \times 375 = \frac{x}{100} \times 375 =3.75x$

So, Amount including tax = 375+3.75x

We are given that Bethany paid $408.75 for a pair of diamond stud earrings.

So, 375+3.75x=408.75

3.75x=408.75-375

$x=\frac{408.75-375}{3.75}$

x=9

Hence She paid 9% of sales tax

5 0

3 years ago

2y= -3x+2 find three points

skelet666 [1.2K]

y=-3x+2/2

if y=0,x=2/3

if x=0,y=1

7 0

3 years ago

Read 2 more answers

Find the absolute maximum and absolute minimum values of the function f(x, y) = x 2 + y 2 − x 2 y + 7 on the set d = {(x, y) : |

dsp73

Looks like $f(x,y)=x^2+y^2-x^2y+7$ .

$f_x=2x-2xy=0\implies2x(1-y)=0\implies x=0\text{ or }y=1$

$f_y=2y-x^2=0\implies2y=x^2$

If $x=0$ , then $y=0$ - critical point at (0, 0).
If $y=1$ , then $x=\pm\sqrt2$ - two critical points at $(-\sqrt2,1)$ and $(\sqrt2,1)$

The latter two critical points occur outside of $D$ since $|\pm\sqrt2|>1$ so we ignore those points.

The Hessian matrix for this function is

$H(x,y)=\begin{bmatrix}f_{xx}&f_{xy}\\f_{yx}&f_{yy}\end{bmatrix}=\begin{bmatrix}2-2y&-2x\\-2x&2\end{bmatrix}$

The value of its determinant at (0, 0) is $\det H(0,0)=4>0$ , which means a minimum occurs at the point, and we have $f(0,0)=7$ .

Now consider each boundary:

If $x=1$ , then

$f(1,y)=8-y+y^2=\left(y-\dfrac12\right)^2+\dfrac{31}4$

which has 3 extreme values over the interval $-1\le y\le1$ of 31/4 = 7.75 at the point (1, 1/2); 8 at (1, 1); and 10 at (1, -1).

If $x=-1$ , then

$f(-1,y)=8-y+y^2$

and we get the same extrema as in the previous case: 8 at (-1, 1), and 10 at (-1, -1).

If $y=1$ , then

$f(x,1)=8$

which doesn't tell us about anything we don't already know (namely that 8 is an extreme value).

If $y=-1$ , then

$f(x,-1)=2x^2+8$

which has 3 extreme values, but the previous cases already include them.

Hence $f(x,y)$ has absolute maxima of 10 at the points (1, -1) and (-1, -1) and an absolute minimum of 0 at (0, 0).

3 0

3 years ago