The stock's value dropped from $38.50 to $34.00.What negative number shows the change in value?

The screen at Dallas' AT&T Stadium is 175 feet wide. A movie theatre screen with

Mazyrski [523]

Answer:

17150 ft²

Step-by-step explanation:

Given the ratio of sides = a : b , then

ratio of areas = a² : b²

Here ratio of sides = 175 : 40 = 35 : 8 , then

ratio of areas = 35² : 8² = 1225 : 64

let x be the area of AT & T Stadium then by proportion

$\frac{x}{1225}$ = $\frac{896}{64}$ ( cross- multiply )

64x = 896 × 1225 = 1097600 ( divide both sides by 64 )

x = 17150

The AT & T screen has an area of 17150 ft²

4 0

2 years ago

A chef is going to use a mixture of two brands of Italian dressing the first brand contains contains 7% vinegar and the second b

Cerrena [4.2K]

Answer:

176 of 12% vinegar

44 of 7% vinegar

Step-by-step explanation:

Percentage vinegar required = 11%

Volume = 220 mL

First brand % = 7%

Second brand % = 12%

Let p = volume of 12% vinegar material

Mixture :

0.12p + 0.07(220 - p) = 220 * 0.11

0.12p + 15.4 - 0.07p = 24.2

0.05p = 8.8

p = 8.8 / 0.05

p = 176

176 of 12% vinegar and

220 - 176 = 44

44 of 7% vinegar

4 0

2 years ago

Solve for p. 9(p-4) = -18 p=

agasfer [191]

Step-by-step explanation:

9(p-4) = -18

9p - 36 = -18

9p = -18 + 36

9p = 18

p= 18/9

p = 2

Check:

9 ( 2 - 4 ) = - 18

9 (-2) = - 18

- 18 = - 18

✓

3 0

2 years ago

Read 2 more answers

Beth made wristbands and belts for a craft sale. She sold 30 of these items. Each wristband sold for $5.50. Each belt sold at $8

baherus [9]

<u>✿</u><u>Answer:</u>

Both sold 18 wristbands and 12 belts

✂----------------------

Let, number of wristbands = x

number of belts = y

$x+y=30$ ➞ $x~5.5$

$5.50x+8.75y=204$ $x1$

$5.50x+5.5y=165\\5.5x+8.75y=204\\---------\\-3.25y=-39$

$y=-39/3.25$

$y=12$

$x=30-12=18$

$So, Both~ sold ~18 ~wristbands ~and ~12~ belts$ .

❄❄❄❄❄❄

<h3>hope it helps...</h3><h3>have a great day!!</h3>

5 0

2 years ago

Initially a tank contains 10 liters of pure water. Brine of unknown (but constant) concentration of salt is flowing in at 1 lite

zhenek [66]

Answer:

Therefore the concentration of salt in the incoming brine is 1.73 g/L.

Step-by-step explanation:

Here the amount of incoming and outgoing of water are equal. Then the amount of water in the tank remain same = 10 liters.

Let the concentration of salt be a gram/L

Let the amount salt in the tank at any time t be Q(t).

$\frac{dQ}{dt} =\textrm {incoming rate - outgoing rate}$

Incoming rate = (a g/L)×(1 L/min)

=a g/min

The concentration of salt in the tank at any time t is = $\frac{Q(t)}{10}$ g/L

Outgoing rate = $(\frac{Q(t)}{10} g/L)(1 L/ min)$ $\frac{Q(t)}{10} g/min$

$\frac{dQ}{dt} = a- \frac{Q(t)}{10}$

$\Rightarrow \frac{dQ}{10a-Q(t)} =\frac{1}{10} dt$

Integrating both sides

$\Rightarrow \int \frac{dQ}{10a-Q(t)} =\int\frac{1}{10} dt$

$\Rightarrow -log|10a-Q(t)|=\frac{1}{10} t +c$ [ where c arbitrary constant]

Initial condition when t= 20 , Q(t)= 15 gram

$\Rightarrow -log|10a-15|=\frac{1}{10}\times 20 +c$

$\Rightarrow -log|10a-15|-2=c$

Therefore ,

$-log|10a-Q(t)|=\frac{1}{10} t -log|10a-15|-2$ .......(1)

In the starting time t=0 and Q(t)=0

Putting t=0 and Q(t)=0 in equation (1) we get

$- log|10a|= -log|10a-15| -2$

$\Rightarrow- log|10a|+log|10a-15|= -2$

$\Rightarrow log|\frac{10a-15}{10a}|= -2$

$\Rightarrow |\frac{10a-15}{10a}|=e ^{-2}$

$\Rightarrow 1-\frac{15}{10a} =e^{-2}$

$\Rightarrow \frac{15}{10a} =1-e^{-2}$

$\Rightarrow \frac{3}{2a} =1-e^{-2}$

$\Rightarrow2a= \frac{3}{1-e^{-2}}$

$\Rightarrow a = 1.73$

Therefore the concentration of salt in the incoming brine is 1.73 g/L

8 0

2 years ago